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BCA1B01T · M1

Matrices, order and notation

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Example 1

Reading order and elements

Question

For $A = 361051 - 2 90 1 - 4 8$ , state the order of $A$ , the number of elements, and the values of $a_{23}$ , $a_{31}$ and $a_{14}$ .

Count rows and columns
There are $3$ horizontal lines and $4$ vertical lines, so the order is $3 \times 4$ .
Count elements
Number of elements $= m \times n = 3 \times 4 = 12$ .
Locate each element (row first, then column)
$a_{23}$ — row $2$ , column $3$ $\Rightarrow 9$ .

$a_{31}$ — row $3$ , column $1$ $\Rightarrow 1$ .

$a_{14}$ — row $1$ , column $4$ $\Rightarrow 1$ .

Answer

Order $3 \times 4$ ; $12$ elements; $a_{23} = 9$ , $a_{31} = 1$ , $a_{14} = 1$ .

NoteThe commonest slip is reading a₂₃ as column 2, row 3. Row index always comes first.

Example 2

Constructing a matrix from a general element

Question

Construct the $2 \times 3$ matrix $A = [a_{ij}]$ where $a_{ij} = \frac{( i + 2 j ) ^{2}}{2}$ .

Set up the empty frame
$i$ runs $1, 2$ (rows) and $j$ runs $1, 2, 3$ (columns):

$A = [a_{11} a_{21} a_{12} a_{22} a_{13} a_{23}]$
Evaluate row 1 ( $i = 1$ )
$a_{11} = \frac{( 1 + 2 ) ^{2}}{2} = \frac{9}{2}, a_{12} = \frac{( 1 + 4 ) ^{2}}{2} = \frac{25}{2}, a_{13} = \frac{( 1 + 6 ) ^{2}}{2} = \frac{49}{2}$
Evaluate row 2 ( $i = 2$ )
$a_{21} = \frac{( 2 + 2 ) ^{2}}{2} = 8, a_{22} = \frac{( 2 + 4 ) ^{2}}{2} = 18, a_{23} = \frac{( 2 + 6 ) ^{2}}{2} = 32$
Assemble
$A = [\frac{9}{2} 8 \frac{25}{2} 18 \frac{49}{2} 32]$

Answer

$A = [9/2 8 25/2 18 49/2 32]$

Example 3

Using equality to find unknowns

Question

Find $x, y, z, w$ if
$[x + 3 z - 1 2 y + x 4 w - 8] = [03 - 7 2 w]$

Check the orders match
Both are $2 \times 2$ , so equality is possible and we may equate element by element.
Equate position (1,1)
$x + 3 = 0 \Rightarrow x = - 3$
Equate position (1,2), substituting $x$
$2 y + x = - 7 \Rightarrow 2 y - 3 = - 7 \Rightarrow 2 y = - 4 \Rightarrow y = - 2$
Equate position (2,1)
$z - 1 = 3 \Rightarrow z = 4$
Equate position (2,2)
$4 w - 8 = 2 w \Rightarrow 2 w = 8 \Rightarrow w = 4$

Answer

$x = - 3, y = - 2, z = 4, w = 4$

NoteWork the positions in an order that lets you substitute — here (1,1) must come before (1,2).

Problem 1Basic

A matrix has $18$ elements. List every possible order it can have. How many of these are square?

Show solution

We need all factor pairs $m \times n = 18$ with $m, n$ positive integers:

$1 \times 18, 2 \times 9, 3 \times 6, 6 \times 3, 9 \times 2, 18 \times 1$

That is $6$ possible orders (the number of divisors of $18$ ).

A square matrix needs $m = n$ , i.e. $m^{2} = 18$ . Since $18$ is not a perfect square, no square order exists.

Answer

6

possible orders:

1 \times 18, 2 \times 9, 3 \times 6, 6 \times 3, 9 \times 2, 18 \times 1

. None are square.

Problem 2Basic

Construct the $3 \times 2$ matrix $A = [a_{ij}]$ with $a_{ij} = ∣ - 3 i + j ∣$ .

Show solution

Run $i = 1, 2, 3$ and $j = 1, 2$ .

Row 1: $a_{11} = ∣ - 3 + 1∣ = 2$ , $a_{12} = ∣ - 3 + 2∣ = 1$

Row 2: $a_{21} = ∣ - 6 + 1∣ = 5$ , $a_{22} = ∣ - 6 + 2∣ = 4$

Row 3: $a_{31} = ∣ - 9 + 1∣ = 8$ , $a_{32} = ∣ - 9 + 2∣ = 7$

$A = 258147$

Answer

A = 258147

Problem 3Exam level

Construct the $3 \times 3$ matrix $A = [a_{ij}]$ where
$a_{ij} = {i + j i - j if i \geq j if i < j$

Show solution

Decide for each cell whether $i \geq j$ (use $i + j$ ) or $i < j$ (use $i - j$ ).

Row 1 ( $i = 1$ ): $j = 1 \Rightarrow i \geq j \Rightarrow 1 + 1 = 2$ ; $j = 2 \Rightarrow i < j \Rightarrow 1 - 2 = - 1$ ; $j = 3 \Rightarrow 1 - 3 = - 2$

Row 2 ( $i = 2$ ): $j = 1 \Rightarrow 2 + 1 = 3$ ; $j = 2 \Rightarrow 2 + 2 = 4$ ; $j = 3 \Rightarrow i < j \Rightarrow 2 - 3 = - 1$

Row 3 ( $i = 3$ ): $j = 1 \Rightarrow 3 + 1 = 4$ ; $j = 2 \Rightarrow 3 + 2 = 5$ ; $j = 3 \Rightarrow 3 + 3 = 6$

$A = 234 - 1 45 - 2 - 1 6$

Note the pattern: on and below the diagonal the rule is $i + j$ ; strictly above it, $i - j$ .

Answer

A = 234 - 1 45 - 2 - 1 6

Problem 4Exam level

Find $a, b, c, d$ if
$[a - b 2 a - b 2 a + c 3 c + d] = [- 1 0 513]$

Show solution

Both matrices are $2 \times 2$ , so equate corresponding elements:

$a - b = - 1 (1) 2 a + c = 5 (2)$
$2 a - b = 0 (3) 3 c + d = 13 (4)$

Subtract $(1)$ from $(3)$ :

$(2 a - b) - (a - b) = 0 - (- 1) \Rightarrow a = 1$

From $(1)$ : $1 - b = - 1 \Rightarrow b = 2$ .

From $(2)$ : $2 (1) + c = 5 \Rightarrow c = 3$ .

From $(4)$ : $3 (3) + d = 13 \Rightarrow d = 13 - 9 = 4$ .

Check in $(3)$ : $2 (1) - 2 = 0$ ✓

Answer

a = 1, b = 2, c = 3, d = 4

Problem 5Challenge

If $A = [a_{ij}]$ is of order $3 \times 3$ with $a_{ij} = \frac{1}{2} (3 i - 2 j)$ , find the sum of all elements of $A$ .

Show solution

Rather than building the whole matrix, sum the general element over all $i$ and $j$ :

$S = i = 1 \sum 3 j = 1 \sum 3 \frac{1}{2} (3 i - 2 j) = \frac{1}{2} [i = 1 \sum 3 j = 1 \sum 3 3 i - i = 1 \sum 3 j = 1 \sum 3 2 j]$

For the first part, each $i$ appears once for each of the $3$ columns:

$i \sum j \sum 3 i = 3 \cdot 3 \cdot (1 + 2 + 3) = 3 \cdot 3 \cdot 6 = 54$

For the second, each $j$ appears once for each of the $3$ rows:

$i \sum j \sum 2 j = 3 \cdot 2 \cdot (1 + 2 + 3) = 3 \cdot 2 \cdot 6 = 36$

$S = \frac{1}{2} (54 - 36) = \frac{18}{2} = 9$

Verification by construction. $A = 1/2 2 7/2 - 1/2 1 5/2 - 3/2 0 3/2$ . Row sums: $- \frac{3}{2}, 3, \frac{15}{2}$ . Total $= - 1.5 + 3 + 7.5 = 9$ ✓

Answer

9

BCA1B01T · M1

Types of matrices

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Example 1

Classifying a matrix under every applicable name

Question

Name every type that $A = 500050005$ belongs to, and state its trace and determinant.

Shape
$3$ rows, $3$ columns $\Rightarrow$ square matrix of order 3.
Position of non-zeros
Every off-diagonal element is $0$ $\Rightarrow$ diagonal. All diagonal elements equal $5$ $\Rightarrow$ scalar, and $A = 5 I_{3}$ .

Since nothing lies below the diagonal and nothing lies above it, $A$ is also both upper triangular and lower triangular.
Transpose test
Transposing a diagonal matrix leaves it unchanged, so $A^{T} = A$ $\Rightarrow$ symmetric. It is not skew-symmetric, because a skew-symmetric matrix needs zero diagonal.
Determinant and singularity
For a triangular matrix, $∣ A ∣$ is the product of the diagonal:

$∣ A ∣ = 5 \times 5 \times 5 = 125 \neq = 0$

so $A$ is non-singular.
Trace
$tr (A) = 5 + 5 + 5 = 15$
Power tests
$A^{2} = 25 I \neq = A$ , so not idempotent. $A^{2} = 25 I \neq = I$ , so not involutory. No power is ever $O$ , so not nilpotent.

Answer

Square, diagonal, scalar, upper triangular, lower triangular, symmetric, non-singular. $tr (A) = 15$ , $∣ A ∣ = 125$ .

NoteOne matrix can carry many names at once. Exam questions asking to “identify the type” expect all applicable names.

Example 2

Splitting a matrix into symmetric and skew-symmetric parts

Question

Express $A = 271 43 - 2 - 6 50$ as the sum of a symmetric and a skew-symmetric matrix.

Write down the transpose
$A^{T} = 24 - 6 735 1 - 2 0$
Form $A + A^{T}$
$A + A^{T} = 411 - 5 1163 - 5 30$
Halve it — this is the symmetric part $P$
$P = \frac{1}{2} (A + A^{T}) = 2 \frac{11}{2} - \frac{5}{2} \frac{11}{2} 3 \frac{3}{2} - \frac{5}{2} \frac{3}{2} 0$

Check: $P^{T} = P$ ✓ (mirror about the diagonal).
Form $A - A^{T}$
$A - A^{T} = 037 - 3 0 - 7 - 7 70$
Halve it — this is the skew-symmetric part $Q$
$Q = \frac{1}{2} (A - A^{T}) = 0 \frac{3}{2} \frac{7}{2} - \frac{3}{2} 0 - \frac{7}{2} - \frac{7}{2} \frac{7}{2} 0$

Check: diagonal is all zero and $Q^{T} = - Q$ ✓
Verify the sum
$P + Q = 2 + 0 \frac{11}{2} + \frac{3}{2} - \frac{5}{2} + \frac{7}{2} \frac{11}{2} - \frac{3}{2} 3 + 0 \frac{3}{2} - \frac{7}{2} - \frac{5}{2} - \frac{7}{2} \frac{3}{2} + \frac{7}{2} 0 = 271 43 - 2 - 6 50 = A ✓$

Answer

$A = P + Q$ with $P = \frac{1}{2} (A + A^{T})$ symmetric and $Q = \frac{1}{2} (A - A^{T})$ skew-symmetric, as computed above.

NoteAlways finish with the P + Q = A check. It costs 30 seconds and catches every sign error.

Example 3

Proving a matrix is orthogonal

Question

Show that $A = \frac{1}{3} 12 - 2 212 2 - 2 - 1$ is orthogonal, and hence write down $A^{- 1}$ .

State what must be shown
$A$ is orthogonal $⟺ A A^{T} = I$ . Let $B = 3 A = 12 - 2 212 2 - 2 - 1$ , so $A = \frac{1}{3} B$ and

$A A^{T} = \frac{1}{3} B \cdot \frac{1}{3} B^{T} = \frac{1}{9} B B^{T}$

so it is enough to show $B B^{T} = 9 I$ .
Write $B^{T}$
$B^{T} = 122 21 - 2 - 2 2 - 1$
Compute $B B^{T}$ row by row
Row 1 of $B$ with the columns of $B^{T}$ (i.e. dotted against the rows of $B$ ):

$(1, 2, 2) \cdot (1, 2, 2) = 1 + 4 + 4 = 9$
$(1, 2, 2) \cdot (2, 1, - 2) = 2 + 2 - 4 = 0$
$(1, 2, 2) \cdot (- 2, 2, - 1) = - 2 + 4 - 2 = 0$

Row 2:
$(2, 1, - 2) \cdot (1, 2, 2) = 2 + 2 - 4 = 0 (2, 1, - 2) \cdot (2, 1, - 2) = 4 + 1 + 4 = 9$
$(2, 1, - 2) \cdot (- 2, 2, - 1) = - 4 + 2 + 2 = 0$

Row 3:
$(- 2, 2, - 1) \cdot (1, 2, 2) = - 2 + 4 - 2 = 0 (- 2, 2, - 1) \cdot (2, 1, - 2) = - 4 + 2 + 2 = 0$
$(- 2, 2, - 1) \cdot (- 2, 2, - 1) = 4 + 4 + 1 = 9$
Assemble
$B B^{T} = 900090009 = 9 I$

Therefore $A A^{T} = \frac{1}{9} \cdot 9 I = I$ , and $A$ is orthogonal.
Read off the inverse
For an orthogonal matrix $A^{- 1} = A^{T}$ , so

$A^{- 1} = \frac{1}{3} 122 21 - 2 - 2 2 - 1$

Answer

$A A^{T} = I$ , so $A$ is orthogonal and $A^{- 1} = A^{T} = \frac{1}{3} 122 21 - 2 - 2 2 - 1$

NoteNotice the shortcut: pulling the 1/3 out front turns fraction arithmetic into integer arithmetic. Do this whenever a common factor is visible.

Example 4

Idempotent, involutory or nilpotent?

Question

Classify each by computing the required power.
(a) $A = 2 - 1 1 - 2 3 - 2 - 4 4 - 3$
(b) $B = [412 - 1 - 3]$ — find $B^{2}$ and comment.
(c) $C = 15 - 2 12 - 1 36 - 3$

(a) Compute $A^{2}$
Row 1 of $A$ against each column:

$(2) (2) + (- 2) (- 1) + (- 4) (1) = 4 + 2 - 4 = 2$
$(2) (- 2) + (- 2) (3) + (- 4) (- 2) = - 4 - 6 + 8 = - 2$
$(2) (- 4) + (- 2) (4) + (- 4) (- 3) = - 8 - 8 + 12 = - 4$

Row 2:
$(- 1) (2) + (3) (- 1) + (4) (1) = - 2 - 3 + 4 = - 1$
$(- 1) (- 2) + (3) (3) + (4) (- 2) = 2 + 9 - 8 = 3$
$(- 1) (- 4) + (3) (4) + (4) (- 3) = 4 + 12 - 12 = 4$

Row 3:
$(1) (2) + (- 2) (- 1) + (- 3) (1) = 2 + 2 - 3 = 1$
$(1) (- 2) + (- 2) (3) + (- 3) (- 2) = - 2 - 6 + 6 = - 2$
$(1) (- 4) + (- 2) (4) + (- 3) (- 3) = - 4 - 8 + 9 = - 3$

$A^{2} = 2 - 1 1 - 2 3 - 2 - 4 4 - 3 = A$

$A$ is idempotent.
(b) Compute $B^{2}$
$B^{2} = [412 - 1 - 3] [412 - 1 - 3] = [16 - 12 48 - 36 - 4 + 3 - 12 + 9] = [412 - 1 - 3] = B$

$B$ is also idempotent (equivalently, periodic with period $1$ ).
(c) Compute $C^{2}$
Row 1: $(1) (1) + (1) (5) + (3) (- 2) = 1 + 5 - 6 = 0$ ; $(1) (1) + (1) (2) + (3) (- 1) = 1 + 2 - 3 = 0$ ; $(1) (3) + (1) (6) + (3) (- 3) = 3 + 6 - 9 = 0$

Row 2: $(5) (1) + (2) (5) + (6) (- 2) = 5 + 10 - 12 = 3$ ; $(5) (1) + (2) (2) + (6) (- 1) = 5 + 4 - 6 = 3$ ; $(5) (3) + (2) (6) + (6) (- 3) = 15 + 12 - 18 = 9$

Row 3: $(- 2) (1) + (- 1) (5) + (- 3) (- 2) = - 2 - 5 + 6 = - 1$ ; $(- 2) (1) + (- 1) (2) + (- 3) (- 1) = - 2 - 2 + 3 = - 1$ ; $(- 2) (3) + (- 1) (6) + (- 3) (- 3) = - 6 - 6 + 9 = - 3$

$C^{2} = 03 - 1 03 - 1 09 - 3$
(c) Go one power further
$C^{3} = C^{2} C = 03 - 1 03 - 1 09 - 3 15 - 2 12 - 1 36 - 3$

Row 2: $3 + 15 - 18 = 0$ ; $3 + 6 - 9 = 0$ ; $9 + 18 - 27 = 0$
Row 3: $- 1 - 5 + 6 = 0$ ; $- 1 - 2 + 3 = 0$ ; $- 3 - 6 + 9 = 0$

$C^{3} = O$

$C$ is nilpotent of index 3.

Answer

(a) idempotent; (b) idempotent; (c) nilpotent of index $3$ .

NoteIf A² is neither A nor I nor O, compute A³ before concluding anything — nilpotency often shows up only at the third power.

Example 5

Hermitian and skew-Hermitian

Question

Verify that $A = 3 2 + i - 1 - 3 i 2 - i 04 - 1 + 3 i 4 - 2$ is Hermitian, and that $i A$ would then be skew-Hermitian.

Take the conjugate $A ˉ$
Flip the sign of every imaginary part:

$A ˉ = 3 2 - i - 1 + 3 i 2 + i 04 - 1 - 3 i 4 - 2$
Transpose it to get $A^{θ} = A ˉ^{T}$
$A^{θ} = 3 2 + i - 1 - 3 i 2 - i 04 - 1 + 3 i 4 - 2$
Compare
$A^{θ} = A$ element for element, so $A$ is Hermitian.

Note the two structural signs of a Hermitian matrix, both present here: the diagonal $3, 0, - 2$ is entirely real, and each pair across the diagonal is a conjugate pair ( $2 - i$ against $2 + i$ ).
Now consider $i A$
Using $(k A)^{θ} = k ˉ A^{θ}$ with $k = i$ , so $k ˉ = - i$ :

$(i A)^{θ} = i ˉ A^{θ} = - i A = - (i A)$

which is exactly the definition of skew-Hermitian.

Answer

$A^{θ} = A$ , so $A$ is Hermitian; and $(i A)^{θ} = - (i A)$ , so $i A$ is skew-Hermitian.

NoteGeneral result: if A is Hermitian then iA is skew-Hermitian, and vice versa. Worth quoting directly in an exam.

Problem 1Basic

State, with a one-line reason, whether each is true or false.
(a) Every scalar matrix is a diagonal matrix.
(b) Every diagonal matrix is a scalar matrix.
(c) Every identity matrix is a scalar matrix.
(d) A skew-symmetric matrix can have a non-zero diagonal element.
(e) A null matrix of order $2 \times 3$ is a square matrix.

Show solution

(a) True. A scalar matrix is defined as a diagonal matrix with all diagonal entries equal, so it satisfies the diagonal condition by definition.

(b) False. $diag (1, 2)$ is diagonal but its diagonal entries differ, so it is not scalar.

(c) True. $I = 1 \cdot I$ , a scalar matrix with $k = 1$ .

(d) False. $A^{T} = - A$ forces $a_{ii} = - a_{ii}$ , hence $2 a_{ii} = 0$ , hence $a_{ii} = 0$ for every $i$ .

(e) False. Square requires $m = n$ ; here $2 \neq = 3$ . A null matrix exists in any order and is square only when the order is square.

Answer(a) T (b) F (c) T (d) F (e) F

Problem 2Basic

Find $x$ and $y$ so that $A = 0 - 3 - 2 30 y x - 4 0$ is skew-symmetric.

Show solution

Skew-symmetric requires $a_{ij} = - a_{j i}$ .

The diagonal is already $0, 0, 0$ ✓

Position $(1, 3)$ and $(3, 1)$ : $a_{13} = - a_{31} \Rightarrow x = - (- 2) = 2$ .

Position $(2, 3)$ and $(3, 2)$ : $a_{23} = - a_{32} \Rightarrow - 4 = - y \Rightarrow y = 4$ .

Check $(1, 2)$ against $(2, 1)$ : $3 = - (- 3)$ ✓

$A = 0 - 3 - 2 304 2 - 4 0$

Answer

x = 2, y = 4

Problem 3Exam level

Express $A = [35 - 1 7]$ as the sum of a symmetric and a skew-symmetric matrix.

Show solution

$A^{T} = [3 - 1 57]$

Symmetric part:
$P = \frac{1}{2} (A + A^{T}) = \frac{1}{2} [64414] = [3227]$

Skew-symmetric part:
$Q = \frac{1}{2} (A - A^{T}) = \frac{1}{2} [06 - 6 0] = [03 - 3 0]$

Check: $P + Q = [35 - 1 7] = A$ ✓, $P^{T} = P$ ✓, $Q^{T} = - Q$ ✓

Answer

A = [3227] + [03 - 3 0]

Problem 4Exam level

Show that $A = [524 - 1 - 5]$ is involutory, and hence find $A^{- 1}$ and $A^{99}$ .

Show solution

$A^{2} = [524 - 1 - 5] [524 - 1 - 5] = [25 - 24 120 - 120 - 5 + 5 - 24 + 25] = [1001] = I$

So $A$ is involutory.

Inverse. $A^{2} = I$ means $A \cdot A = I$ , so by definition $A^{- 1} = A = [524 - 1 - 5]$ .

High power. Since $A^{2} = I$ , even powers give $I$ and odd powers give $A$ :

$A^{99} = A^{98} \cdot A = (A^{2})^{49} A = I^{49} A = A = [524 - 1 - 5]$

Answer

A^{- 1} = A

and

A^{99} = A = [524 - 1 - 5]

Problem 5Exam level

Show that $A = 15 - 2 12 - 1 36 - 3$ is nilpotent and state its index. Deduce $∣ A ∣$ without expanding a determinant.

Show solution

Compute $A^{2}$ .

Row 1: $1 + 5 - 6 = 0$ , $1 + 2 - 3 = 0$ , $3 + 6 - 9 = 0$
Row 2: $5 + 10 - 12 = 3$ , $5 + 4 - 6 = 3$ , $15 + 12 - 18 = 9$
Row 3: $- 2 - 5 + 6 = - 1$ , $- 2 - 2 + 3 = - 1$ , $- 6 - 6 + 9 = - 3$

$A^{2} = 03 - 1 03 - 1 09 - 3 \neq = O$

Compute $A^{3} = A^{2} A$ .

Row 1 is zero, so stays zero.
Row 2: $3 + 15 - 18 = 0$ , $3 + 6 - 9 = 0$ , $9 + 18 - 27 = 0$
Row 3: $- 1 - 5 + 6 = 0$ , $- 1 - 2 + 3 = 0$ , $- 3 - 6 + 9 = 0$

$A^{3} = O$

$A$ is nilpotent of index 3.

Determinant. Taking determinants of $A^{3} = O$ :

$∣ A ∣^{3} = ∣ A^{3} ∣ = ∣ O ∣ = 0 \Rightarrow ∣ A ∣ = 0$

So $A$ is singular — as every nilpotent matrix must be.

AnswerNilpotent of index

3

;

∣ A ∣ = 0

Problem 6Exam level

Prove that for any square matrix $A$ , both $A + A^{T}$ and $A A^{T}$ are symmetric, while $A - A^{T}$ is skew-symmetric.

Show solution

Use the two transpose rules $(X + Y)^{T} = X^{T} + Y^{T}$ , $(X Y)^{T} = Y^{T} X^{T}$ , and $(X^{T})^{T} = X$ .

$A + A^{T}$ is symmetric:
$(A + A^{T})^{T} = A^{T} + (A^{T})^{T} = A^{T} + A = A + A^{T}$
Equal to itself, so symmetric. ∎

$A - A^{T}$ is skew-symmetric:
$(A - A^{T})^{T} = A^{T} - (A^{T})^{T} = A^{T} - A = - (A - A^{T})$
Equal to its own negative, so skew-symmetric. ∎

$A A^{T}$ is symmetric:
$(A A^{T})^{T} = (A^{T})^{T} A^{T} = A A^{T}$
Equal to itself, so symmetric. ∎

(The same argument gives $A^{T} A$ symmetric, and it holds even for non-square $A$ , where $A A^{T}$ is $m \times m$ and $A^{T} A$ is $n \times n$ .)

AnswerShown by applying

(X Y)^{T} = Y^{T} X^{T}

and

(X^{T})^{T} = X

to each expression.

Problem 7Challenge

If $A$ is a skew-symmetric matrix of odd order $n$ , prove $∣ A ∣ = 0$ .

Show solution

Start from the definition and take determinants.

$A^{T} = - A$
$\Rightarrow ∣ A^{T} ∣ = ∣ - A ∣$

Now use two determinant properties:

$∣ A^{T} ∣ = ∣ A ∣$ — transposing does not change the determinant.
$∣ k A ∣ = k^{n} ∣ A ∣$ for a matrix of order $n$ — each of the $n$ rows contributes a factor $k$ .

With $k = - 1$ :

$∣ A ∣ = (- 1)^{n} ∣ A ∣$

Since $n$ is odd, $(- 1)^{n} = - 1$ :

$∣ A ∣ = - ∣ A ∣ \Rightarrow 2∣ A ∣ = 0 \Rightarrow ∣ A ∣ = 0 ■$

Every skew-symmetric matrix of odd order is therefore singular. (For even $n$ the argument gives $∣ A ∣ = ∣ A ∣$ , which says nothing — and indeed $[0 - 2 20]$ has determinant $4 \neq = 0$ .)

Answer

∣ A ∣ = (- 1)^{n} ∣ A ∣ = - ∣ A ∣ \Rightarrow ∣ A ∣ = 0

for odd

n

Problem 8Challenge

If $A$ is idempotent, prove that $B = I - A$ is also idempotent, and that $A B = B A = O$ .

Show solution

Given: $A^{2} = A$ .

$B$ is idempotent.
$B^{2} = (I - A) (I - A) = I \cdot I - I \cdot A - A \cdot I + A^{2}$
$= I - A - A + A^{2}$
Substituting $A^{2} = A$ :
$= I - A - A + A = I - A = B ■$

$A B = O$ .
$A B = A (I - A) = A I - A^{2} = A - A = O ■$

$B A = O$ .
$B A = (I - A) A = I A - A^{2} = A - A = O ■$

Interpretation: $A$ and $I - A$ are complementary projections. Whatever $A$ keeps, $I - A$ discards, so applying one after the other destroys everything.

Answer

B^{2} = I - A = B

, and

A B = B A = O

BCA1B01T · M1

Matrix operations

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Example 1

Checking conformability before computing

Question

$A$ is $3 \times 4$ , $B$ is $4 \times 2$ , $C$ is $2 \times 3$ . Which of $A B$ , $B A$ , $B C$ , $C A$ , $A B C$ , $A (B C)$ are defined, and of what order?

$A B$
$(3 \times 4) (4 \times 2)$ — inner dimensions match. Defined, order $3 \times 2$ .
$B A$
$(4 \times 2) (3 \times 4)$ — $2 \neq = 3$ . Not defined.
$B C$
$(4 \times 2) (2 \times 3)$ — match. Defined, order $4 \times 3$ .
$C A$
$(2 \times 3) (3 \times 4)$ — match. Defined, order $2 \times 4$ .
$A B C$
$A B$ is $3 \times 2$ ; then $(3 \times 2) (2 \times 3)$ — match. Defined, order $3 \times 3$ .
$A (B C)$
$B C$ is $4 \times 3$ ; then $(3 \times 4) (4 \times 3)$ — match. Defined, order $3 \times 3$ — the same as $A B C$ , as associativity demands.

Answer

$A B (3 \times 2)$ , $B C (4 \times 3)$ , $C A (2 \times 4)$ , $A B C = A (B C) (3 \times 3)$ are defined. $B A$ is not.

NoteWrite orders under the letters before doing any arithmetic. Two seconds here saves a whole wrong page.

Example 2

A full 3 × 3 product, position by position

Question

Find $A B$ where $A = 147258369$ and $B = 2 - 1 4 031120$ .

Order check
$(3 \times 3) (3 \times 3)$ — defined, result is $3 \times 3$ .
Row 1 of $A$ = $(1, 2, 3)$
Against column 1 $(2, - 1, 4)$ : $1 (2) + 2 (- 1) + 3 (4) = 2 - 2 + 12 = 12$

Against column 2 $(0, 3, 1)$ : $1 (0) + 2 (3) + 3 (1) = 0 + 6 + 3 = 9$

Against column 3 $(1, 2, 0)$ : $1 (1) + 2 (2) + 3 (0) = 1 + 4 + 0 = 5$
Row 2 of $A$ = $(4, 5, 6)$
Column 1: $4 (2) + 5 (- 1) + 6 (4) = 8 - 5 + 24 = 27$

Column 2: $4 (0) + 5 (3) + 6 (1) = 0 + 15 + 6 = 21$

Column 3: $4 (1) + 5 (2) + 6 (0) = 4 + 10 + 0 = 14$
Row 3 of $A$ = $(7, 8, 9)$
Column 1: $7 (2) + 8 (- 1) + 9 (4) = 14 - 8 + 36 = 42$

Column 2: $7 (0) + 8 (3) + 9 (1) = 0 + 24 + 9 = 33$

Column 3: $7 (1) + 8 (2) + 9 (0) = 7 + 16 + 0 = 23$
Assemble
$A B = 1227429213351423$

Answer

$A B = 1227429213351423$

Example 3

Demonstrating that AB ≠ BA

Question

For $A = [1324]$ and $B = [0110]$ , compute $A B$ and $B A$ and comment.

Compute $A B$
$A B = [1 (0) + 2 (1) 3 (0) + 4 (1) 1 (1) + 2 (0) 3 (1) + 4 (0)] = [2413]$
Compute $B A$
$B A = [0 (1) + 1 (3) 1 (1) + 0 (3) 0 (2) + 1 (4) 1 (2) + 0 (4)] = [3142]$
Compare
$[2413] \neq = [3142] \Rightarrow A B \neq = B A$

Reading the result: $B$ is a swap matrix. Multiplying on the right by $B$ swapped the columns of $A$ ; multiplying on the left by $B$ swapped the rows. Side of multiplication decides what happens — this is exactly why order matters.

Answer

$A B = [2413]$ , $B A = [3142]$ , so $A B \neq = B A$ .

Example 4

Expanding a matrix expression correctly

Question

If $A = [2013]$ and $B = [12 - 1 0]$ , verify that $(A + B)^{2} \neq = A^{2} + 2 A B + B^{2}$ .

Compute $A + B$ and its square
$A + B = [3203]$

$(A + B)^{2} = [3203] [3203] = [9 + 0 6 + 6 0 + 0 0 + 9] = [91209]$
Compute the pieces on the right
$A^{2} = [2013] [2013] = [4059]$

$B^{2} = [12 - 1 0] [12 - 1 0] = [1 - 2 2 + 0 - 1 - 0 - 2 + 0] = [- 1 2 - 1 - 2]$

$A B = [2 + 2 0 + 6 - 2 + 0 0 + 0] = [46 - 2 0] B A = [2 - 0 4 + 0 1 - 3 2 + 0] = [24 - 2 2]$

Note immediately that $A B \neq = B A$ .
Evaluate the false right-hand side
$A^{2} + 2 A B + B^{2} = [4059] + [812 - 4 0] + [- 1 2 - 1 - 2] = [111407]$

This is not $[91209]$ , so the identity fails.
Confirm the correct expansion
$A^{2} + A B + B A + B^{2} = [4059] + [46 - 2 0] + [24 - 2 2] + [- 1 2 - 1 - 2] = [91209] ✓$

which matches $(A + B)^{2}$ exactly.

Answer

$(A + B)^{2} = [91209]$ but $A^{2} + 2 A B + B^{2} = [111407]$ . The correct expansion is $A^{2} + A B + B A + B^{2}$ .

NoteNever write 2AB when expanding (A+B)² for matrices unless you have first shown AB = BA.

Problem 1Basic

Given $A = [23 - 1 0]$ and $B = [1 - 2 45]$ , find $3 A - 2 B$ .

Show solution

$3 A = [69 - 3 0] 2 B = [2 - 4 810]$

$3 A - 2 B = [6 - 2 9 - (- 4) - 3 - 8 0 - 10] = [413 - 11 - 10]$

Answer

[413 - 11 - 10]

Problem 2Basic

Find $A B$ and $B A$ where $A = [123]$ and $B = 456$ .

Show solution

$A B$ : $(1 \times 3) (3 \times 1)$ gives a $1 \times 1$ matrix.

$A B = [1 (4) + 2 (5) + 3 (6)] = [4 + 10 + 18] = [32]$

$B A$ : $(3 \times 1) (1 \times 3)$ gives a $3 \times 3$ matrix.

$B A = 456 [123] = 45681012121518$

Both products exist but they are not even the same order — the sharpest possible illustration that $A B \neq = B A$ .

Answer

A B = [32]

(order

1 \times 1

);

B A = 45681012121518

(order

3 \times 3

Problem 3Exam level

If $A = 102020213$ , show that $A^{3} - 6 A^{2} + 7 A + 2 I = O$ .

Show solution

Step 1 — compute $A^{2}$ .

Row 1 $(1, 0, 2)$ : $1 + 0 + 4 = 5$ , $0 + 0 + 0 = 0$ , $2 + 0 + 6 = 8$
Row 2 $(0, 2, 1)$ : $0 + 0 + 2 = 2$ , $0 + 4 + 0 = 4$ , $0 + 2 + 3 = 5$
Row 3 $(2, 0, 3)$ : $2 + 0 + 6 = 8$ , $0 + 0 + 0 = 0$ , $4 + 0 + 9 = 13$

$A^{2} = 5280408513$

Step 2 — compute $A^{3} = A^{2} A$ .

Row 1 $(5, 0, 8)$ : $5 + 0 + 16 = 21$ , $0 + 0 + 0 = 0$ , $10 + 0 + 24 = 34$
Row 2 $(2, 4, 5)$ : $2 + 0 + 10 = 12$ , $0 + 8 + 0 = 8$ , $4 + 4 + 15 = 23$
Row 3 $(8, 0, 13)$ : $8 + 0 + 26 = 34$ , $0 + 0 + 0 = 0$ , $16 + 0 + 39 = 55$

$A^{3} = 211234080342355$

Step 3 — assemble term by term.

$6 A^{2} = 30124802404830787 A = 70140140147212 I = 200020002$

Step 4 — add position by position.

$(1, 1)$ : $21 - 30 + 7 + 2 = 0$ $(1, 2)$ : $0 - 0 + 0 + 0 = 0$ $(1, 3)$ : $34 - 48 + 14 + 0 = 0$

$(2, 1)$ : $12 - 12 + 0 + 0 = 0$ $(2, 2)$ : $8 - 24 + 14 + 2 = 0$ $(2, 3)$ : $23 - 30 + 7 + 0 = 0$

$(3, 1)$ : $34 - 48 + 14 + 0 = 0$ $(3, 2)$ : $0$ $(3, 3)$ : $55 - 78 + 21 + 2 = 0$

$A^{3} - 6 A^{2} + 7 A + 2 I = O ■$

(This is the Cayley–Hamilton theorem in action — the polynomial is the characteristic polynomial of $A$ .)

AnswerAll nine positions evaluate to

0

, so the expression equals

O

Problem 4Exam level

Find a matrix $X$ such that $2 A + 3 X = 5 B$ , where $A = [- 3 5 21]$ and $B = [41 0 - 2]$ .

Show solution

Rearrange first, then substitute — never the other way round.

$3 X = 5 B - 2 A \Rightarrow X = \frac{1}{3} (5 B - 2 A)$

$5 B = [205 0 - 10] 2 A = [- 6 10 42]$

$5 B - 2 A = [20 - (- 6) 5 - 10 0 - 4 - 10 - 2] = [26 - 5 - 4 - 12]$

$X = \frac{1}{3} [26 - 5 - 4 - 12] = [\frac{26}{3} - \frac{5}{3} - \frac{4}{3} - 4]$

Check: $2 A + 3 X = [- 6 10 42] + [26 - 5 - 4 - 12] = [205 0 - 10] = 5 B$ ✓

Answer

X = \frac{1}{3} [26 - 5 - 4 - 12]

Problem 5Exam level

Verify $(A B)^{T} = B^{T} A^{T}$ for $A = 135240$ and $B = [21 - 1 3 04]$ .

Show solution

Left side. $A$ is $3 \times 2$ , $B$ is $2 \times 3$ , so $A B$ is $3 \times 3$ .

Row 1 $(1, 2)$ : $2 + 2 = 4$ , $- 1 + 6 = 5$ , $0 + 8 = 8$
Row 2 $(3, 4)$ : $6 + 4 = 10$ , $- 3 + 12 = 9$ , $0 + 16 = 16$
Row 3 $(5, 0)$ : $10 + 0 = 10$ , $- 5 + 0 = - 5$ , $0 + 0 = 0$

$A B = 41010 59 - 5 8160 \Rightarrow (A B)^{T} = 45810916 10 - 5 0$

Right side. $B^{T}$ is $3 \times 2$ , $A^{T}$ is $2 \times 3$ , so $B^{T} A^{T}$ is $3 \times 3$ ✓

$B^{T} = 2 - 1 0 134 A^{T} = [123450]$

Row 1 $(2, 1)$ : $2 + 2 = 4$ , $6 + 4 = 10$ , $10 + 0 = 10$
Row 2 $(- 1, 3)$ : $- 1 + 6 = 5$ , $- 3 + 12 = 9$ , $- 5 + 0 = - 5$
Row 3 $(0, 4)$ : $0 + 8 = 8$ , $0 + 16 = 16$ , $0 + 0 = 0$

$B^{T} A^{T} = 45810916 10 - 5 0$

Both sides agree. ∎ (Note $A^{T} B^{T}$ is not even defined here: $(2 \times 3) (3 \times 2)$ would be $2 \times 2$ — wrong order. The reversal is not optional.)

Answer

(A B)^{T} = B^{T} A^{T} = 45810916 10 - 5 0

Problem 6Challenge

Find all matrices that commute with $A = [1011]$ .

Show solution

Let $X = [a c b d]$ and impose $A X = X A$ .

$A X = [1011] [a c b d] = [a + c c b + d d]$

$X A = [a c b d] [1011] = [a c a + b c + d]$

Equate corresponding elements:

$(1, 1)$ : $a + c = a \Rightarrow c = 0$
$(1, 2)$ : $b + d = a + b \Rightarrow d = a$
$(2, 1)$ : $c = c$ — no information
$(2, 2)$ : $d = c + d \Rightarrow c = 0$ — consistent

So $c = 0$ , $d = a$ , and $b$ is free:

$X = [a 0 b a], a, b \in R$

Equivalently $X = a I + b N$ where $N = [0010]$ . Every such matrix commutes with $A$ ; nothing else does.

Sanity check with $a = 2, b = 3$ : $[1011] [2032] = [2052]$ and $[2032] [1011] = [2052]$ ✓

Answer

X = [a 0 b a]

for any scalars

a, b

BCA1B01T · M1

Determinants

Read the topic →

Example 1

Order 3 by expansion along the first row

Question

Evaluate $215 - 1 40 3 - 2 1$ .

Write the expansion with the checkerboard signs
$Δ = 2 40 - 2 1 - (- 1) 15 - 2 1 + 31540$
Evaluate the three 2 × 2 determinants
$40 - 2 1 = 4 (1) - (- 2) (0) = 4$
$15 - 2 1 = 1 (1) - (- 2) (5) = 1 + 10 = 11$
$1540 = 1 (0) - 4 (5) = - 20$
Combine
$Δ = 2 (4) + 1 (11) + 3 (- 20) = 8 + 11 - 60 = - 41$
Cross-check by expanding along column 2
Column 2 is $(- 1, 4, 0)$ with signs $(-, +, -)$ :

$Δ = - (- 1) 15 - 2 1 + 42531 - 0 = 1 (11) + 4 (2 - 15) = 11 - 52 = - 41 ✓$

Answer

$Δ = - 41$

NoteWatch the sign on the middle term: it is −a₁₂ × M₁₂, so a negative a₁₂ produces a positive contribution.

Example 2

Exploiting zeros — expanding along the best line

Question

Evaluate $3271016000204513$ .

Find the emptiest line
Column 3 is $(0, 0, 2, 0)$ — three zeros. Expanding along it leaves a single term.
Expand along column 3
The surviving element is $a_{33} = 2$ , with sign $(- 1)^{3 + 3} = +$ :

$Δ = + 2321010453$
Repeat the trick on the 3 × 3
Its column 2 is $(0, 1, 0)$ — expand along it. The surviving element is at position $(2, 2)$ , sign $+$ :

$321010453 = + 13143 = 9 - 4 = 5$
Combine
$Δ = 2 \times 5 = 10$

Answer

$Δ = 10$

NoteA 4×4 that would take six 3×3 expansions collapsed to one 2×2. Always scan for the emptiest row or column first.

Example 3

Using row operations to manufacture zeros

Question

Evaluate $1472583610$ using properties.

Clear column 1 using row 1
Apply $R_{2} \to R_{2} - 4 R_{1}$ and $R_{3} \to R_{3} - 7 R_{1}$ . By P7 the determinant is unchanged.

$R_{2}$ : $(4 - 4, 5 - 8, 6 - 12) = (0, - 3, - 6)$
$R_{3}$ : $(7 - 7, 8 - 14, 10 - 21) = (0, - 6, - 11)$

$Δ = 100 2 - 3 - 6 3 - 6 - 11$
Expand along the new column 1
Only $a_{11} = 1$ survives, with sign $+$ :

$Δ = 1 - 3 - 6 - 6 - 11$
Evaluate
$= (- 3) (- 11) - (- 6) (- 6) = 33 - 36 = - 3$
Check directly
Expanding the original along row 1:

$1 (50 - 48) - 2 (40 - 42) + 3 (32 - 35) = 2 + 4 - 9 = - 3 ✓$

Answer

$Δ = - 3$

Example 4

A determinant with symbols — factorising by properties

Question

Prove that $1 a a^{2} 1 b b^{2} 1 c c^{2} = (a - b) (b - c) (c - a)$ .

Create zeros with column operations
Apply $C_{1} \to C_{1} - C_{2}$ and $C_{2} \to C_{2} - C_{3}$ (both allowed by P7, applied to columns via P1):

$Δ = 0 a - b a^{2} - b^{2} 0 b - c b^{2} - c^{2} 1 c c^{2}$
Take common factors out of columns 1 and 2
Column 1 has the common factor $(a - b)$ , since $a^{2} - b^{2} = (a - b) (a + b)$ . Column 2 has $(b - c)$ . By P4:

$Δ = (a - b) (b - c) 01 a + b 01 b + c 1 c c^{2}$
Expand along row 1
Only the entry $1$ at position $(1, 3)$ survives, with sign $(- 1)^{1 + 3} = +$ :

$Δ = (a - b) (b - c) 1 a + b 1 b + c$
Evaluate and simplify
$1 a + b 1 b + c = (b + c) - (a + b) = c - a$

$Δ = (a - b) (b - c) (c - a) ■$

Answer

$Δ = (a - b) (b - c) (c - a)$

NoteThis is the 3×3 Vandermonde determinant. The pattern — subtract adjacent columns, then pull out the differences — solves most symbolic determinant questions.

Problem 1Basic

Evaluate (a) $73 - 2 5$ (b) $cos θ sin θ - sin θ cos θ$

Show solution

(a) $7 (5) - (- 2) (3) = 35 + 6 = 41$

(b) $cos θ cos θ - (- sin θ) (sin θ) = cos^{2} θ + sin^{2} θ = 1$

Result (b) says every rotation matrix has determinant $1$ — rotation preserves area, which is exactly what a determinant of $1$ means geometrically.

Answer(a)

41

(b)

1

Problem 2Basic

Find the minor $M_{23}$ and the cofactor $A_{23}$ of $14 - 1 306 - 2 52$ .

Show solution

Minor. Delete row $2$ and column $3$ :

$M_{23} = 1 - 1 36 = 1 (6) - 3 (- 1) = 6 + 3 = 9$

Cofactor. Attach the sign $(- 1)^{2 + 3} = (- 1)^{5} = - 1$ :

$A_{23} = (- 1)^{2 + 3} M_{23} = - 9$

Answer

M_{23} = 9

A_{23} = - 9

Problem 3Exam level

Evaluate $213 3 - 2 1 - 1 42$ by (a) expansion along row 1 and (b) Sarrus' rule.

Show solution

(a) Expansion along row 1.

$Δ = 2 - 2 1 42 - 31342 + (- 1) 13 - 2 1$

$- 2 1 42 = - 4 - 4 = - 81342 = 2 - 12 = - 10 13 - 2 1 = 1 + 6 = 7$

$Δ = 2 (- 8) - 3 (- 10) - 1 (7) = - 16 + 30 - 7 = 7$

(b) Sarrus' rule.

Down-right diagonals:
$2 (- 2) (2) + 3 (4) (3) + (- 1) (1) (1) = - 8 + 36 - 1 = 27$

Down-left diagonals:
$(- 1) (- 2) (3) + 2 (4) (1) + 3 (1) (2) = 6 + 8 + 6 = 20$

$Δ = 27 - 20 = 7 ✓$

Both methods agree.

Answer

Δ = 7

Problem 4Exam level

Find $x$ if $x 13 2 x 2 31 x = 0$ .

Show solution

Expand along row 1:

$x x 2 1 x - 2 13 1 x + 3 13 x 2 = 0$

$x (x^{2} - 2) - 2 (x - 3) + 3 (2 - 3 x) = 0$

$x^{3} - 2 x - 2 x + 6 + 6 - 9 x = 0$

$x^{3} - 13 x + 12 = 0$

Try $x = 1$ : $1 - 13 + 12 = 0$ ✓, so $(x - 1)$ is a factor.

Dividing: $x^{3} - 13 x + 12 = (x - 1) (x^{2} + x - 12) = (x - 1) (x + 4) (x - 3)$

$x = 1, x = 3, x = - 4$

At each of these values the matrix is singular.

Answer

x = 1, 3, - 4

Problem 5Exam level

Without expanding, show that $111 a b c b + c c + a a + b = 0$ .

Show solution

Apply the column operation $C_{3} \to C_{3} + C_{2}$ , which by P7 leaves the determinant unchanged:

$Δ = 111 a b c a + b + c a + b + c a + b + c$

Column 3 now has the common factor $(a + b + c)$ . Take it out by P4:

$Δ = (a + b + c) 111 a b c 111$

Columns $1$ and $3$ are now identical, so by P3 the determinant is zero:

$Δ = (a + b + c) \times 0 = 0 ■$

No expansion was needed at any point.

Answer

Δ = 0

, since a column operation produces two identical columns.

Problem 6Exam level

If $A$ is a $3 \times 3$ matrix with $∣ A ∣ = 5$ , find (a) $∣3 A ∣$ , (b) $∣ A^{T} ∣$ , (c) $∣ A^{2} ∣$ , (d) $∣ A^{- 1} ∣$ , (e) $∣2 A^{- 1} ∣$ .

Show solution

Here $n = 3$ throughout.

(a) $∣ k A ∣ = k^{n} ∣ A ∣$ , so $∣3 A ∣ = 3^{3} \times 5 = 27 \times 5 = 135$

(b) $∣ A^{T} ∣ = ∣ A ∣ = 5$

(c) $∣ A^{2} ∣ = ∣ A ∣^{2} = 5^{2} = 25$

(d) $∣ A^{- 1} ∣ = \frac{1}{∣ A ∣} = \frac{1}{5}$

(e) $∣2 A^{- 1} ∣ = 2^{3} ∣ A^{- 1} ∣ = 8 \times \frac{1}{5} = \frac{8}{5}$

The trap in (a) and (e) is writing $∣ k A ∣ = k ∣ A ∣$ . The scalar comes out once per row, and there are $n$ rows.

Answer(a)

135

(b)

5

(c)

25

(d)

1/5

(e)

8/5

Problem 7Challenge

Evaluate $a b c b c a c a b$ and state when it is zero.

Show solution

Apply $R_{1} \to R_{1} + R_{2} + R_{3}$ . Every entry of the new row 1 becomes $a + b + c$ :

$Δ = a + b + c b c a + b + c c a a + b + c a b = (a + b + c) 1 b c 1 c a 1 a b$

Now clear row 1 with $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ :

$= (a + b + c) 1 b c 0 c - b a - c 0 a - b b - c$

Expand along row 1:

$= (a + b + c) [(c - b) (b - c) - (a - b) (a - c)]$

$= (a + b + c) [- (b - c)^{2} - (a^{2} - a c - ab + b c)]$

$= (a + b + c) [- (b^{2} - 2 b c + c^{2}) - a^{2} + a c + ab - b c]$

$= (a + b + c) [- a^{2} - b^{2} - c^{2} + ab + b c + c a]$

$Δ = - (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$

Using the identity $a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$ :

$Δ = - (a^{3} + b^{3} + c^{3} - 3 ab c) = 3 ab c - a^{3} - b^{3} - c^{3}$

When is it zero? Either $a + b + c = 0$ , or $a^{2} + b^{2} + c^{2} - ab - b c - c a = 0$ . The second factor equals $\frac{1}{2} [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}]$ , which for real numbers vanishes only when $a = b = c$ .

So $Δ = 0 ⟺ a + b + c = 0$ or $a = b = c$ .

Answer

Δ = 3 ab c - a^{3} - b^{3} - c^{3}

; zero exactly when

a + b + c = 0

a = b = c

BCA1B01T · M1

Adjoint and inverse

Read the topic →

Example 1

Inverse of a 2 × 2 matrix

Question

Find $A^{- 1}$ for $A = [4276]$ and verify.

Determinant
$∣ A ∣ = 4 (6) - 7 (2) = 24 - 14 = 10 \neq = 0$

Non-singular, so the inverse exists.
Adjoint by swap-and-negate
Swap $4$ and $6$ ; negate $7$ and $2$ :

$adj A = [6 - 2 - 7 4]$
Divide
$A^{- 1} = \frac{1}{10} [6 - 2 - 7 4] = [0.6 - 0.2 - 0.7 0.4]$
Verify $A A^{- 1} = I$
$A A^{- 1} = \frac{1}{10} [4276] [6 - 2 - 7 4] = \frac{1}{10} [24 - 14 12 - 12 - 28 + 28 - 14 + 24] = \frac{1}{10} [100010] = I ✓$

Answer

$A^{- 1} = \frac{1}{10} [6 - 2 - 7 4]$

Example 2

Inverse of a 3 × 3 by the adjoint method

Question

Find $A^{- 1}$ for $A = 105216340$ .

Determinant — expand along row 2 (it has a zero)
Row 2 is $(0, 1, 4)$ with signs $(-, +, -)$ :

$∣ A ∣ = - 0 + 11530 - 41526 = 1 (0 - 15) - 4 (6 - 10) = - 15 + 16 = 1$

$∣ A ∣ = 1 \neq = 0$ , so $A^{- 1}$ exists — and the division at the end will be free.
Cofactors, row 1
$A_{11} = + 1640 = 0 - 24 = - 24$
$A_{12} = - 0540 = - (0 - 20) = 20$
$A_{13} = + 0516 = 0 - 5 = - 5$
Cofactors, row 2
$A_{21} = - 2630 = - (0 - 18) = 18$
$A_{22} = + 1530 = 0 - 15 = - 15$
$A_{23} = - 1526 = - (6 - 10) = 4$
Cofactors, row 3
$A_{31} = + 2134 = 8 - 3 = 5$
$A_{32} = - 1034 = - (4 - 0) = - 4$
$A_{33} = + 1021 = 1 - 0 = 1$
Cofactor matrix, then transpose it
$C = - 24 185 20 - 15 - 4 - 5 41 adj A = C^{T} = - 24 20 - 5 18 - 15 4 5 - 4 1$
Divide by $∣ A ∣ = 1$
$A^{- 1} = \frac{1}{1} - 24 20 - 5 18 - 15 4 5 - 4 1 = - 24 20 - 5 18 - 15 4 5 - 4 1$
Verify one row
Row 1 of $A$ against column 1 of $A^{- 1}$ :

$1 (- 24) + 2 (20) + 3 (- 5) = - 24 + 40 - 15 = 1 ✓$

Row 1 of $A$ against column 2 of $A^{- 1}$ :

$1 (18) + 2 (- 15) + 3 (4) = 18 - 30 + 12 = 0 ✓$

Answer

$A^{- 1} = - 24 20 - 5 18 - 15 4 5 - 4 1$

NoteLay the nine cofactors out in a grid as you compute them, then transpose once at the end. Trying to write the adjoint directly is where sign errors creep in.

Example 3

The same inverse by Gauss–Jordan

Question

Find $A^{- 1}$ for $A = [2714]$ using elementary row operations.

Set up the augmented array
$[27141001]$
Make a leading 1 — $R_{1} \to \frac{1}{2} R_{1}$
$[17 \frac{1}{2} 4 \frac{1}{2} 0 01]$
Clear below — $R_{2} \to R_{2} - 7 R_{1}$
$[10 \frac{1}{2} \frac{1}{2} \frac{1}{2} - \frac{7}{2} 01]$
Make the second leading 1 — $R_{2} \to 2 R_{2}$
$[10 \frac{1}{2} 1 \frac{1}{2} - 7 02]$
Clear above — $R_{1} \to R_{1} - \frac{1}{2} R_{2}$
$[1001 \frac{1}{2} + \frac{7}{2} - 7 - 1 2] = [1001 4 - 7 - 1 2]$
Read off the right block
$A^{- 1} = [4 - 7 - 1 2]$

Cross-check with the shortcut: $∣ A ∣ = 8 - 7 = 1$ , and swap-and-negate gives $[4 - 7 - 1 2]$ ✓

Answer

$A^{- 1} = [4 - 7 - 1 2]$

Problem 1Basic

Find the inverse of $A = [3152]$ , or show it does not exist.

Show solution

$∣ A ∣ = 3 (2) - 5 (1) = 6 - 5 = 1 \neq = 0$

Non-singular, so the inverse exists.

$adj A = [2 - 1 - 5 3]$

$A^{- 1} = \frac{1}{1} [2 - 1 - 5 3] = [2 - 1 - 5 3]$

Check: $[3152] [2 - 1 - 5 3] = [6 - 5 2 - 2 - 15 + 15 - 5 + 6] = I$ ✓

Answer

A^{- 1} = [2 - 1 - 5 3]

Problem 2Basic

Show that $A = [2163]$ has no inverse.

Show solution

$∣ A ∣ = 2 (3) - 6 (1) = 6 - 6 = 0$

Since $∣ A ∣ = 0$ , the matrix is singular and $A^{- 1}$ does not exist.

The reason is visible in the matrix: row 1 is exactly $2 \times$ row 2, so the rows are proportional and property P6 forces the determinant to zero.

Answer

∣ A ∣ = 0

, so

A

is singular and has no inverse.

Problem 3Exam level

Find $adj A$ and $A^{- 1}$ for $A = 112 12 - 1 1 - 3 3$ .

Show solution

Determinant (row 1):

$∣ A ∣ = 1 (6 - 3) - 1 (3 + 6) + 1 (- 1 - 4) = 3 - 9 - 5 = - 11 \neq = 0$

Cofactors.

$A_{11} = + 2 - 1 - 3 3 = 6 - 3 = 3 A_{12} = - 12 - 3 3 = - (3 + 6) = - 9$
$A_{13} = + 12 2 - 1 = - 1 - 4 = - 5$

$A_{21} = - 1 - 1 13 = - (3 + 1) = - 4 A_{22} = + 1213 = 3 - 2 = 1$
$A_{23} = - 12 1 - 1 = - (- 1 - 2) = 3$

$A_{31} = + 12 1 - 3 = - 3 - 2 = - 5 A_{32} = - 11 1 - 3 = - (- 3 - 1) = 4$
$A_{33} = + 1112 = 2 - 1 = 1$

Cofactor matrix and adjoint.

$C = 3 - 4 - 5 - 9 14 - 5 31 adj A = C^{T} = 3 - 9 - 5 - 4 13 - 5 41$

Inverse.

$A^{- 1} = - \frac{1}{11} 3 - 9 - 5 - 4 13 - 5 41 = \frac{1}{11} - 3 95 4 - 1 - 3 5 - 4 - 1$

Check (row 1 of $A$ × column 1 of $adj A$ ): $1 (3) + 1 (- 9) + 1 (- 5) = - 11 = ∣ A ∣$ ✓

Answer

adj A = 3 - 9 - 5 - 4 13 - 5 41

A^{- 1} = \frac{1}{11} - 3 95 4 - 1 - 3 5 - 4 - 1

Problem 4Exam level

Find $A^{- 1}$ for $A = 124 0 - 1 1 238$ by Gauss–Jordan.

Show solution

$124 0 - 1 1 238100010001$

$R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - 4 R_{1}$ :

$100 0 - 1 1 2 - 1 0 1 - 2 - 4 010001$

$R_{2} \to - R_{2}$ :

$100011210 12 - 4 0 - 1 0 001$

$R_{3} \to R_{3} - R_{2}$ :

$100010 21 - 1 12 - 6 0 - 1 1 001$

$R_{3} \to - R_{3}$ :

$100010211126 0 - 1 - 1 00 - 1$

$R_{1} \to R_{1} - 2 R_{3}$ , $R_{2} \to R_{2} - R_{3}$ :

$100010001 - 11 - 4 6 20 - 1 21 - 1$

$A^{- 1} = - 11 - 4 6 20 - 1 21 - 1$

Check (row 1 of $A$ × column 1 of $A^{- 1}$ ): $1 (- 11) + 0 (- 4) + 2 (6) = - 11 + 12 = 1$ ✓
(row 2 of $A$ × column 1): $2 (- 11) + (- 1) (- 4) + 3 (6) = - 22 + 4 + 18 = 0$ ✓

Answer

A^{- 1} = - 11 - 4 6 20 - 1 21 - 1

Problem 5Exam level

If $A = [34 - 2 - 2]$ , verify $A (adj A) = ∣ A ∣ I$ .

Show solution

$∣ A ∣ = 3 (- 2) - (- 2) (4) = - 6 + 8 = 2$

$adj A = [- 2 - 4 23]$

$A (adj A) = [34 - 2 - 2] [- 2 - 4 23] = [- 6 + 8 - 8 + 8 6 - 6 8 - 6] = [2002] = 2 I = ∣ A ∣ I ✓$

Multiplying on the other side gives the same:

$(adj A) A = [- 2 - 4 23] [34 - 2 - 2] = [- 6 + 8 - 12 + 12 4 - 4 8 - 6] = 2 I ✓$

AnswerBoth products equal

2 I = ∣ A ∣ I

Problem 6Challenge

$A$ is a non-singular matrix of order $3$ with $∣ A ∣ = 4$ . Find $∣ adj A ∣$ , $∣ adj (adj A) ∣$ and $∣ A^{- 1} ∣$ .

Show solution

Here $n = 3$ .

$∣ adj A ∣$ . Take determinants of the identity $A (adj A) = ∣ A ∣ I$ :

$∣ A ∣ \cdot ∣ adj A ∣ = ∣ A ∣ I = ∣ A ∣^{n} = ∣ A ∣^{3}$

$\Rightarrow ∣ adj A ∣ = ∣ A ∣^{n - 1} = 4^{2} = 16$

$∣ adj (adj A) ∣$ . Apply the same rule to the matrix $adj A$ , whose determinant is $16$ :

$∣ adj (adj A) ∣ = ∣ adj A ∣^{n - 1} = 1 6^{2} = 256$

(Equivalently $∣ A ∣^{(n - 1)^{2}} = 4^{4} = 256$ .)

$∣ A^{- 1} ∣$ . From $A A^{- 1} = I$ , taking determinants gives $∣ A ∣ ∣ A^{- 1} ∣ = 1$ :

$∣ A^{- 1} ∣ = \frac{1}{4}$

Answer

∣ adj A ∣ = 16

∣ adj (adj A) ∣ = 256

∣ A^{- 1} ∣ = 1/4

Problem 7Challenge

If $A$ and $B$ are non-singular matrices of the same order, prove $(A B)^{- 1} = B^{- 1} A^{- 1}$ .

Show solution

To prove $X$ is the inverse of $A B$ , it is enough to show $(A B) X = I$ and $X (A B) = I$ . Take $X = B^{- 1} A^{- 1}$ .

First product, using associativity throughout:

$(A B) (B^{- 1} A^{- 1}) = A (B B^{- 1}) A^{- 1} = A I A^{- 1} = A A^{- 1} = I$

Second product:

$(B^{- 1} A^{- 1}) (A B) = B^{- 1} (A^{- 1} A) B = B^{- 1} I B = B^{- 1} B = I$

Both products give $I$ , and the inverse is unique, so

$(A B)^{- 1} = B^{- 1} A^{- 1} ■$

Why the order reverses. $B^{- 1}$ must be adjacent to $B$ for the cancellation to happen. Writing $A^{- 1} B^{- 1}$ would leave $A B A^{- 1} B^{- 1}$ , which does not simplify unless $A$ and $B$ happen to commute.

Answer

(A B) (B^{- 1} A^{- 1}) = (B^{- 1} A^{- 1}) (A B) = I

, so

(A B)^{- 1} = B^{- 1} A^{- 1}

BCA1B01T · M1

Elementary operations, echelon form and rank

Read the topic →

Example 1

Rank by reduction to echelon form

Question

Find the rank of $A = 1232463710$ .

Use the pivot $a_{11} = 1$ to clear column 1
$R_{2} \to R_{2} - 2 R_{1}$ : $(2 - 2, 4 - 4, 7 - 6) = (0, 0, 1)$

$R_{3} \to R_{3} - 3 R_{1}$ : $(3 - 3, 6 - 6, 10 - 9) = (0, 0, 1)$

$A \sim 100200311$
Clear below the next pivot
The next pivot is the $1$ in row 2, column 3. Apply $R_{3} \to R_{3} - R_{2}$ :

$A \sim 100200310$

This is echelon form: pivots at $(1, 1)$ and $(2, 3)$ , staircase descending, zero row at the bottom.
Count non-zero rows
Two non-zero rows, so $ρ (A) = 2$ .
Cross-check with minors
Rank $2$ predicts $∣ A ∣ = 0$ and some non-zero $2 \times 2$ minor.

$∣ A ∣ = 1 (40 - 42) - 2 (20 - 21) + 3 (12 - 12) = - 2 + 2 + 0 = 0 ✓$

$1237 = 7 - 6 = 1 \neq = 0 ✓$

Both conditions hold, confirming $ρ (A) = 2$ .

Answer

$ρ (A) = 2$

NoteColumn 2 is exactly 2 × column 1 — that dependence is what costs the matrix a unit of rank.

Example 2

Rank of a rectangular matrix

Question

Find the rank of $A = 123246 - 1 13 3 - 2 - 7$ .

Note the ceiling
$A$ is $3 \times 4$ , so $ρ (A) \leq min (3, 4) = 3$ .
Clear column 1
$R_{2} \to R_{2} - 2 R_{1}$ : $(0, 0, 1 - (- 2), - 2 - 6) = (0, 0, 3, - 8)$

$R_{3} \to R_{3} - 3 R_{1}$ : $(0, 0, 3 - (- 3), - 7 - 9) = (0, 0, 6, - 16)$

$A \sim 100200 - 1 36 3 - 8 - 16$
Clear below the second pivot
Pivot is the $3$ at position $(2, 3)$ . Apply $R_{3} \to R_{3} - 2 R_{2}$ :

$(0, 0, 6 - 6, - 16 + 16) = (0, 0, 0, 0)$

$A \sim 100200 - 1 30 3 - 8 0$
Count
Two non-zero rows: $ρ (A) = 2$ .

Answer

$ρ (A) = 2$

Example 3

Rank depending on a parameter

Question

Find the values of $k$ for which $A = 123 2 k 6 369$ has rank $1$ , rank $2$ , and rank $3$ .

Test for rank 3 first
Rank $3$ requires $∣ A ∣ \neq = 0$ . But row 3 $= 3 \times$ row 1 regardless of $k$ , so by the proportional-rows property

$∣ A ∣ = 0 for every k$

Rank $3$ is impossible.
Reduce
$R_{2} \to R_{2} - 2 R_{1}$ and $R_{3} \to R_{3} - 3 R_{1}$ :

$A \sim 100 2 k - 4 0 300$
Split on the pivot in row 2
If $k - 4 \neq = 0$ , i.e. $k \neq = 4$ : two non-zero rows, so $ρ (A) = 2$ .

If $k = 4$ : row 2 vanishes, leaving one non-zero row, so $ρ (A) = 1$ .

Answer

$ρ (A) = 1$ when $k = 4$ ; $ρ (A) = 2$ for every $k \neq = 4$ ; rank $3$ never occurs.

NoteWith a parameter, always reduce first and split into cases only at the point where a pivot could vanish.

Problem 1Basic

Find the rank of $A = [1236]$ by both methods.

Show solution

Echelon. $R_{2} \to R_{2} - 2 R_{1}$ gives $[1030]$ — one non-zero row.

$ρ (A) = 1$

Minors. The only order- $2$ minor is $∣ A ∣ = 1 (6) - 3 (2) = 0$ . There is a non-zero order- $1$ minor (for instance $∣1∣ = 1$ ).

Largest non-zero minor has order $1$ , so $ρ (A) = 1$ ✓

Answer

ρ (A) = 1

Problem 2Basic

Reduce $A = 021142310$ to echelon form and state its rank.

Show solution

The $(1, 1)$ entry is $0$ , so swap first: $R_{1} \leftrightarrow R_{3}$ .

$120241013$

$R_{2} \to R_{2} - 2 R_{1}$ :

$100201013$

The pivot pattern is out of order, so swap $R_{2} \leftrightarrow R_{3}$ :

$100210031$

This is echelon form with three pivots.

$ρ (A) = 3$

Being $3 \times 3$ of full rank, $A$ is non-singular. (Indeed the echelon form is upper triangular with determinant $1 \cdot 1 \cdot 1 = 1$ , and the two swaps make $∣ A ∣ = (- 1)^{2} \cdot 1 = 1$ .)

Answer

ρ (A) = 3

(full rank, non-singular)

Problem 3Exam level

Find the rank of $A = 12312461369148121$ .

Show solution

Rows $2$ and $3$ are $2 \times$ and $3 \times$ row $1$ , so they will vanish immediately.

$R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - 3 R_{1}$ , $R_{4} \to R_{4} - R_{1}$ :

$A \sim 1000 200 - 1 300 - 2 400 - 3$

Move the non-zero row up: $R_{2} \leftrightarrow R_{4}$ .

$A \sim 1000 2 - 1 00 3 - 2 00 4 - 3 00$

Echelon form with two non-zero rows.

$ρ (A) = 2$

Answer

ρ (A) = 2

Problem 4Exam level

Find the rank of $A = 2022 - 1 335 347114156$ .

Show solution

Pivot $a_{11} = 2$ . Apply $R_{3} \to R_{3} - R_{1}$ and $R_{4} \to R_{4} - R_{1}$ :

$R_{3}$ : $(0, 3 - (- 1), 7 - 3, 5 - 4) = (0, 4, 4, 1)$
$R_{4}$ : $(0, 5 - (- 1), 11 - 3, 6 - 4) = (0, 6, 8, 2)$

$A \sim 2000 - 1 346 34484112$

Pivot is now the $3$ at $(2, 2)$ . Clear below: $R_{3} \to 3 R_{3} - 4 R_{2}$ , $R_{4} \to R_{4} - 2 R_{2}$ .

$R_{3}$ : $(0, 12 - 12, 12 - 16, 3 - 4) = (0, 0, - 4, - 1)$
$R_{4}$ : $(0, 6 - 6, 8 - 8, 2 - 2) = (0, 0, 0, 0)$

$A \sim 2000 - 1 300 34 - 4 0 41 - 1 0$

Echelon form with three non-zero rows.

$ρ (A) = 3$

(Scaling $R_{3}$ by $3$ is legitimate — multiplying a row by a non-zero scalar preserves rank. It keeps the arithmetic in integers.)

Answer

ρ (A) = 3

Problem 5Exam level

For what value of $λ$ does $A = 30 - 3 15 λ 48 - 4$ have rank less than $3$ ?

Show solution

Rank $< 3$ for a $3 \times 3$ matrix means $∣ A ∣ = 0$ .

Expand along column 1:

$∣ A ∣ = 3 5 λ 8 - 4 - 0 + (- 3) 1548$

$= 3 (- 20 - 8 λ) - 3 (8 - 20)$

$= - 60 - 24 λ + 36 = - 24 - 24 λ$

Setting $∣ A ∣ = 0$ :

$- 24 - 24 λ = 0 \Rightarrow λ = - 1$

Confirm the rank at $λ = - 1$ . Apply $R_{3} \to R_{3} + R_{1}$ :

$300150480$

Two non-zero rows, so $ρ (A) = 2$ when $λ = - 1$ , and $ρ (A) = 3$ otherwise.

Answer

λ = - 1

, giving

ρ (A) = 2

Problem 6Challenge

Reduce $A = 110 12 - 1 23 - 1$ to normal form $[I_{r} O O O]$ and state its rank.

Show solution

Row operations first. $R_{2} \to R_{2} - R_{1}$ :

$100 11 - 1 21 - 1$

$R_{3} \to R_{3} + R_{2}$ :

$100110210$

$R_{1} \to R_{1} - R_{2}$ :

$100010110$

Now column operations. $C_{3} \to C_{3} - C_{1} - C_{2}$ :

Column 3 becomes $(1 - 1 - 0, 1 - 0 - 1, 0) = (0, 0, 0)$ .

$100010000 = [I_{2} O O O]$

The identity block has order $2$ , so

$ρ (A) = 2$

AnswerNormal form

[I_{2} O O O]

, so

ρ (A) = 2

BCA1B01T · M1

Systems of linear equations

Read the topic →

Example 1

Three unknowns by Cramer's rule

Question

Solve by Cramer's rule:
$x + y + z x - y + z 2 x + y - z = 6 = 2 = 1$

Write the coefficient determinant
$Δ = 112 1 - 1 1 11 - 1$

Expanding along row 1:

$= 1 [(- 1) (- 1) - (1) (1)] - 1 [(1) (- 1) - (1) (2)] + 1 [(1) (1) - (- 1) (2)]$
$= 1 (1 - 1) - 1 (- 1 - 2) + 1 (1 + 2) = 0 + 3 + 3 = 6$

$Δ = 6 \neq = 0$ , so a unique solution exists.
$Δ_{1}$ — replace column 1 with $B = (6, 2, 1)$
$Δ_{1} = 621 1 - 1 1 11 - 1 = 6 (1 - 1) - 1 (- 2 - 1) + 1 (2 + 1) = 0 + 3 + 3 = 6$
$Δ_{2}$ — replace column 2
$Δ_{2} = 112621 11 - 1 = 1 (- 2 - 1) - 6 (- 1 - 2) + 1 (1 - 4) = - 3 + 18 - 3 = 12$
$Δ_{3}$ — replace column 3
$Δ_{3} = 112 1 - 1 1 621 = 1 (- 1 - 2) - 1 (1 - 4) + 6 (1 + 2) = - 3 + 3 + 18 = 18$
Divide
$x = \frac{6}{6} = 1, y = \frac{12}{6} = 2, z = \frac{18}{6} = 3$
Check in all three equations
$1 + 2 + 3 = 6$ ✓ $1 - 2 + 3 = 2$ ✓ $2 + 2 - 3 = 1$ ✓

Answer

$x = 1, y = 2, z = 3$

Example 2

The same system by Gauss elimination

Question

Solve by Gauss elimination:
$x + y + z x - y + z 2 x + y - z = 6 = 2 = 1$

Form the augmented matrix
$112 1 - 1 1 11 - 1 621$
Clear column 1
$R_{2} \to R_{2} - R_{1}$ : $(0, - 2, 0 ∣ - 4)$

$R_{3} \to R_{3} - 2 R_{1}$ : $(0, - 1, - 3 ∣ - 11)$

$100 1 - 2 - 1 10 - 3 6 - 4 - 11$
Clear column 2
$R_{3} \to 2 R_{3} - R_{2}$ : $(0, - 2 + 2, - 6 - 0 ∣ - 22 + 4) = (0, 0, - 6 ∣ - 18)$

$100 1 - 2 0 10 - 6 6 - 4 - 18$

Upper triangular — elimination is complete.
Back-substitute
From row 3: $- 6 z = - 18 \Rightarrow z = 3$

From row 2: $- 2 y = - 4 \Rightarrow y = 2$

From row 1: $x + y + z = 6 \Rightarrow x + 2 + 3 = 6 \Rightarrow x = 1$
Note the ranks
$ρ (A) = 3$ and $ρ ([A ∣ B]) = 3$ , with $n = 3$ unknowns. Equal ranks, both equal to $n$ — consistent with a unique solution, exactly as found.

Answer

$x = 1, y = 2, z = 3$ — the same as by Cramer's rule, with far less arithmetic.

NoteScaling R₃ by 2 before subtracting avoids fractions. Legal, and worth doing every time.

Example 3

Solving by the inverse method

Question

Solve using $X = A^{- 1} B$ :
$2 x - 3 y 4 x + y = 5 = 3$

Write in matrix form
$A [24 - 3 1] X [x y] = B [53]$
Invert $A$
$∣ A ∣ = 2 (1) - (- 3) (4) = 2 + 12 = 14 \neq = 0$

$A^{- 1} = \frac{1}{14} [1 - 4 32]$
Multiply
$X = A^{- 1} B = \frac{1}{14} [1 - 4 32] [53] = \frac{1}{14} [5 + 9 - 20 + 6] = \frac{1}{14} [14 - 14] = [1 - 1]$
Check
$2 (1) - 3 (- 1) = 2 + 3 = 5$ ✓ $4 (1) + (- 1) = 3$ ✓

Answer

$x = 1, y = - 1$

NoteKeep the 1/14 outside until the very last step — it turns the whole computation into integer arithmetic.

Example 4

Testing consistency by rank

Question

Test for consistency and solve if possible:
$x + 2 y + z 2 x + 3 y + 2 z 3 x + 5 y + 3 z = 3 = 5 = 9$

Form the augmented matrix
$123235123359$
Reduce
$R_{2} \to R_{2} - 2 R_{1}$ : $(0, - 1, 0 ∣ - 1)$

$R_{3} \to R_{3} - 3 R_{1}$ : $(0, - 1, 0 ∣ 0)$

$100 2 - 1 - 1 100 3 - 1 0$

$R_{3} \to R_{3} - R_{2}$ : $(0, 0, 0 ∣ 1)$

$100 2 - 1 0 100 3 - 1 1$
Read the ranks
Coefficient block: two non-zero rows, so $ρ (A) = 2$ .

Augmented matrix: three non-zero rows, so $ρ ([A ∣ B]) = 3$ .
Conclude
$ρ (A) = 2 \neq = 3 = ρ ([A ∣ B])$

The system is inconsistent — no solution.

The last row says $0 x + 0 y + 0 z = 1$ , i.e. $0 = 1$ , which is impossible. Geometrically the three planes have no common point.

Answer

Inconsistent: $ρ (A) = 2 \neq = ρ ([A ∣ B]) = 3$ . No solution exists.

Example 5

A homogeneous system with non-trivial solutions

Question

Solve
$x + 2 y - z 2 x + y + z 3 x + 3 y = 0 = 0 = 0$

Check the determinant
$∣ A ∣ = 123213 - 1 10 = 1 (0 - 3) - 2 (0 - 3) + (- 1) (6 - 3) = - 3 + 6 - 3 = 0$

$∣ A ∣ = 0$ , so non-trivial solutions exist.
Row-reduce (no $B$ column needed — it stays zero)
$R_{2} \to R_{2} - 2 R_{1}$ : $(0, - 3, 3)$

$R_{3} \to R_{3} - 3 R_{1}$ : $(0, - 3, 3)$

$100 2 - 3 - 3 - 1 33$

$R_{3} \to R_{3} - R_{2}$ :

$100 2 - 3 0 - 1 30$
Count the free variables
$ρ (A) = 2$ , $n = 3$ unknowns, so there are $n - ρ (A) = 1$ free variable — a one-parameter family of solutions.
Parametrise and back-substitute
Let $z = k$ .

Row 2: $- 3 y + 3 z = 0 \Rightarrow y = z = k$

Row 1: $x + 2 y - z = 0 \Rightarrow x + 2 k - k = 0 \Rightarrow x = - k$
Write the solution
$x y z = k - 1 11, k \in R$

Check with $k = 1$ , i.e. $(x, y, z) = (- 1, 1, 1)$ : $- 1 + 2 - 1 = 0$ ✓, $- 2 + 1 + 1 = 0$ ✓, $- 3 + 3 = 0$ ✓

Setting $k = 0$ recovers the trivial solution.

Answer

$(x, y, z) = k (- 1, 1, 1)$ for any real $k$ .

NoteAlways present a homogeneous answer as a multiple of a fixed vector. Giving one particular solution loses marks — the question asks for all of them.

Problem 1Basic

Solve by Cramer's rule: $3 x + 2 y = 7, x - y = 1$ .

Show solution

$Δ = 31 2 - 1 = - 3 - 2 = - 5 \neq = 0$

$Δ_{1} = 71 2 - 1 = - 7 - 2 = - 9 Δ_{2} = 3171 = 3 - 7 = - 4$

$x = \frac{- 9}{- 5} = \frac{9}{5}, y = \frac{- 4}{- 5} = \frac{4}{5}$

Check: $3 (9/5) + 2 (4/5) = 27/5 + 8/5 = 35/5 = 7$ ✓ $9/5 - 4/5 = 1$ ✓

Answer

x = 9/5, y = 4/5

Problem 2Basic

Write $2 x - y + 3 z = 9, x + y + z = 6, x - y + z = 2$ in the form $A X = B$ and state the augmented matrix.

Show solution

$A = 211 - 1 1 - 1 311, X = x y z, B = 962$

$A X = B : 211 - 1 1 - 1 311 x y z = 962$

Augmented matrix:

$[A ∣ B] = 211 - 1 1 - 1 311962$

Answer

[A ∣ B] = 211 - 1 1 - 1 311962

Problem 3Exam level

Solve by Gauss–Jordan:
$x + y + z 2 x + 5 y + 7 z 2 x + y - z = 9 = 52 = 0$

Show solution

$122151 17 - 1 9520$

$R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - 2 R_{1}$ :

$100 13 - 1 15 - 3 934 - 18$

$R_{2} \leftrightarrow R_{3}$ , then $R_{2} \to - R_{2}$ to get a leading $1$ :

$10011313591834$

$R_{1} \to R_{1} - R_{2}$ , $R_{3} \to R_{3} - 3 R_{2}$ :

$100010 - 2 3 - 4 - 9 18 - 20$

$R_{3} \to - \frac{1}{4} R_{3}$ :

$100010 - 2 31 - 9 185$

$R_{1} \to R_{1} + 2 R_{3}$ , $R_{2} \to R_{2} - 3 R_{3}$ :

$100010001135$

Read off directly:

$x = 1, y = 3, z = 5$

Check: $1 + 3 + 5 = 9$ ✓ $2 + 15 + 35 = 52$ ✓ $2 + 3 - 5 = 0$ ✓

Answer

x = 1, y = 3, z = 5

Problem 4Exam level

Test for consistency and solve:
$x + y + z x + 2 y + 3 z x + 4 y + 7 z = 6 = 14 = 30$

Show solution

$11112413761430$

$R_{2} \to R_{2} - R_{1}$ , $R_{3} \to R_{3} - R_{1}$ :

$1001131266824$

$R_{3} \to R_{3} - 3 R_{2}$ :

$100110120680$

Ranks. $ρ (A) = 2$ and $ρ ([A ∣ B]) = 2$ . Equal, so the system is consistent. But $n = 3$ unknowns and $r = 2$ , so there are $n - r = 1$ free variable and infinitely many solutions.

Parametrise. Let $z = k$ .

Row 2: $y + 2 k = 8 \Rightarrow y = 8 - 2 k$

Row 1: $x + (8 - 2 k) + k = 6 \Rightarrow x = k - 2$

$x y z = - 2 80 + k 1 - 2 1, k \in R$

Check with $k = 2$ , giving $(0, 4, 2)$ : $0 + 4 + 2 = 6$ ✓, $0 + 8 + 6 = 14$ ✓, $0 + 16 + 14 = 30$ ✓

AnswerConsistent with infinitely many solutions:

x = k - 2, y = 8 - 2 k, z = k

Problem 5Exam level

Find the values of $λ$ for which the system has (i) a unique solution, (ii) no solution, (iii) infinitely many solutions:
$x + y + z x + 2 y + 4 z x + 4 y + 10 z = 1 = λ = λ^{2}$

Show solution

$1111241410 1 λ λ^{2}$

$R_{2} \to R_{2} - R_{1}$ , $R_{3} \to R_{3} - R_{1}$ :

$100113139 1 λ - 1 λ^{2} - 1$

$R_{3} \to R_{3} - 3 R_{2}$ :

$100110130 1 λ - 1 λ^{2} - 3 λ + 2$

since $λ^{2} - 1 - 3 (λ - 1) = λ^{2} - 3 λ + 2 = (λ - 1) (λ - 2)$ .

(i) Unique solution. This would need $ρ (A) = 3$ , but the coefficient block has a zero row for every $λ$ , so $ρ (A) = 2$ always. A unique solution never occurs.

(ii) No solution. Need $(λ - 1) (λ - 2) \neq = 0$ , which makes $ρ ([A ∣ B]) = 3 \neq = 2 = ρ (A)$ :

$λ \neq = 1 and λ \neq = 2$

(iii) Infinitely many. Need $(λ - 1) (λ - 2) = 0$ , giving $ρ (A) = ρ ([A ∣ B]) = 2 < 3$ :

$λ = 1 or λ = 2$

Solution when $λ = 1$ : row 2 becomes $y + 3 z = 0$ . With $z = k$ : $y = - 3 k$ , and $x = 1 - y - z = 1 + 3 k - k = 1 + 2 k$ .

$(x, y, z) = (1 + 2 k, - 3 k, k)$

Solution when $λ = 2$ : row 2 becomes $y + 3 z = 1$ . With $z = k$ : $y = 1 - 3 k$ , and $x = 1 - (1 - 3 k) - k = 2 k$ .

$(x, y, z) = (2 k, 1 - 3 k, k)$

Answer(i) never; (ii)

λ \neq = 1, 2

; (iii)

λ = 1

λ = 2

Problem 6Exam level

Show that the system $x + 2 y + 3 z = 0, 3 x + 4 y + 4 z = 0, 7 x + 10 y + 12 z = 0$ has only the trivial solution.

Show solution

Homogeneous, so it is automatically consistent. Test $∣ A ∣$ .

$∣ A ∣ = 13724103412$

Expand along row 1:

$= 1 (48 - 40) - 2 (36 - 28) + 3 (30 - 28)$
$= 1 (8) - 2 (8) + 3 (2) = 8 - 16 + 6 = - 2$

$∣ A ∣ = - 2 \neq = 0$

Since the coefficient matrix is non-singular, $ρ (A) = 3 = n$ , and the only solution of $A X = O$ is

$X = A^{- 1} O = O$

that is, $x = y = z = 0$ — the trivial solution only. ∎

Answer

∣ A ∣ = - 2 \neq = 0

, so

x = y = z = 0

is the only solution.

Problem 7Challenge

Find $k$ so that $k x + 3 y - z = 0, 4 x + k y - 2 z = 0, 2 x + y - z = 0$ has a non-trivial solution, and find that solution for the smaller value of $k$ .

Show solution

A homogeneous system has non-trivial solutions exactly when $∣ A ∣ = 0$ .

$∣ A ∣ = k 42 3 k 1 - 1 - 2 - 1$

Expand along row 1:

$= k (- k + 2) - 3 (- 4 + 4) + (- 1) (4 - 2 k)$
$= - k^{2} + 2 k - 0 - 4 + 2 k$
$= - k^{2} + 4 k - 4 = - (k^{2} - 4 k + 4) = - (k - 2)^{2}$

$∣ A ∣ = 0 ⟺ (k - 2)^{2} = 0 ⟺ k = 2$

Only one value, $k = 2$ (a repeated root).

Solving at $k = 2$ .

$242321 - 1 - 2 - 1$

$R_{2} \to R_{2} - 2 R_{1}$ : $(0, - 4, 0)$
$R_{3} \to R_{3} - R_{1}$ : $(0, - 2, 0)$

$200 3 - 4 - 2 - 1 00 R_{3} \to 2 R_{3} - R_{2} 200 3 - 4 0 - 1 00$

$ρ (A) = 2$ , $n = 3$ , so one free parameter.

Row 2: $- 4 y = 0 \Rightarrow y = 0$ .

Row 1: $2 x + 0 - z = 0 \Rightarrow z = 2 x$ . Let $x = t$ .

$(x, y, z) = t (1, 0, 2), t \in R$

Check with $t = 1$ , i.e. $(1, 0, 2)$ : $2 + 0 - 2 = 0$ ✓, $4 + 0 - 4 = 0$ ✓, $2 + 0 - 2 = 0$ ✓

Answer

k = 2

; solution

(x, y, z) = t (1, 0, 2)

BCA1B01T · M1

Eigenvalues and eigenvectors

Read the topic →

Example 1

Eigenvalues and eigenvectors of a 2 × 2

Question

Find the eigenvalues and corresponding eigenvectors of $A = [4213]$ .

Write the characteristic equation
$tr A = 4 + 3 = 7, ∣ A ∣ = 12 - 2 = 10$

$λ^{2} - 7 λ + 10 = 0$
Solve for $λ$
$(λ - 5) (λ - 2) = 0 ⟹ λ = 5, 2$

Sanity check: $5 + 2 = 7 = tr A$ ✓ and $5 \times 2 = 10 = ∣ A ∣$ ✓
Eigenvector for $λ = 5$
$A - 5 I = [- 1 2 1 - 2]$

Row 2 is $- 2 \times$ row 1, as it must be (the matrix is singular). The single independent equation is

$- x_{1} + x_{2} = 0 ⟹ x_{2} = x_{1}$

Taking $x_{1} = 1$ :

$X_{1} = [11]$
Eigenvector for $λ = 2$
$A - 2 I = [2211]$

The equation is $2 x_{1} + x_{2} = 0 \Rightarrow x_{2} = - 2 x_{1}$ .

Taking $x_{1} = 1$ :

$X_{2} = [1 - 2]$
Verify both
$A X_{1} = [4213] [11] = [55] = 5 [11] ✓$

$A X_{2} = [4213] [1 - 2] = [2 - 4] = 2 [1 - 2] ✓$

Answer

$λ = 5$ with $X_{1} = [11]$ ; $λ = 2$ with $X_{2} = [1 - 2]$ .

Example 2

A 3 × 3 with distinct eigenvalues

Question

Find the eigenvalues and eigenvectors of $A = 210120111$ .

Compute $S_{1}$ , $S_{2}$ , $S_{3}$
$S_{1} = tr A = 2 + 2 + 1 = 5$

Principal minors of order $2$ (delete row $i$ and column $i$ ):

$2011 = 2, 2011 = 2, 2112 = 3$

$S_{2} = 2 + 2 + 3 = 7$

$S_{3} = ∣ A ∣ = 2 (2 - 0) - 1 (1 - 0) + 1 (0 - 0) = 4 - 1 = 3$
Write and solve the characteristic equation
$λ^{3} - 5 λ^{2} + 7 λ - 3 = 0$

Try $λ = 1$ : $1 - 5 + 7 - 3 = 0$ ✓, so $(λ - 1)$ is a factor.

$λ^{3} - 5 λ^{2} + 7 λ - 3 = (λ - 1) (λ^{2} - 4 λ + 3) = (λ - 1) (λ - 1) (λ - 3)$

$λ = 1, 1, 3$

Check: $1 + 1 + 3 = 5 = S_{1}$ ✓ and $1 \times 1 \times 3 = 3 = ∣ A ∣$ ✓
Eigenvector for $λ = 3$
$A - 3 I = - 1 10 1 - 1 0 11 - 2$

Row 3: $- 2 x_{3} = 0 \Rightarrow x_{3} = 0$ .

Row 1: $- x_{1} + x_{2} + 0 = 0 \Rightarrow x_{2} = x_{1}$ .

Taking $x_{1} = 1$ :

$X_{1} = 110$

Check: $A X_{1} = (2 + 1 + 0, 1 + 2 + 0, 0) = (3, 3, 0) = 3 X_{1}$ ✓
Eigenvectors for the repeated $λ = 1$
$A - I = 110110110$

All rows reduce to the single equation

$x_{1} + x_{2} + x_{3} = 0$

Rank is $1$ , so $n - ρ = 3 - 1 = 2$ free variables — this repeated eigenvalue supplies two independent eigenvectors.

Choose $x_{2} = 1, x_{3} = 0 \Rightarrow x_{1} = - 1$ :
$X_{2} = - 1 10$

Choose $x_{2} = 0, x_{3} = 1 \Rightarrow x_{1} = - 1$ :
$X_{3} = - 1 01$

Check $X_{2}$ : $A X_{2} = (- 2 + 1 + 0, - 1 + 2 + 0, 0) = (- 1, 1, 0) = 1 \cdot X_{2}$ ✓

Answer

$λ = 3$ : $X_{1} = 110$ . $λ = 1$ (twice): $X_{2} = - 1 10$ , $X_{3} = - 1 01$ .

NoteA repeated eigenvalue does not always give two independent eigenvectors. Count them by computing the rank of A − λI: the number is n − ρ(A − λI).

Example 3

Reading eigenvalues off a triangular matrix

Question

Find the eigenvalues of $A = 300 7 - 1 0 - 2 54$ and hence the eigenvalues of $A^{2}$ , $A^{- 1}$ and $A + 2 I$ .

Use the triangular shortcut
$A$ is upper triangular, so $∣ A - λ I ∣$ is also triangular and equals

$(3 - λ) (- 1 - λ) (4 - λ) = 0$

$λ = 3, - 1, 4$

No expansion needed.
Confirm with the trace and determinant
$tr A = 3 - 1 + 4 = 6$ and $3 + (- 1) + 4 = 6$ ✓

$∣ A ∣ = 3 (- 1) (4) = - 12$ and $3 \times (- 1) \times 4 = - 12$ ✓
Apply the property rules
$A^{2}$ : square each eigenvalue — $9, 1, 16$

$A^{- 1}$ : reciprocate each — $\frac{1}{3}, - 1, \frac{1}{4}$ (valid since no eigenvalue is $0$ )

$A + 2 I$ : add $2$ to each — $5, 1, 6$

Answer

$A$ : $3, - 1, 4$ . $A^{2}$ : $9, 1, 16$ . $A^{- 1}$ : $1/3, - 1, 1/4$ . $A + 2 I$ : $5, 1, 6$ .

NoteQuestions about A², A⁻¹ or A + kI never require recomputing a characteristic equation. Transform the eigenvalues you already have.

Problem 1Basic

Find the eigenvalues of $A = [5142]$ .

Show solution

$tr A = 5 + 2 = 7, ∣ A ∣ = 10 - 4 = 6$

$λ^{2} - 7 λ + 6 = 0 ⟹ (λ - 6) (λ - 1) = 0$

$λ = 6, 1$

Check: $6 + 1 = 7$ ✓, $6 \times 1 = 6$ ✓

Answer

λ = 6, 1

Problem 2Basic

Find the eigenvalues and eigenvectors of $A = [1322]$ .

Show solution

$tr A = 3, ∣ A ∣ = 2 - 6 = - 4$

$λ^{2} - 3 λ - 4 = 0 ⟹ (λ - 4) (λ + 1) = 0 ⟹ λ = 4, - 1$

For $λ = 4$ :
$A - 4 I = [- 3 3 2 - 2] ⟹ - 3 x_{1} + 2 x_{2} = 0 ⟹ x_{2} = \frac{3}{2} x_{1}$

Take $x_{1} = 2$ : $X_{1} = [23]$

For $λ = - 1$ :
$A + I = [2323] ⟹ x_{1} + x_{2} = 0 ⟹ x_{2} = - x_{1}$

Take $x_{1} = 1$ : $X_{2} = [1 - 1]$

Verify: $A X_{1} = (2 + 6, 6 + 6) = (8, 12) = 4 X_{1}$ ✓ $A X_{2} = (1 - 2, 3 - 2) = (- 1, 1) = - X_{2}$ ✓

Answer

λ = 4

with

[23]

;

λ = - 1

with

[1 - 1]

Problem 3Exam level

Find the characteristic equation and eigenvalues of $A = 113151311$ .

Show solution

$S_{1} = 1 + 5 + 1 = 7$

Principal $2 \times 2$ minors:

$5111 = 4, 1331 = - 8, 1115 = 4$

$S_{2} = 4 - 8 + 4 = 0$

$S_{3} = ∣ A ∣ = 1 (5 - 1) - 1 (1 - 3) + 3 (1 - 15) = 4 + 2 - 42 = - 36$

Characteristic equation:

$λ^{3} - 7 λ^{2} + 0 λ + 36 = 0 ⟹ λ^{3} - 7 λ^{2} + 36 = 0$

Try $λ = - 2$ : $- 8 - 28 + 36 = 0$ ✓

Divide: $λ^{3} - 7 λ^{2} + 36 = (λ + 2) (λ^{2} - 9 λ + 18) = (λ + 2) (λ - 3) (λ - 6)$

$λ = - 2, 3, 6$

Check: $- 2 + 3 + 6 = 7 = S_{1}$ ✓ $(- 2) (3) (6) = - 36 = ∣ A ∣$ ✓

(All eigenvalues are real, as they must be — $A$ is symmetric.)

Answer

λ^{3} - 7 λ^{2} + 36 = 0

;

λ = - 2, 3, 6

Problem 4Exam level

Find the eigenvalues and eigenvectors of $A = 211232112$ .

Show solution

Characteristic equation.

$S_{1} = 2 + 3 + 2 = 7$

$S_{2} = 3212 + 2112 + 2123 = 4 + 3 + 4 = 11$

$S_{3} = ∣ A ∣ = 2 (6 - 2) - 2 (2 - 1) + 1 (2 - 3) = 8 - 2 - 1 = 5$

$λ^{3} - 7 λ^{2} + 11 λ - 5 = 0$

Try $λ = 1$ : $1 - 7 + 11 - 5 = 0$ ✓

$(λ - 1) (λ^{2} - 6 λ + 5) = (λ - 1) (λ - 1) (λ - 5)$

$λ = 1, 1, 5$

Eigenvector for $λ = 5$ .

$A - 5 I = - 3 11 2 - 2 2 11 - 3$

$R_{1} \leftrightarrow R_{2}$ , then $R_{2} \to R_{2} + 3 R_{1}$ and $R_{3} \to R_{3} - R_{1}$ :

$100 - 2 - 4 4 14 - 4 \sim 100 - 2 - 4 0 140$

Row 2: $- 4 x_{2} + 4 x_{3} = 0 \Rightarrow x_{2} = x_{3}$ .
Row 1: $x_{1} - 2 x_{2} + x_{3} = 0 \Rightarrow x_{1} = 2 x_{2} - x_{3} = x_{3}$ .

Take $x_{3} = 1$ : $X_{1} = 111$

Check: $A X_{1} = (2 + 2 + 1, 1 + 3 + 1, 1 + 2 + 2) = (5, 5, 5) = 5 X_{1}$ ✓

Eigenvectors for $λ = 1$ .

$A - I = 111222111$

Rank $1$ , so $3 - 1 = 2$ independent eigenvectors. The single equation is

$x_{1} + 2 x_{2} + x_{3} = 0$

Take $x_{2} = 1, x_{3} = 0 \Rightarrow x_{1} = - 2$ : $X_{2} = - 2 10$

Take $x_{2} = 0, x_{3} = 1 \Rightarrow x_{1} = - 1$ : $X_{3} = - 1 01$

Check $X_{2}$ : $A X_{2} = (- 4 + 2 + 0, - 2 + 3 + 0, - 2 + 2 + 0) = (- 2, 1, 0) = X_{2}$ ✓

Answer

λ = 5

111

;

λ = 1

(twice):

- 2 10

and

- 1 01

Problem 5Exam level

If the eigenvalues of a $3 \times 3$ matrix $A$ are $2, - 3, 4$ , find $tr A$ , $∣ A ∣$ , the eigenvalues of $A^{3}$ , of $A^{- 1}$ , and of $3 A - I$ .

Show solution

Trace = sum of eigenvalues:
$tr A = 2 + (- 3) + 4 = 3$

Determinant = product of eigenvalues:
$∣ A ∣ = 2 \times (- 3) \times 4 = - 24$

Non-zero, so $A^{- 1}$ exists.

$A^{3}$ — cube each eigenvalue:
$8, - 27, 64$

$A^{- 1}$ — reciprocate each:
$\frac{1}{2}, - \frac{1}{3}, \frac{1}{4}$

$3 A - I$ — apply $λ \mapsto 3 λ - 1$ :
$3 (2) - 1 = 5, 3 (- 3) - 1 = - 10, 3 (4) - 1 = 11$

(In general, if $p$ is any polynomial then the eigenvalues of $p (A)$ are $p (λ_{i})$ .)

Answer

tr A = 3

∣ A ∣ = - 24

;

A^{3}

8, - 27, 64

;

A^{- 1}

1/2, - 1/3, 1/4

;

3 A - I

5, - 10, 11

Problem 6Challenge

Prove that if $λ$ is an eigenvalue of a non-singular matrix $A$ , then $1/ λ$ is an eigenvalue of $A^{- 1}$ with the same eigenvector.

Show solution

Start from the definition, with $X \neq = O$ :

$A X = λ X$

First, $λ \neq = 0$ . If $λ$ were $0$ then $A X = O$ with $X \neq = O$ , which would make $A$ singular — contradicting the hypothesis. So dividing by $λ$ is legal.

Multiply both sides on the left by $A^{- 1}$ :

$A^{- 1} (A X) = A^{- 1} (λ X)$

$(A^{- 1} A) X = λ A^{- 1} X$

$I X = λ A^{- 1} X ⟹ X = λ A^{- 1} X$

Divide by $λ$ :

$A^{- 1} X = \frac{1}{λ} X$

This is exactly the eigenvalue equation for $A^{- 1}$ , with eigenvalue $1/ λ$ and the same eigenvector $X$ . ∎

The same argument applied repeatedly gives the general rule: $A^{m} X = λ^{m} X$ for every integer $m$ , positive or negative.

Answer

A X = λ X \Rightarrow A^{- 1} X = (1/ λ) X

, same eigenvector.

BCA1B01T · M1

Cayley–Hamilton theorem

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Example 1

Verifying the theorem for a 2 × 2 and finding A⁻¹

Question

Verify the Cayley–Hamilton theorem for $A = [1423]$ and use it to find $A^{- 1}$ .

Characteristic equation
$tr A = 1 + 3 = 4, ∣ A ∣ = 3 - 8 = - 5$

$λ^{2} - 4 λ - 5 = 0$
Compute $A^{2}$
$A^{2} = [1423] [1423] = [1 + 8 4 + 12 2 + 6 8 + 9] = [916817]$
Substitute into $A^{2} - 4 A - 5 I$
$4 A = [416812], 5 I = [5005]$

$A^{2} - 4 A - 5 I = [9 - 4 - 5 16 - 16 - 0 8 - 8 - 0 17 - 12 - 5] = [0000] = O ✓$

The theorem is verified.
Multiply the identity by $A^{- 1}$
$A^{2} - 4 A - 5 I = O$

$\times A^{- 1} A - 4 I - 5 A^{- 1} = O$
Solve for $A^{- 1}$
$5 A^{- 1} = A - 4 I ⟹ A^{- 1} = \frac{1}{5} (A - 4 I)$

$A - 4 I = [1 - 4 4 2 3 - 4] = [- 3 4 2 - 1]$

$A^{- 1} = \frac{1}{5} [- 3 4 2 - 1]$
Cross-check with the adjoint formula
$A^{- 1} = \frac{1}{∣ A ∣} adj A = \frac{1}{- 5} [3 - 4 - 2 1] = \frac{1}{5} [- 3 4 2 - 1] ✓$

Answer

$A^{2} - 4 A - 5 I = O$ , and $A^{- 1} = \frac{1}{5} [- 3 4 2 - 1]$ .

Example 2

The theorem for a 3 × 3, and the inverse from it

Question

Verify the Cayley–Hamilton theorem for $A = 123 2 - 1 1 34 - 1$ and hence find $A^{- 1}$ .

Compute $S_{1}, S_{2}, S_{3}$
$S_{1} = 1 + (- 1) + (- 1) = - 1$

Principal $2 \times 2$ minors:
$- 1 1 4 - 1 = 1 - 4 = - 3, 13 3 - 1 = - 1 - 9 = - 10, 12 2 - 1 = - 1 - 4 = - 5$

$S_{2} = - 3 - 10 - 5 = - 18$

$S_{3} = ∣ A ∣ = 1 (1 - 4) - 2 (- 2 - 12) + 3 (2 + 3) = - 3 + 28 + 15 = 40$
Write the characteristic equation
$λ^{3} - S_{1} λ^{2} + S_{2} λ - S_{3} = 0$

$λ^{3} + λ^{2} - 18 λ - 40 = 0$

By the theorem: $A^{3} + A^{2} - 18 A - 40 I = O$ .
Compute $A^{2}$
Row 1 $(1, 2, 3)$ : $1 + 4 + 9 = 14$ , $2 - 2 + 3 = 3$ , $3 + 8 - 3 = 8$
Row 2 $(2, - 1, 4)$ : $2 - 2 + 12 = 12$ , $4 + 1 + 4 = 9$ , $6 - 4 - 4 = - 2$
Row 3 $(3, 1, - 1)$ : $3 + 2 - 3 = 2$ , $6 - 1 - 1 = 4$ , $9 + 4 + 1 = 14$

$A^{2} = 14122394 8 - 2 14$
Compute $A^{3} = A^{2} A$
Row 1 $(14, 3, 8)$ : $14 + 6 + 24 = 44$ , $28 - 3 + 8 = 33$ , $42 + 12 - 8 = 46$
Row 2 $(12, 9, - 2)$ : $12 + 18 - 6 = 24$ , $24 - 9 - 2 = 13$ , $36 + 36 + 2 = 74$
Row 3 $(2, 4, 14)$ : $2 + 8 + 42 = 52$ , $4 - 4 + 14 = 14$ , $6 + 16 - 14 = 8$

$A^{3} = 44245233131446748$
Substitute and verify
$18 A = 183654 36 - 18 18 5472 - 18, 40 I = 400004000040$

Position $(1, 1)$ : $44 + 14 - 18 - 40 = 0$ ✓
Position $(1, 2)$ : $33 + 3 - 36 - 0 = 0$ ✓
Position $(1, 3)$ : $46 + 8 - 54 - 0 = 0$ ✓
Position $(2, 1)$ : $24 + 12 - 36 - 0 = 0$ ✓
Position $(2, 2)$ : $13 + 9 + 18 - 40 = 0$ ✓
Position $(2, 3)$ : $74 - 2 - 72 - 0 = 0$ ✓
Position $(3, 1)$ : $52 + 2 - 54 - 0 = 0$ ✓
Position $(3, 2)$ : $14 + 4 - 18 - 0 = 0$ ✓
Position $(3, 3)$ : $8 + 14 + 18 - 40 = 0$ ✓

$A^{3} + A^{2} - 18 A - 40 I = O ✓$
Derive $A^{- 1}$
Multiply through by $A^{- 1}$ :

$A^{2} + A - 18 I - 40 A^{- 1} = O$

$A^{- 1} = \frac{1}{40} (A^{2} + A - 18 I)$
Evaluate
$A^{2} + A = 1514558511213$

$A^{2} + A - 18 I = - 3 145 5 - 10 5 112 - 5$

$A^{- 1} = \frac{1}{40} - 3 145 5 - 10 5 112 - 5$

Check (row 1 of $A$ × column 1 of the bracket): $1 (- 3) + 2 (14) + 3 (5) = - 3 + 28 + 15 = 40$ , and $40/40 = 1$ ✓

Answer

$A^{3} + A^{2} - 18 A - 40 I = O$ ; $A^{- 1} = \frac{1}{40} - 3 145 5 - 10 5 112 - 5$

NoteCompute S₁, S₂, S₃ first and write the equation before touching A² or A³. It tells you exactly which powers you need.

Problem 1Basic

Verify the Cayley–Hamilton theorem for $A = [23 - 1 1]$ .

Show solution

$tr A = 3, ∣ A ∣ = 2 + 3 = 5$

Characteristic equation: $λ^{2} - 3 λ + 5 = 0$

$A^{2} = [23 - 1 1] [23 - 1 1] = [4 - 3 6 + 3 - 2 - 1 - 3 + 1] = [19 - 3 - 2]$

$A^{2} - 3 A + 5 I = [19 - 3 - 2] - [69 - 3 3] + [5005]$

$= [1 - 6 + 5 9 - 9 + 0 - 3 + 3 + 0 - 2 - 3 + 5] = [0000] = O ✓$

Answer

A^{2} - 3 A + 5 I = O

✓

Problem 2Exam level

Use the Cayley–Hamilton theorem to find $A^{- 1}$ for $A = [3 - 1 12]$ .

Show solution

$tr A = 5, ∣ A ∣ = 6 + 1 = 7$

$λ^{2} - 5 λ + 7 = 0 ⟹ A^{2} - 5 A + 7 I = O$

Multiply by $A^{- 1}$ :

$A - 5 I + 7 A^{- 1} = O ⟹ A^{- 1} = \frac{1}{7} (5 I - A)$

$5 I - A = [5 - 3 1 - 1 5 - 2] = [21 - 1 3]$

$A^{- 1} = \frac{1}{7} [21 - 1 3]$

Check: $\frac{1}{7} [3 - 1 12] [21 - 1 3] = \frac{1}{7} [7007] = I$ ✓

Answer

A^{- 1} = \frac{1}{7} [21 - 1 3]

Problem 3Exam level

Verify the Cayley–Hamilton theorem for $A = 102020213$ and find $A^{- 1}$ .

Show solution

Coefficients.

$S_{1} = 1 + 2 + 3 = 6$

$S_{2} = 2013 + 1223 + 1002 = 6 + (- 1) + 2 = 7$

$S_{3} = ∣ A ∣ = 1 (6 - 0) - 0 + 2 (0 - 4) = 6 - 8 = - 2$

Characteristic equation: $λ^{3} - 6 λ^{2} + 7 λ + 2 = 0$

Compute $A^{2}$ .

Row 1 $(1, 0, 2)$ : $1 + 0 + 4 = 5$ , $0 + 0 + 0 = 0$ , $2 + 0 + 6 = 8$
Row 2 $(0, 2, 1)$ : $0 + 0 + 2 = 2$ , $0 + 4 + 0 = 4$ , $0 + 2 + 3 = 5$
Row 3 $(2, 0, 3)$ : $2 + 0 + 6 = 8$ , $0$ , $4 + 0 + 9 = 13$

$A^{2} = 5280408513$

Compute $A^{3} = A^{2} A$ .

Row 1 $(5, 0, 8)$ : $5 + 0 + 16 = 21$ , $0$ , $10 + 0 + 24 = 34$
Row 2 $(2, 4, 5)$ : $2 + 0 + 10 = 12$ , $0 + 8 + 0 = 8$ , $4 + 4 + 15 = 23$
Row 3 $(8, 0, 13)$ : $8 + 0 + 26 = 34$ , $0$ , $16 + 0 + 39 = 55$

$A^{3} = 211234080342355$

Verify $A^{3} - 6 A^{2} + 7 A + 2 I$ :

$(1, 1)$ : $21 - 30 + 7 + 2 = 0$ $(1, 2)$ : $0$ $(1, 3)$ : $34 - 48 + 14 = 0$
$(2, 1)$ : $12 - 12 + 0 = 0$ $(2, 2)$ : $8 - 24 + 14 + 2 = 0$ $(2, 3)$ : $23 - 30 + 7 = 0$
$(3, 1)$ : $34 - 48 + 14 = 0$ $(3, 2)$ : $0$ $(3, 3)$ : $55 - 78 + 21 + 2 = 0$

$A^{3} - 6 A^{2} + 7 A + 2 I = O ✓$

Inverse. Multiply by $A^{- 1}$ :

$A^{2} - 6 A + 7 I + 2 A^{- 1} = O ⟹ A^{- 1} = - \frac{1}{2} (A^{2} - 6 A + 7 I)$

$6 A = 6012012012618$

$A^{2} - 6 A + 7 I = 5 - 6 + 7 2 - 0 8 - 12 0 4 - 12 + 7 0 8 - 12 5 - 6 13 - 18 + 7 = 62 - 4 0 - 1 0 - 4 - 1 2$

$A^{- 1} = - \frac{1}{2} 62 - 4 0 - 1 0 - 4 - 1 2 = - 3 - 1 2 0 \frac{1}{2} 0 2 \frac{1}{2} - 1$

Check (row 1 of $A$ × column 1 of $A^{- 1}$ ): $1 (- 3) + 0 (- 1) + 2 (2) = 1$ ✓

Answer

A^{3} - 6 A^{2} + 7 A + 2 I = O

;

A^{- 1} = - 3 - 1 2 0 1/2 0 2 1/2 - 1

Problem 4Challenge

If $A = [1221]$ , use the Cayley–Hamilton theorem to express $A^{5}$ in the form $p A + q I$ .

Show solution

$tr A = 2, ∣ A ∣ = 1 - 4 = - 3$

$λ^{2} - 2 λ - 3 = 0 ⟹ A^{2} = 2 A + 3 I$

Now climb the powers, reducing at each stage.

$A^{3} = A \cdot A^{2} = A (2 A + 3 I) = 2 A^{2} + 3 A = 2 (2 A + 3 I) + 3 A = 7 A + 6 I$

$A^{4} = A \cdot A^{3} = A (7 A + 6 I) = 7 A^{2} + 6 A = 7 (2 A + 3 I) + 6 A = 20 A + 21 I$

$A^{5} = A \cdot A^{4} = A (20 A + 21 I) = 20 A^{2} + 21 A = 20 (2 A + 3 I) + 21 A = 61 A + 60 I$

$A^{5} = 61 A + 60 I$

Verification. $61 A + 60 I = [61 + 60 122 122 61 + 60] = [121122122121]$

Directly: the eigenvalues of $A$ are $3$ and $- 1$ , so those of $A^{5}$ are $243$ and $- 1$ . Then $tr (A^{5}) = 242 = 121 + 121$ ✓ and $∣ A^{5} ∣ = - 243 = 12 1^{2} - 12 2^{2} = (121 - 122) (121 + 122) = - 243$ ✓

Answer

A^{5} = 61 A + 60 I = [121122122121]

BCA1B02T · M1

What a computer is, and how it is built

Read the topic →

Example 1

Tracing one instruction through the machine cycle

Question

A program contains the instruction ADD the number in location 200 to the accumulator. Describe what each functional unit does, step by step.

Fetch — the control unit collects the instruction
The PC holds the address of this instruction. The CU copies that address into the MAR, memory returns the instruction into the MBR, and it is moved into the IR. The PC is then incremented so it points at the next instruction.
Decode — the control unit works out what is being asked
The CU examines the instruction in the IR and separates it into two parts: the operation code (ADD) and the operand address (200). It now knows it must fetch a value and hand it to the ALU.
Fetch the operand
Address $200$ is placed on the address bus, the control bus signals read, and the value stored there travels back along the data bus into the MBR.
Execute — the ALU does the arithmetic
The CU signals the ALU to add the value in the MBR to the value already in the accumulator. The addition itself is the only step in which data is actually changed.
Store the result
The sum stays in the accumulator, ready for the next instruction. Had the instruction been STORE, the result would travel back over the data bus to a memory location instead.
Repeat
Control returns to the CU, which reads the PC and begins the next fetch. The machine does nothing else, ever — it simply repeats this cycle billions of times a second.

Answer

Fetch (CU + memory + buses) → decode (CU) → fetch operand (memory) → execute (ALU) → store (accumulator/memory), then the cycle repeats.

NoteExam questions on the block diagram are really questions about this cycle. Learn the cycle and the diagram explains itself.

Example 2

Reading a specification

Question

A processor has a $32$ -bit address bus and a $64$ -bit data bus. How much memory can it address, and how many bytes move in one transfer?

Address bus width fixes the number of locations
$n$ address lines can select $2^{n}$ distinct locations. With $n = 32$ :

$2^{32} = 4, 294, 967, 296 locations$
Convert to familiar units
Each location holds one byte, so

$2^{32} bytes = 2^{2} \times 2^{30} bytes = 4 GB$
Data bus width fixes the transfer size
$64$ lines carry $64$ bits at once:

$\frac{64 bits}{8 bits per byte} = 8 bytes per transfer$
Read the meaning
This is exactly why 32-bit systems could not use more than 4 GB of RAM — not a licensing limit but an arithmetic one. Widening the address bus to $64$ lines raises the ceiling to $2^{64}$ bytes, or $16$ exbibytes.

Answer

$2^{32}$ bytes $= 4$ GB addressable; $8$ bytes per transfer.

Problem 1Basic

Define a computer and list its five characteristics with one line each.

Show solution

Definition. A computer is an electronic device that accepts data and instructions as input, processes the data according to a stored program, produces output, and can store data and results for later use.

Characteristics.

Speed — performs millions to billions of operations per second; timings are quoted in milli-, micro-, nano- and picoseconds.
Accuracy — arithmetic is always correct; errors trace back to wrong input or wrong instructions (GIGO).
Diligence — works indefinitely without tiredness, boredom or loss of concentration.
Versatility — the same hardware performs completely different tasks because only the program changes.
Storage capacity — stores very large volumes of data permanently and retrieves any item quickly.

Full marks usually need the limitations too: no intelligence of its own, no intuition, and no ability to handle a case the program does not cover.

AnswerSpeed, accuracy, diligence, versatility, storage capacity.

Problem 2Basic

Distinguish between hardware, software and firmware, with one example of each.

Show solution

	Hardware	Software	Firmware
Nature	Physical, tangible	Logical, intangible	Program held permanently in hardware
Made by	Manufacturing	Programming	Programming, then burnt into a chip
Changed by	Replacing a part	Installing a new version	Flashing the chip
Wears out?	Yes	No	No
Example	Keyboard, RAM, hard disk	Windows, MS Word, a C compiler	BIOS/UEFI, the program inside a washing machine's controller

Firmware is the interesting case: it is written like software but delivered inside a hardware component, usually in ROM, and it is what allows the machine to start up before any real software has loaded.

AnswerHardware is physical; software is the programs; firmware is software permanently stored in a hardware chip, e.g. the BIOS.

Problem 3Exam level

Draw the block diagram of a digital computer and explain the function of each unit.

Show solution

Block diagram.

$Input unit ⟶ CPU ALU ∣ Control Unit ↕ Memory unit ⟶ Output unit$

Solid arrows show the flow of data; the control unit additionally sends control signals to every other unit.

1. Input unit. Accepts data and instructions from the user and converts them into binary form. Keyboard, mouse, scanner, microphone.

2. Memory unit. Primary storage. Holds the program being executed, the data it needs, intermediate results and the output waiting to be sent.

3. Arithmetic and Logic Unit. Carries out all arithmetic ( $+, -, \times, \div$ ) and all logical operations (comparison, AND, OR, NOT). It is the only unit that changes data values.

4. Control unit. Directs and coordinates the whole system. It fetches each instruction, decodes it, and issues the control signals that cause it to be executed. It processes no data itself.

5. Output unit. Converts binary results into human-readable form — display, printer, speaker.

The ALU and CU together with the registers form the CPU.

AnswerFive units: input, memory, ALU, control unit, output — with ALU + CU forming the CPU.

Problem 4Exam level

What is the machine cycle? Name the registers involved and state what each holds.

Show solution

The machine cycle (instruction cycle) is the sequence the control unit repeats for every instruction in a program:

$Fetch \to Decode \to Execute \to Store$

Fetch — bring the next instruction from memory into the CPU.
Decode — separate it into operation code and operand, and work out what is required.
Execute — the ALU performs the operation.
Store — put the result in the accumulator or back into memory.

A program is nothing more than this cycle repeated once per instruction.

Registers involved.

Register	Holds
Program Counter (PC)	Address of the next instruction
Instruction Register (IR)	The instruction currently being executed
Memory Address Register (MAR)	The address being read from or written to
Memory Buffer Register (MBR)	The data travelling to or from memory
Accumulator (AC)	The result of the current ALU operation

Note that the PC is incremented during the fetch phase, not at the end — which is why a jump instruction works by overwriting the PC during execute.

AnswerFetch → decode → execute → store, using PC, IR, MAR, MBR and the accumulator.

Problem 5Exam level

What is a bus? Describe the three types and explain what the width of each determines.

Show solution

A bus is a group of parallel conducting lines that carries signals between the functional units of a computer. Instead of a separate wire between every pair of components, all units share one set of lines.

1. Data bus. Carries the data and instructions themselves, in both directions (memory → CPU and CPU → memory). Its width is the number of bits transferred simultaneously — a $64$ -bit data bus moves $8$ bytes per transfer. A wider data bus therefore raises throughput.

2. Address bus. Carries the address of the memory location or device being accessed. It is unidirectional — only the CPU puts addresses on it. Its width fixes the addressable memory: $n$ lines give $2^{n}$ locations.

$16 lines \to 2^{16} = 64 KB, 32 lines \to 4 GB, 64 lines \to 2^{64} bytes$

3. Control bus. Carries the command and timing signals that say what is to happen and when — read, write, interrupt, clock, reset. It is bidirectional, since devices also signal back to the CPU.

Summary: the data bus decides how much moves at a time, the address bus decides how much memory can exist, and the control bus decides when and what kind of transfer happens.

AnswerData bus (width = bits per transfer), address bus (width =

2^{n}

addressable locations), control bus (timing and command signals).

Problem 6Challenge

"A computer is fast and accurate but has no intelligence." Justify this statement with reference to GIGO.

Show solution

The statement separates two different things: capability and judgement.

Fast and accurate. The circuits perform arithmetic in nanoseconds, and they perform it correctly every time. A computer will not misadd two numbers, will not lose concentration on the ten-millionth record, and will not produce a different answer on a Monday. This is a property of the hardware.

No intelligence. The machine has no idea what its numbers mean. It cannot notice that a date of birth in the year 2190 is impossible, that a salary of $- 50, 000$ is nonsense, or that the user meant to type 100 rather than 1000 — unless a programmer has explicitly written an instruction to check for exactly that. It follows the instructions it is given, faithfully and literally, including instructions that are wrong.

GIGO — Garbage In, Garbage Out. This is the practical consequence. The accuracy of the output depends entirely on the accuracy of the input and of the program:

$correct data + correct program \to correct result$
$wrong data + correct program \to wrong result, delivered confidently$

The second line is the danger. A wrong answer arrives with exactly the same speed, formatting and apparent authority as a right one. The machine offers no warning because it cannot tell the difference.

Consequence for practice. This is why real systems spend so much effort on validation (is this value of a plausible type and range?) and verification (was it entered as intended?), and why the responsibility for a wrong result always sits with the people who supplied the data or wrote the program — never with the computer.

AnswerSpeed and accuracy are hardware properties; correctness of meaning depends wholly on the input and the program, so wrong input yields wrong output delivered with full confidence — GIGO.

BCA1B02T · M1

Generations and classification of computers

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Example 1

Placing a machine in its generation

Question

A computer uses a single silicon chip containing several million transistors, runs a graphical operating system, and connects to a network. Which generation is it, and what evidence supports the answer?

Identify the switching technology
Millions of components on one chip is VLSI — very large scale integration. That gives the microprocessor.
Match the technology to a generation
Vacuum tube → 1st, transistor → 2nd, IC (SSI/MSI) → 3rd, microprocessor (VLSI) → 4th, ULSI with parallel processing and AI → 5th.
3
Confirm with the secondary evidence
Two further details agree:
- a graphical operating system is a fourth-generation development (third-generation machines had command-driven operating systems)
- networking and the Internet belong to the fourth generation
Check what would move it to the fifth
The fifth generation is characterised not by being newer but by inference — natural language processing, expert systems, learning. Nothing in the description suggests that, so it stays in the fourth.

Answer

Fourth generation — VLSI microprocessor, GUI operating system and networking are all fourth-generation characteristics.

NoteAnswer generation questions from the switching component first. Dates and features are supporting evidence, not the definition.

Example 2

Choosing the right class of computer

Question

Recommend a class of computer for each, with a reason: (a) the Indian Railways reservation system, (b) forecasting the monsoon, (c) a college computer laboratory, (d) a hotel lift controller.

(a) Railway reservation
Thousands of clerks and websites booking at once, every transaction must be recorded reliably, and the system may never lose a booking. The requirement is transaction throughput and reliability, not calculation speed.

$\Rightarrow$ Mainframe.
(b) Monsoon forecasting
One single enormous numerical model of the atmosphere, needing as many floating-point operations per second as can be bought, with very few users.

$\Rightarrow$ Supercomputer, working by parallel processing.
(c) College laboratory
Many independent single users each doing general-purpose work — programming, documents, browsing — with cost per seat mattering.

$\Rightarrow$ Microcomputers (desktop PCs), probably networked to a server.
(d) Lift controller
One fixed task, forever: read the buttons and floor sensors, drive the motor, open the doors. The program never changes and is held in firmware.

$\Rightarrow$ Special-purpose (embedded) computer.
Note the distinction being tested
(a)–(c) are answered on the size and capacity classification; (d) is answered on the purpose classification. A question can ask for either, so read which axis is wanted.

Answer

(a) mainframe (b) supercomputer (c) microcomputers (d) special-purpose embedded computer.

Problem 1Basic

Name the switching technology of each of the five generations of computers.

Show solution

Generation	Switching technology
First	Vacuum tubes
Second	Transistors
Third	Integrated circuits (SSI and MSI)
Fourth	Microprocessors — VLSI
Fifth	ULSI with parallel processing (and artificial intelligence as the goal)

The examinable point behind the table: each step packs more switches into less space, and every other improvement — speed, size, cost, reliability, heat — follows from that.

AnswerVacuum tube, transistor, IC, microprocessor (VLSI), ULSI.

Problem 2Basic

Differentiate between analog and digital computers, giving two examples of each.

Show solution

	Analog computer	Digital computer
Data handled	Continuously varying physical quantities	Discrete values, ultimately $0$ and $1$
Basic action	Measures	Counts
Accuracy	Limited by the measuring instrument	Limited only by the number of digits used
Speed	Lower	Very high
Output	A reading on a scale, or a graph	Numbers, text, images
Storage	Very limited	Very large
Programmability	Usually fixed by its wiring	Fully programmable
Examples	Mercury thermometer, speedometer, analog voltmeter, slide rule	Personal computer, mainframe, calculator, smartphone

A hybrid computer contains both: an ICU patient monitor measures pulse and temperature as analog signals and then processes them digitally to display readings and raise alarms.

AnswerAnalog measures continuous quantities (thermometer, speedometer); digital counts discrete values (PC, calculator).

Problem 3Exam level

Explain the classification of computers based on size and capacity.

Show solution

1. Microcomputer. Built around a single microprocessor and intended for one user at a time. Smallest, cheapest and most numerous. Includes desktops, laptops, notebooks, tablets, palmtops, smartphones and workstations. A workstation is a high-end microcomputer for engineering, design or graphics work.

2. Minicomputer. Medium-sized and multi-user, serving a department or small organisation with anything from a few to a few hundred terminals. More powerful and more expensive than a microcomputer, less so than a mainframe. Historic examples: PDP-11, VAX-8000. The category has largely been replaced by powerful servers.

3. Mainframe computer. Very large, supporting thousands of simultaneous users. Designed for extremely high transaction throughput and for continuous, reliable operation over years rather than for raw arithmetic speed. Requires a controlled environment and specialist staff. Used by banks, railways, insurance companies and tax departments. Example: IBM Z series.

4. Supercomputer. The fastest and most expensive class, built to run one enormous calculation as quickly as possible rather than to serve many users. Achieves its speed by massive parallel processing across thousands of processors. Performance is measured in FLOPS. Used for weather forecasting, climate modelling, molecular and nuclear simulation and aerodynamic design. Indian examples: PARAM series, AIRAWAT.

The key contrast to state explicitly: a mainframe is optimised for many transactions, a supercomputer for one calculation. Both are large, but for opposite reasons.

AnswerMicrocomputer (single user), minicomputer (departmental multi-user), mainframe (thousands of users, transaction throughput), supercomputer (parallel, raw speed).

Problem 4Exam level

Compare the first and fourth generations of computers under at least six headings.

Show solution

Heading	First generation (1940–56)	Fourth generation (1971–)
Component	Vacuum tubes	Microprocessor (VLSI)
Memory	Magnetic drum	Semiconductor RAM, very large
Size	Filled a whole room	Fits on a lap or in a pocket
Speed	Milliseconds per operation	Nanoseconds per operation
Power and heat	Enormous; needed heavy air conditioning	Low; runs on a battery
Reliability	Very poor — tubes failed constantly	Very high
Cost	Affordable only to governments	Affordable to individuals
Language	Machine language only	All high-level languages, GUI
Input/output	Punched cards, printouts	Keyboard, mouse, touch, display
Users	One operator, batch jobs	Personal and networked, interactive
Example	ENIAC, UNIVAC-I	IBM PC, Pentium, modern laptops

Every difference in the table traces to a single cause: the number of switches that can be placed in a given volume rose by many orders of magnitude, so the machine became smaller, cooler, cheaper, faster and more reliable all at once.

AnswerCompared on component, memory, size, speed, power, reliability, cost, language, I/O and users — see table.

Problem 5Exam level

What is a hybrid computer? Explain with an example why both analog and digital sections are needed.

Show solution

A hybrid computer contains both an analog and a digital section, using each for the part of the job it does best. The analog section measures a continuously varying physical quantity; the digital section then processes, stores and displays that measurement.

Standard example — an ICU patient monitor.

Analog section. The patient's temperature, pulse rate, blood pressure and blood-oxygen level are continuous physical quantities. Sensors convert each into a continuously varying electrical voltage. Only an analog device can pick up a continuous signal in the first place.
Analog-to-digital conversion. The voltage is sampled and converted into a number.
Digital section. The numbers are displayed as readings, plotted as a trace, stored as a record of the last several hours, compared against safe limits, and used to trigger an alarm when a limit is crossed.

Why both are needed. The quantities being observed exist in the physical world as continuous signals, so measurement must be analog. But storing a history, comparing against thresholds, raising an alarm and printing a report are all discrete decisions on stored data — work only a digital computer can do accurately and repeatably.

Other examples: a petrol pump (analog flow measurement, digital price calculation), and industrial process controllers.

AnswerA computer with both analog and digital sections — e.g. an ICU monitor measures vital signs as analog signals and processes, stores and alarms on them digitally.

Problem 6Challenge

A mainframe and a supercomputer are both very large and very expensive. Explain why one cannot simply be used in place of the other.

Show solution

They are large for opposite reasons, and each is poor at the other's job.

What a mainframe optimises. Throughput of many small independent transactions, and uninterrupted availability. A booking system might handle tens of thousands of requests per second, each of which does almost no arithmetic — read a record, check a seat, write a record. The design effort goes into:

very high input/output bandwidth, because the work is dominated by moving records, not computing
transaction integrity, so a booking is never half-completed
redundancy and hot-swappable parts, so the machine runs for years without stopping
serving thousands of concurrent users fairly

What a supercomputer optimises. The wall-clock time of one enormous calculation. A weather model divides the atmosphere into millions of cells and solves equations for each, repeatedly. The design effort goes into:

the largest possible number of floating-point operations per second, across thousands of processors working on one problem
extremely fast interconnect between processors, since they must exchange partial results constantly
vector and parallel arithmetic units

Why they are not interchangeable.

Put the reservation system on a supercomputer and most of its arithmetic units sit idle while it waits on disk and network, and it has neither the transaction machinery nor the year-round reliability the job requires — an expensive machine doing an I/O-bound job badly.

Put the weather model on a mainframe and the single calculation runs far too slowly to be a forecast; a prediction of tomorrow's rain that takes three days to compute has no value at all.

The general principle. "Powerful" is not one quantity. A machine is fast at the thing it was designed to be fast at — here, transactions per second versus floating-point operations per second — and choosing hardware means matching the design to which of those the workload actually needs.

AnswerA mainframe is built for transaction throughput, I/O bandwidth and continuous reliability; a supercomputer for the FLOPS of a single parallel calculation. Each performs the other's workload badly.

BCA1B02T · M1

Memory and storage

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Example 1

Calculating storage capacity

Question

A page of text holds $40$ lines of $70$ characters. How many such pages fit on a $700$ MB CD? How many fit on a $64$ GB pen drive?

Find the size of one page
One character occupies one byte, so

$40 \times 70 = 2800 bytes per page$
Convert the CD capacity to bytes
$700 MB = 700 \times 1024 \times 1024 = 734, 003, 200 bytes$
Divide
$\frac{734 , 003 , 200}{2800} \approx 262, 144 pages$

Over a quarter of a million pages of plain text on one CD.
Repeat for the pen drive
$64 GB = 64 \times 102 4^{3} = 68, 719, 476, 736 bytes$

$\frac{68 , 719 , 476 , 736}{2800} \approx 24, 542, 670 pages$

Roughly $24.5$ million pages — about $94$ times the CD.
Read the result
Plain text is tiny. Storage fills up with images, audio and video, not with words — a single photograph from a phone camera occupies more space than a thousand pages of text.

Answer

About $262, 144$ pages on a 700 MB CD and about $24.5$ million on a 64 GB pen drive.

Example 2

Why unsaved work disappears

Question

A student types two pages in a word processor without saving, and the power fails. Explain exactly what is lost and why, in terms of the memory hierarchy.

Locate the document while it is being typed
Every keystroke is placed in RAM — main memory. That is where the running copy of the document lives, together with the word-processor program itself.
State the property of RAM that matters
RAM is volatile: it stores each bit as a charge that must be continuously refreshed by the power supply. Remove power and the charge drains within moments, taking the contents with it.
Explain what saving does
Saving copies the document from RAM to secondary memory — an SSD or hard disk. Those are non-volatile: an SSD stores each bit as trapped charge in a flash cell and a hard disk as a magnetised region, neither of which needs power to persist.
Conclude
Everything typed since the last save existed only in volatile RAM and is unrecoverable. Anything saved earlier is intact on disk. Autosave features work by silently performing this copy every few minutes.

Answer

Everything since the last save is lost, because the working copy was in volatile RAM; saving copies it to non-volatile secondary memory.

NoteThis is the standard exam illustration of volatility. Name RAM, say volatile, and say what saving actually copies.

Problem 1Basic

Arrange in increasing order of size: bit, terabyte, byte, kilobyte, nibble, gigabyte, megabyte. State how many bytes each of the last four contains.

Show solution

Increasing order:

$bit < nibble < byte < kilobyte < megabyte < gigabyte < terabyte$

$1$ bit $= 1$ binary digit ( $0$ or $1$ )
$1$ nibble $= 4$ bits
$1$ byte $= 8$ bits $=$ one character
$1$ KB $= 1024$ bytes $= 2^{10}$ bytes
$1$ MB $= 1024$ KB $= 2^{20} = 1, 048, 576$ bytes
$1$ GB $= 1024$ MB $= 2^{30} = 1, 073, 741, 824$ bytes
$1$ TB $= 1024$ GB $= 2^{40} \approx 1.1 \times 1 0^{12}$ bytes

Answerbit < nibble < byte < KB (

2^{10}

) < MB (

2^{20}

) < GB (

2^{30}

) < TB (

2^{40}

) bytes.

Problem 2Basic

Distinguish between primary and secondary memory under at least five headings.

Show solution

Heading	Primary memory	Secondary memory
CPU access	Directly accessible by the CPU	Not direct — data must be copied into primary memory first
Volatility	RAM is volatile; contents lost on power off	Non-volatile; contents persist
Speed	Very fast (nanoseconds)	Much slower (microseconds for SSD, milliseconds for HDD)
Capacity	Smaller — gigabytes	Much larger — terabytes
Cost per byte	High	Low
Technology	Semiconductor	Magnetic, optical or flash
Purpose	Holds what is in use now	Stores data and programs long term
Examples	RAM, ROM, cache, registers	Hard disk, SSD, CD/DVD, pen drive, magnetic tape

The single sentence that carries the distinction: primary memory holds what the computer is working on, secondary memory holds what it will need later.

AnswerPrimary is CPU-accessible, fast, small, expensive and mostly volatile; secondary is not directly accessible, slower, large, cheap and non-volatile.

Problem 3Exam level

Explain RAM and ROM, and describe the four types of ROM.

Show solution

RAM — Random Access Memory. The computer's working memory, holding the operating system, running programs and their data. Two defining properties: any location can be accessed directly in the same time as any other (random access), and the contents are lost when power is removed (volatile). It is read/write. Two kinds:

SRAM — stores each bit in a flip-flop, needs no refreshing, fast but expensive and bulky; used for cache.
DRAM — stores each bit as charge in a capacitor, must be refreshed thousands of times per second, slower but cheap and dense; used for main memory.

ROM — Read Only Memory. Holds data written once and thereafter only read, and is non-volatile. It contains what the machine needs before any software is available: the BIOS/UEFI and the bootstrap loader that locates and starts the operating system.

Types of ROM.

MROM (Masked ROM). Contents fixed during manufacture using a mask. Cannot be changed at all. Cheapest in very large quantities.
PROM (Programmable ROM). Supplied blank; the user writes it once with a special device called a PROM programmer or burner, after which it is permanent. Write once, read many.
EPROM (Erasable PROM). Can be erased and rewritten, but only wholesale: the whole chip is cleared by exposing it to ultraviolet light through a quartz window, which takes several minutes and requires removing the chip.
EEPROM (Electrically Erasable PROM). Erased and rewritten electrically, byte by byte, in place, without removing the chip. Flash memory is a form of EEPROM, and is what pen drives, memory cards and SSDs are made of — which is why a modern BIOS can be updated by software.

The progression across the four is simply increasing ease of rewriting.

AnswerRAM is volatile read/write working memory (SRAM and DRAM); ROM is non-volatile read-only memory holding firmware, in the forms MROM, PROM, EPROM and EEPROM/flash.

Problem 4Exam level

What is cache memory? Explain cache hit, cache miss and locality of reference, and say why cache improves performance.

Show solution

Cache memory is a small block of very fast SRAM placed between the CPU and main memory. It holds copies of the data and instructions the CPU has used most recently or uses most often.

How a request is served. When the CPU needs an item it looks in cache first.

Cache hit — the item is present, and is supplied at cache speed. No RAM access is needed.
Cache miss — the item is absent, so it must be fetched from the slower main memory. On the way it is also copied into cache, on the expectation that it will be wanted again.

The hit ratio is the fraction of requests that are hits, and it is the figure that decides how much the cache is helping.

Why it works — locality of reference. Real programs do not access memory at random. They exhibit:

temporal locality — an item used now is likely to be used again shortly (a loop counter, a running total)
spatial locality — an item near one just used is likely to be needed next (the following instruction, the next element of an array)

Because of this, a small cache holding recent items captures a large share of all accesses.

Why performance improves. RAM access costs the CPU many idle clock cycles; cache access costs very few. If most accesses are hits, the CPU spends most of its time computing rather than waiting. Without cache, a fast processor would sit idle waiting on comparatively slow DRAM, and its speed would be largely wasted.

Levels. L1 is smallest and fastest, inside each core; L2 is larger and slightly slower; L3 is largest and shared between cores — a hierarchy within the hierarchy, for exactly the same reason as the main one.

AnswerFast SRAM between CPU and RAM. A hit is served at cache speed, a miss must go to RAM. Locality of reference (temporal and spatial) makes a small cache capture most accesses, so the CPU waits far less.

Problem 5Exam level

Compare a hard disk drive with a solid state drive, and explain the terms seek time and rotational latency.

Show solution

	Hard disk drive (HDD)	Solid state drive (SSD)
Technology	Magnetic platters and moving heads	Flash memory (EEPROM)
Moving parts	Yes	None
Speed	Slower (milliseconds)	Much faster (microseconds)
Noise and heat	Audible, warmer	Silent, cooler
Shock resistance	Poor — a knock can crash the heads	Good
Power use	Higher	Lower
Cost per byte	Low	Higher
Lifetime limit	Mechanical wear	Limited number of write cycles per cell

Organisation of a hard disk. Data sits in concentric tracks on each platter surface; each track is divided into sectors, the smallest addressable unit; the same numbered track on every platter together forms a cylinder.

Access time on an HDD is the sum of three parts:

$access time = seek time + rotational latency + transfer time$

Seek time — the time for the read/write head to move radially to the correct track. Mechanical, and the largest component.
Rotational latency — having reached the track, the time waiting for the required sector to rotate under the head. On average half a revolution: at $7200$ rpm one revolution takes $8.33$ ms, so average latency is about $4.17$ ms.
Transfer time — the time to actually read or write the data once positioned.

Why an SSD is so much faster. It has no head to move and no platter to rotate, so both seek time and rotational latency are zero. This is also why an SSD's performance barely changes when files are scattered, whereas a fragmented HDD slows noticeably.

AnswerHDD is magnetic and mechanical, cheap and slower; SSD is flash-based with no moving parts, faster and dearer. Seek time = head movement to the track; rotational latency = wait for the sector to arrive; both are zero on an SSD.

Problem 6Challenge

Explain the memory hierarchy. Why not simply build a computer with a very large amount of the fastest memory?

Show solution

The hierarchy. From fastest and smallest to slowest and largest:

$registers \to cache (L1, L2, L3) \to main memory (RAM) \to secondary (SSD, HDD) \to tertiary (tape, optical)$

Moving down the hierarchy, capacity rises and cost per byte falls; access time gets worse. Moving up, everything gets faster and more expensive per byte.

Why not build the whole machine from the fastest memory? Three reasons, and all three would still apply even with unlimited money.

1. Cost. SRAM, the memory used for cache, costs orders of magnitude more per byte than DRAM, which in turn costs far more than flash or magnetic storage. Building a machine with several terabytes of SRAM would put its price beyond any conceivable use.

2. Physics — density and heat. An SRAM cell needs six transistors; a DRAM cell needs one transistor and one capacitor. SRAM is therefore several times larger per bit, so far less of it fits on a chip. It also draws more power and produces more heat, which limits how much can be packed together and cooled.

3. Physics — distance. Even with free, cool, dense fast memory, signals travel at a finite speed. A larger memory occupies more physical space, so its far corners are further from the CPU and take longer to reach. Big memory is slow memory, unavoidably. This is why L1 cache is deliberately kept tiny — a larger L1 would be a slower L1.

Why the hierarchy works anyway. Because of locality of reference, a program at any moment uses only a small fraction of its data. The hierarchy keeps that active fraction near the top and everything else lower down, and moves items up as they are needed. The result is a system whose average access time approaches that of the fast levels while its capacity is that of the slow ones.

Volatility adds a second reason. The fast levels are all volatile. Even if a machine could hold everything in RAM, it would still need non-volatile secondary storage so that data survives being switched off.

AnswerBecause fast memory is far costlier per byte, much less dense and hotter, and because a physically larger memory is inherently slower to reach. Locality of reference lets a small fast level serve most accesses, so the hierarchy gives nearly the speed of the top with the capacity of the bottom — and volatility means non-volatile storage is needed regardless.

BCA1B02T · M1

Input and output devices

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Example 1

Choosing the right recognition device

Question

Name the input device best suited to each, with a reason: (a) processing 40,000 multiple-choice answer sheets, (b) clearing bank cheques, (c) converting a printed book into an editable file, (d) billing at a supermarket counter.

(a) Answer sheets
What must be detected is only whether a bubble is filled at a known position — no characters are involved.

$\Rightarrow$ OMR (Optical Mark Recognition). Fast, and reliable because it looks only for the presence of a mark.
(b) Cheques
The code line must be readable even though the cheque may be folded, stamped, signed over or smudged. Ink containing iron oxide can be read magnetically regardless of what is printed on top of it.

$\Rightarrow$ MICR (Magnetic Ink Character Recognition), using the E-13B font.
(c) A printed book
Two stages are needed. A scanner produces an image of each page — but an image of text cannot be edited. OCR software then identifies the characters in that image and produces editable text.

$\Rightarrow$ Scanner + OCR. Naming only the scanner is the common half-answer.
(d) Supermarket billing
Each product carries a printed pattern of bars encoding its product number, which is looked up in the shop's database for name and price.

$\Rightarrow$ Barcode reader.

Answer

(a) OMR (b) MICR (c) scanner followed by OCR (d) barcode reader.

NoteA scanner alone gives you a picture of text. OCR is what makes it text. Exam answers must say both.

Example 2

Choosing a printer from the requirement

Question

Recommend a printer for each, with justification: (a) a railway booking counter printing a ticket and two carbon copies, (b) a college office printing 500 pages of notices a day, (c) a photographer printing colour photographs, (d) an architect printing an A0 building plan.

(a) Ticket with carbon copies
Only an impact printer can transfer an image through carbon paper to sheets underneath, because the printing action is a physical strike.

$\Rightarrow$ Dot matrix printer. Print quality hardly matters; the multiple-copy requirement decides it, and no non-impact printer can meet it.
(b) 500 pages a day of plain text
High volume, sharp text, low cost per page and speed all matter; colour and photographic quality do not.

$\Rightarrow$ Laser printer. A page printer, so it is fast, and toner is far cheaper per page than ink.
(c) Colour photographs
What matters is continuous-tone colour on glossy paper; speed and cost per page do not.

$\Rightarrow$ Inkjet printer. Liquid ink blends smoothly and reproduces photographs better than toner.
(d) An A0 building plan
The output is a very large, dimensionally accurate line drawing, far bigger than any sheet a normal printer accepts.

$\Rightarrow$ Plotter, which draws with pens on large paper and holds accuracy across the whole sheet.
The reasoning pattern
In every case the requirement names one constraint that only one technology satisfies: carbon copies → impact; volume and cost → laser; photographic colour → inkjet; very large accurate lines → plotter. Identify that constraint first and the answer follows.

Answer

(a) dot matrix (impact, for carbon copies) (b) laser (c) inkjet (d) plotter.

Problem 1Basic

List the five groups of keys on a standard keyboard, with two examples from each.

Show solution

Group	Purpose	Examples
Alphanumeric	Text entry, in the QWERTY layout	A–Z, 0–9, punctuation
Numeric keypad	Fast entry of numbers and arithmetic	The right-hand digit block, $+$ and $-$
Function keys	Program-specific commands	F1 (help), F5 (refresh)
Cursor control	Moving the insertion point	Arrow keys, Home, End, Page Up/Down
Special / modifier	Control the effect of other keys or the system	Enter, Shift, Ctrl, Alt, Esc, Tab, Caps Lock, Backspace, Delete

A standard keyboard has about 104 keys. Note that Backspace deletes to the left of the cursor and Delete to the right — a distinction questions sometimes ask for.

AnswerAlphanumeric, numeric keypad, function, cursor control, and special/modifier keys.

Problem 2Basic

Define resolution, pixel, refresh rate and dot pitch.

Show solution

Pixel — picture element: the smallest individually addressable dot on a display. Every image is built from pixels.
Resolution — the total number of pixels on the screen, written $width \times height$ , for example $1920 \times 1080$ (Full HD). Higher resolution means finer detail and more content visible at once.
Refresh rate — the number of times per second the whole image is redrawn, measured in hertz (Hz). $60$ Hz is standard; higher rates give smoother motion. On old CRT monitors a low refresh rate produced visible flicker.
Dot pitch — the distance between the centres of two adjacent pixels, in millimetres. Smaller is better: a smaller dot pitch packs pixels closer, giving a sharper image.

Note that resolution alone does not fix sharpness — $1920 \times 1080$ looks razor sharp on a phone and coarse on a large television, because the same pixel count is spread over a larger area, giving a larger dot pitch.

AnswerPixel = smallest addressable dot; resolution = pixel count as width × height; refresh rate = redraws per second in Hz; dot pitch = distance between adjacent pixels, smaller being sharper.

Problem 3Exam level

Distinguish between impact and non-impact printers, and describe two examples of each.

Show solution

	Impact printer	Non-impact printer
Mechanism	Strikes an inked ribbon against the paper	Never touches the paper
Noise	Loud	Quiet
Print quality	Lower	High to very high
Speed	Lower	Higher
Multiple (carbon) copies	Possible	Not possible
Graphics	Limited	Full
Cost per page	Low	Low (laser) to high (inkjet)

Impact — two examples.

Dot matrix printer. The print head carries a vertical column of pins; selected pins are driven forward to strike the ribbon, so each character is formed from a matrix of dots. 9-pin heads give draft quality, 24-pin heads near-letter quality. A character printer. Still used where carbon copies are needed — railway tickets, invoices, bank passbooks.
Line printer (drum or chain). Prints an entire line at a time rather than a character, making it very fast. A drum printer carries the character set on a rotating drum; a chain printer on a revolving chain. Used with mainframes for very high volume listings.

Non-impact — two examples.

Inkjet printer. Sprays extremely fine droplets of liquid ink through nozzles onto the paper. Inexpensive to buy, excellent colour and photographic reproduction, but high cost per page and ink that can smudge or dry in the nozzles.
Laser printer. A laser beam writes the page image as a pattern of electrostatic charge on a rotating drum; dry toner powder clings to the charged regions, is transferred to the paper and fused by heat. A page printer — it assembles and prints a complete page at a time. Fast, very sharp, and cheap per page, making it the office standard.

The decisive difference to state in an answer: only an impact printer can produce carbon copies, because only it applies physical pressure to the paper.

AnswerImpact printers strike a ribbon against paper — noisy, lower quality, but can make carbon copies (dot matrix, line printer). Non-impact printers never touch the paper — quiet and high quality but no carbon copies (inkjet, laser).

Problem 4Exam level

Explain OCR, OMR, MICR and barcode readers, and state one application of each.

Show solution

All four read data from a printed document, but each recognises something different.

1. OCR — Optical Character Recognition. Software that examines a scanned image and identifies the shapes in it as characters, converting a picture of text into editable, searchable text. Works on printed and, less reliably, handwritten text.
Application: digitising printed books, records and forms; reading vehicle number plates.

2. OMR — Optical Mark Recognition. Detects only the presence or absence of a mark at a fixed, predetermined position on a specially printed sheet. It does not identify characters at all, which is exactly why it is so fast and so accurate.
Application: marking multiple-choice examination answer sheets; ballot papers; survey forms.

3. MICR — Magnetic Ink Character Recognition. Reads characters printed in ink containing iron oxide, in the standard E-13B font, by magnetising the ink and sensing the pattern. Because the reading is magnetic rather than optical, it succeeds even when the characters are overprinted, stamped, folded or smudged, and it is very hard to forge.
Application: the code line at the foot of bank cheques, giving cheque number, bank and branch code and account number.

4. Barcode reader. Scans a printed pattern of parallel bars of varying thickness that encodes a number, and passes that number to a program which looks up the corresponding record. A QR code reader does the same for a two-dimensional code, which holds much more data and can be read at any angle.
Application: retail billing and stock control; library book issue; parcel tracking.

How to keep them apart: OCR reads characters, OMR reads marks, MICR reads magnetic characters, and a barcode reader reads a bar pattern.

AnswerOCR = characters from an image (digitising books); OMR = marks at fixed positions (answer sheets); MICR = magnetic-ink characters (cheques); barcode reader = bar patterns encoding a number (retail billing).

Problem 5Exam level

What is a plotter? How does it differ from a printer, and what are its two types?

Show solution

A plotter is a hard-copy output device that draws with one or more pens moved mechanically across the paper under program control. Rather than depositing dots to build an image, it draws continuous lines from point to point.

Differences from a printer.

	Printer	Plotter
Forms the image from	Dots	Continuous drawn lines
Best at	Text and photographs	Accurate line drawings
Paper size	Usually up to A4/A3	Very large — A1, A0 and wider
Accuracy	Adequate for documents	Dimensionally precise
Speed	Fast	Slow
Cost	Low to moderate	High

Two types.

Drum plotter. The paper is wrapped round a rotating drum. The drum turns the paper along one axis while the pen carriage moves along the other, so the two motions together trace any line. Because the paper can keep rolling, a drum plotter can produce a drawing of almost unlimited length while occupying little floor space.
Flatbed plotter. The paper lies flat and stationary on a bed, and the pen moves over it in both directions. Accuracy is higher, since the paper never moves, but the machine must be at least as large as the biggest sheet it can take.

Applications. Engineering and architectural drawings, circuit layouts, maps and survey plans, and graph plotting — all cases where the correctness of a line matters more than the appearance of text.

AnswerA plotter draws with moving pens to produce large, dimensionally accurate line drawings; printers build images from dots. The two types are drum (paper rotates) and flatbed (paper stationary).

Problem 6Challenge

A scanned page of a book cannot be edited in a word processor, but a typed page can. Explain the difference, and describe what OCR does about it.

Show solution

The two files look similar on screen but contain fundamentally different things.

A typed page is stored as character codes. The letter A is held as the number $65$ (in ASCII), and the file records a sequence of such codes together with formatting instructions. Because the file knows which characters are present, the program can search for a word, change a font, correct a spelling, count words, or read the text aloud.

A scanned page is stored as an image — a grid of pixels, each with a colour value. The file records that a certain pixel is black and its neighbour is white. It contains no characters at all. The shapes only look like letters to a human reader; to the program they are indistinguishable from a photograph of a tree. Nothing can be searched, corrected or restyled, because there is nothing there to search.

$typed page = sequence of character codes scanned page = grid of pixel values$

What OCR does. Optical Character Recognition bridges the two. It:

pre-processes the image — straightens it, removes speckles, and reduces it to black and white
segments it, locating blocks of text, then lines, then words, then individual characters
recognises each isolated shape, by comparing it against stored patterns or by measuring its features, and decides which character it most probably is
post-processes using a dictionary and language rules, so that a shape read ambiguously as rn can be corrected to m when the surrounding letters make a real word

The output is a file of character codes — now genuinely editable, searchable and reformattable.

Why OCR is imperfect. The recognition step is a decision under uncertainty. Poor scan quality, unusual fonts, handwriting, small print, tables, multiple columns and mixed languages all raise the error rate, and the classic confusions ( $0$ /O, $1$ /l/I, rn/m) are between shapes that genuinely differ by very few pixels. This is why OCR output always needs proofreading, and why a scanned archive is usually stored as image and OCR text together — the image as the faithful record, the text so it can be searched.

AnswerA typed page stores character codes, so the program knows which characters exist; a scan stores only pixel values and contains no characters. OCR analyses the image, segments and recognises each shape, and outputs character codes — imperfectly, so the result needs proofreading.

BCA1B02T · M1

Number systems and computer codes

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Example 1

Converting a binary number with a fraction to decimal

Question

Convert $(1101.101)_{2}$ to decimal.

Write the place value above each digit
$2^{3} 1 2^{2} 1 2^{1} 0 2^{0} 1 . . 2^{- 1} 1 2^{- 2} 0 2^{- 3} 1$
Expand the integer part
$1 \times 8 + 1 \times 4 + 0 \times 2 + 1 \times 1 = 8 + 4 + 0 + 1 = 13$
Expand the fractional part
$1 \times \frac{1}{2} + 0 \times \frac{1}{4} + 1 \times \frac{1}{8} = 0.5 + 0 + 0.125 = 0.625$
Add
$(1101.101)_{2} = 13 + 0.625 = (13.625)_{10}$

Answer

$(13.625)_{10}$

Example 2

Converting a decimal number with a fraction to binary

Question

Convert $(45.6875)_{10}$ to binary.

Integer part — divide by 2 repeatedly, recording remainders
$2222224522115210101101$

Read the remainders bottom to top: $101101$ .

Check: $32 + 8 + 4 + 1 = 45$ ✓
Fractional part — multiply by 2 repeatedly, recording carries
$0.6875 \times 2 0.375 \times 2 0.75 \times 2 0.5 \times 2 = 1 .375 = 0 .75 = 1 .5 = 1 .0 \leftarrow fraction is now 0, stop$

Read the bold integer parts top to bottom: $.1011$ .

Check: $0.5 + 0 + 0.125 + 0.0625 = 0.6875$ ✓
Combine the two parts
$(45.6875)_{10} = (101101.1011)_{2}$

Answer

$(101101.1011)_{2}$

NoteRemainders are read upward, carries downward. Writing both checks costs a few seconds and guarantees the marks.

Example 3

Converting between octal and hexadecimal

Question

Convert $(576)_{8}$ to hexadecimal.

There is no direct route — go through binary
Octal and hexadecimal have no simple digit correspondence, because $8$ is not a power of $16$ . Binary is the bridge: $3$ bits per octal digit, $4$ bits per hex digit.
Expand each octal digit into 3 bits
$5 \to 101, 7 \to 111, 6 \to 110$

$(576)_{8} = (101111110)_{2}$
Regroup the same bits into fours, from the right
$101111110 ⟶ 000101111110$

A leading zero is added at the left to complete the first group of four.
Replace each group of four with its hex digit
$0001 = 1, 0111 = 7, 1110 = E$

$(576)_{8} = (17 E)_{16}$
Verify through decimal
$(576)_{8} = 5 \times 64 + 7 \times 8 + 6 = 320 + 56 + 6 = 382$
$(17 E)_{16} = 1 \times 256 + 7 \times 16 + 14 = 256 + 112 + 14 = 382 ✓$

Answer

$(17 E)_{16}$

Example 4

Subtraction using 2's complement

Question

Using 8-bit 2's complement arithmetic, evaluate $(45)_{10} - (28)_{10}$ .

Write both numbers in 8-bit binary
$45 = 32 + 8 + 4 + 1 = 00101101$
$28 = 16 + 8 + 4 = 00011100$
Find the 2's complement of the number being subtracted
Invert every bit of $28$ :

$00011100 invert 11100011$

Add $1$ :

$11100011 + 1 = 11100100$

So $- 28$ is represented by $11100100$ .
Add, instead of subtracting
$00101101 + 11100100 100010001$
Discard the carry out of the leftmost column
A carry out of the most significant bit is simply thrown away in 2's complement addition. The result is

$00010001$

The leading bit is $0$ , so the answer is positive.
Convert back to decimal
$00010001 = 16 + 1 = 17$

And indeed $45 - 28 = 17$ ✓
Note what this buys the hardware
No subtractor circuit was used anywhere. The machine inverted, added one, and then added — which is why a CPU needs only an adder and an inverter to do both addition and subtraction.

Answer

$00010001 = (17)_{10}$

NoteIf the leading bit of the result is 1, the answer is negative — take its 2's complement to read the magnitude.

Problem 1Basic

Convert to decimal: (a) $(11011)_{2}$ (b) $(347)_{8}$ (c) $(2 AF)_{16}$

Show solution

(a) $(11011)_{2}$

$1 \times 16 + 1 \times 8 + 0 \times 4 + 1 \times 2 + 1 \times 1 = 16 + 8 + 0 + 2 + 1 = 27$

(b) $(347)_{8}$

$3 \times 8^{2} + 4 \times 8^{1} + 7 \times 8^{0} = 3 \times 64 + 4 \times 8 + 7 = 192 + 32 + 7 = 231$

(c) $(2 AF)_{16}$ — with A $= 10$ and F $= 15$

$2 \times 256 + 10 \times 16 + 15 \times 1 = 512 + 160 + 15 = 687$

Answer(a)

27

(b)

231

(c)

687

Problem 2Basic

Convert $(93)_{10}$ to (a) binary (b) octal (c) hexadecimal.

Show solution

(a) Binary — divide by $2$ :

$22222229346231152101011101$

Reading upward: $(1011101)_{2}$ . Check: $64 + 16 + 8 + 4 + 1 = 93$ ✓

(b) Octal — group the binary in threes from the right:

$1011101 \to 001011101 \to (135)_{8}$

Check: $1 \times 64 + 3 \times 8 + 5 = 64 + 24 + 5 = 93$ ✓

(c) Hexadecimal — group the same bits in fours from the right:

$1011101 \to 01011101 \to (5 D)_{16}$

Check: $5 \times 16 + 13 = 80 + 13 = 93$ ✓

Converting once to binary and then regrouping is far quicker than dividing three separate times.

Answer(a)

(1011101)_{2}

(b)

(135)_{8}

(c)

(5 D)_{16}

Problem 3Exam level

Convert $(0.8125)_{10}$ to binary, and $(110.011)_{2}$ to octal and hexadecimal.

Show solution

Part 1 — $(0.8125)_{10}$ to binary. Multiply by $2$ repeatedly:

$0.8125 \times 2 0.625 \times 2 0.25 \times 2 0.5 \times 2 = 1 .625 = 1 .25 = 0 .5 = 1 .0 \leftarrow stop$

Reading the integer parts downward: $(0.1101)_{2}$ .

Check: $0.5 + 0.25 + 0 + 0.0625 = 0.8125$ ✓

Part 2 — $(110.011)_{2}$ to octal. Group in threes outward from the binary point:

$6110.3011 = (6.3)_{8}$

Part 3 — the same number to hexadecimal. Group in fours outward from the point, padding at both outer ends:

$60110.60110 = (6.6)_{16}$

Note the padding directions: to the left of the point pad on the far left; to the right of the point pad on the far right. Padding the fraction on the wrong side changes its value.

Check: $(110.011)_{2} = 6.375$ ; $(6.3)_{8} = 6 + 3/8 = 6.375$ ✓; $(6.6)_{16} = 6 + 6/16 = 6.375$ ✓

Answer

(0.1101)_{2}

;

(110.011)_{2} = (6.3)_{8} = (6.6)_{16}

Problem 4Exam level

Perform in binary: (a) $1011 + 1101$ (b) $11010 - 1011$ (c) $101 \times 110$

Show solution

(a) Addition $1011 + 1101$

$carry: 11110 1011 + 1101 11000$

Column by column from the right: $1 + 1 = 10$ (write $0$ , carry $1$ ); $1 + 0 + 1 = 10$ (write $0$ , carry $1$ ); $0 + 1 + 1 = 10$ (write $0$ , carry $1$ ); $1 + 1 + 1 = 11$ (write $1$ , carry $1$ ); the final carry $1$ is written down.

Result: $(11000)_{2}$ . Check: $11 + 13 = 24 = 16 + 8$ ✓

(b) Subtraction $11010 - 1011$

$11010 - 01011 01111$

From the right: $0 - 1$ needs a borrow, giving $1$ ; the next column, now reduced, gives $1$ ; then $1$ ; then $1$ ; then $0$ .

Result: $(1111)_{2}$ . Check: $26 - 11 = 15$ ✓

(c) Multiplication $101 \times 110$

$101 \times 110 000 (\times 0) 1010 (\times 1) 10100 (\times 1) 11110$

Each partial product is either $000$ or a copy of $101$ , shifted left one place each time.

Result: $(11110)_{2}$ . Check: $5 \times 6 = 30 = 16 + 8 + 4 + 2$ ✓

Answer(a)

11000

(b)

1111

(c)

11110

Problem 5Exam level

Find the 1's and 2's complement of $(01101001)_{2}$ , and explain why computers use 2's complement rather than signed magnitude.

Show solution

1's complement — invert every bit:

$01101001 ⟶ 10010110$

2's complement — add $1$ to the 1's complement:

$10010110 + 1 = 10010111$

Shortcut check. Copy bits from the right up to and including the first $1$ , then invert the rest. The original ends in $\dots 001$ ; keeping the final $1$ and inverting everything left of it gives $10010111$ ✓

Why 2's complement is preferred. Three reasons.

1. One zero, not two. In signed magnitude, $00000000$ is $+ 0$ and $10000000$ is $- 0$ — two bit patterns for the same value. Comparisons must then test for both, and one of the $256$ available patterns is wasted. In 2's complement there is exactly one zero, and the 8-bit range is the clean $- 128$ to $+ 127$ .

2. Subtraction becomes addition. In 2's complement,

$A - B = A + (2’s complement of B)$

so no subtractor circuit is needed at all — an inverter and the existing adder suffice. In signed magnitude the hardware must compare magnitudes, decide which is larger, subtract the smaller from the larger, and then work out the sign of the answer separately. That is far more circuitry for the same job.

3. Carries need no special handling. Any carry out of the most significant bit is simply discarded and the answer is still correct. The 1's complement scheme instead requires an end-around carry — the carry out must be added back into the least significant bit — which is an extra step and an extra piece of hardware.

Simpler circuitry that is also faster is why every general-purpose processor represents signed integers in 2's complement.

Answer1's complement

= 10010110

; 2's complement

= 10010111

. 2's complement is used because it has a single zero, turns subtraction into addition, and needs no end-around carry.

Problem 6Exam level

What is ASCII? Given that 'A' is $65$ and 'a' is $97$ , find the codes for 'E' and 'e', and explain how a program converts a letter to uppercase.

Show solution

ASCII — American Standard Code for Information Interchange — is a character encoding that assigns a unique number to every letter, digit, punctuation mark and control character. It uses 7 bits, so it defines $2^{7} = 128$ characters, normally stored one per byte. Extended ASCII uses the eighth bit as well, for $256$ characters.

Codes for 'E' and 'e'. The letters are consecutive, so 'E' is the 5th letter:

$’E’ = 65 + 4 = 69 ’e’ = 97 + 4 = 101$

The relationship between cases.

$97 - 65 = 32$

Every lowercase letter is exactly $32$ more than its uppercase counterpart, and this holds for the whole alphabet because both runs are contiguous and in the same order.

Converting case. A program therefore does arithmetic, not a table lookup:

$uppercase = lowercase - 32 lowercase = uppercase + 32$

Since $32 = 2^{5}$ , the difference is a single bit — bit 5. So the conversion is really one bit operation: clear bit 5 for uppercase, set it for lowercase.

A caution. The program must first check that the character actually is a letter. Blindly subtracting $32$ from a digit or a punctuation mark produces a different, wrong character — subtracting $32$ from '5' ( $53$ ) gives $21$ , a control character.

Other useful ASCII facts. Digits '0' to '9' occupy $48$ – $57$ , so the numeric value of a digit character is its code minus $48$ ; space is $32$ ; and carriage return is $13$ .

AnswerASCII is a 7-bit, 128-character encoding. 'E'

= 69

, 'e'

= 101

. Lowercase is always

32

(i.e. bit 5) above uppercase, so case conversion is one subtraction or addition — applied only after checking the character is a letter.

Problem 7Challenge

Distinguish between BCD and pure binary representation of a decimal number, and explain where BCD is preferred despite being less efficient.

Show solution

Pure binary treats the number as a single quantity and converts the whole of it. BCD codes each decimal digit separately in 4 bits.

Take $(59)_{10}$ .

Pure binary:
$59 = 32 + 16 + 8 + 2 + 1 = (111011)_{2} — 6 bits$

BCD:
$5 \to 0101, 9 \to 1001$
$(59)_{10} = (01011001)_{BCD} — 8 bits$

The two bit patterns are completely different, and BCD needed two more bits for the same number.

Why BCD is less efficient. Four bits can represent $16$ patterns, but a decimal digit needs only $10$ . The six patterns $1010$ to $1111$ are invalid in BCD and simply wasted. The waste grows with the size of the number: representing a 10-digit value takes 40 bits in BCD but only 34 in pure binary. BCD arithmetic is also more complicated, because after adding two BCD digits the hardware must check whether the result exceeded $1001$ and, if so, add $6$ to correct it.

Where BCD is preferred anyway. Whenever the number spends most of its life being displayed or entered as decimal, and comparatively little of it being calculated with.

Digital displays, calculators and instruments. Each 4-bit group drives one digit of a seven-segment display directly. In pure binary, showing $59$ on a display would first require dividing by $10$ to extract each digit — a division per digit, per refresh.
Exact decimal fractions in financial work. Values such as $0.1$ have no terminating representation in binary, so repeated pure-binary arithmetic on money accumulates rounding error. BCD (and the decimal types built on it) keeps decimal fractions exact, which is why currency is often held in decimal rather than binary floating point.
Frequent conversion at the boundary. If a value is read from a keypad and written to a display many times but added only rarely, the cost of conversion dominates the cost of arithmetic — and BCD makes conversion nearly free.

The general principle. Pure binary optimises storage and calculation; BCD optimises conversion to and from decimal, and decimal exactness. Which is more efficient depends entirely on which of those the application does more of.

AnswerPure binary converts the whole number (

59 = 111011

); BCD codes each digit in 4 bits (

59 = 01011001

), wasting the six patterns above

1001

. BCD is preferred where numbers are mainly displayed or entered in decimal, or where decimal fractions must stay exact — calculators, instrument displays and financial arithmetic.

BCA1B02T · M1

Types of software

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Example 1

Classifying software correctly

Question

Classify each as system or application software, and name the sub-type: (a) Windows 11, (b) MS Excel, (c) a printer driver, (d) a Python interpreter, (e) a college's fee-collection program, (f) WinZip.

Apply the test — who does it serve?
If it manages or supports the machine, it is system software. If it does the user's job, it is application software.
(a) Windows 11 and (c) a printer driver
Windows manages memory, files, processes and devices — system software, specifically the operating system.

A printer driver lets the operating system control one piece of hardware — system software, a device driver (a utility).
(d) A Python interpreter and (f) WinZip
The interpreter translates a high-level program into machine code, statement by statement — system software, a language translator.

WinZip compresses files to save space; it maintains the system rather than doing a user's task — system software, a utility.
(b) MS Excel and (e) the fee-collection program
Excel does a user's job — calculation and analysis — and is sold ready-made to everyone: application software, general-purpose (packaged).

The fee-collection program does a user's job and was written for this one college's rules: application software, custom (tailor-made).
Watch the trap
A Python interpreter is system software, but a program written in Python to manage a library is application software. The classification follows the purpose, not the language.

Answer

System: (a) operating system, (c) device driver/utility, (d) language translator, (f) utility. Application: (b) general-purpose, (e) custom.

Example 2

Compiler or interpreter — reading the symptoms

Question

A student reports: "My program printed three lines of output and then stopped with an error on line 12." What does this tell you about how the program was translated?

Note what actually happened
Some output appeared before the error was reported. So the first eleven lines were executed, and only then was the fault on line 12 discovered.
Rule out a compiler
A compiler translates the entire program before any of it runs. A syntax error on line 12 would be reported at compile time, together with every other error found, and no executable would be produced — so nothing could have been printed at all.
Conclude it was interpreted
An interpreter translates and executes one statement at a time. Lines 1–11 were translated and executed, producing output, and the error surfaced only when line 12 was reached.

$\Rightarrow$ The program was interpreted — consistent with Python, BASIC or JavaScript.
State the general rule the question is testing
Partial output followed by an error is the signature of interpretation. Errors reported before any output, with several listed at once, is the signature of compilation.

Answer

It was interpreted. A compiler would have found the error before producing an executable, so no output could have appeared.

NoteThis also explains why interpreted languages feel easier to learn on: you see results from the working part of a program even while a later part is still broken.

Problem 1Basic

Differentiate between system software and application software, with three examples of each.

Show solution

	System software	Application software
Purpose	Manages and controls the computer itself	Performs a task for the user
Written for	The hardware	A user's requirement
Necessity	Essential — the machine cannot work without it	Optional — installed as needed
Runs	From the moment the machine starts	When the user opens it
Depends on	The hardware	The system software beneath it
Written in	Usually low-level or system languages	Usually high-level languages
Examples	Windows, Linux, a compiler, a device driver, antivirus	MS Word, MS Excel, Chrome, a payroll program

The one-line distinction: system software serves the machine, application software serves the user.

AnswerSystem software (Windows, compilers, drivers) manages the machine; application software (Word, Excel, a payroll program) does the user's work.

Problem 2Basic

What is a utility program? List four with their functions.

Show solution

A utility program is a small system program that maintains, analyses or supports the computer, rather than doing a user's productive work. Utilities keep the system healthy and usable.

Antivirus — scans files and memory for malicious software, quarantines or removes what it finds, and monitors the system continuously. Examples: Windows Defender, Avast.
Disk defragmenter — over time a file's blocks become scattered across a hard disk, so the heads must move repeatedly to read one file. The defragmenter rearranges blocks so each file is contiguous, reducing seek time. (It is not used on SSDs, which have no seek time and whose cells have a limited number of writes.)
File compression tool — reduces file size for storage or transfer, and combines many files into one archive. Examples: WinZip, 7-Zip, WinRAR.
Backup and restore — copies data to another medium on a schedule, so that it can be recovered after deletion, corruption or hardware failure.

Others: disk cleanup, device drivers, disk formatting tools, screen savers, file managers.

AnswerA small system program that maintains or supports the computer — e.g. antivirus, disk defragmenter, file compression, backup software.

Problem 3Exam level

Compare a compiler and an interpreter under at least six headings. Which would you prefer while learning to program, and why?

Show solution

Heading	Compiler	Interpreter
Unit translated	The whole program at once	One statement at a time
Output produced	A separate executable file	None; nothing is saved
When translation happens	Once, before execution	On every run
Error reporting	All errors listed together, before execution	Stops at the first error found
Execution speed	Fast — machine code runs directly	Slow — translation repeats each time
Memory at run time	Lower; the compiler is not needed	Higher; the interpreter must stay in memory
Ease of debugging	Harder — errors are reported away from the running program	Easier — execution stops at the offending line
Distribution	Ship the executable; source can stay private	The source must be supplied
Examples	C, C++, Rust	Python, BASIC, JavaScript

Which is better while learning — an interpreter. Three reasons:

The feedback loop is shorter. Type a statement, run it, see the result. There is no separate compile step between writing and seeing what happens.
Errors point straight at the mistake. Execution stops at the line that failed, with the values in place, so the cause is usually visible. A compiler's list of forty cascading errors from one missing brace is far harder for a beginner to read.
A partly working program still runs. The working part executes and produces output even though a later part is broken — which is encouraging, and useful for testing an idea in pieces.

Why production systems still favour compilation: execution speed, lower memory use, and the ability to distribute a program without its source. In practice many modern languages do both — Java compiles to bytecode which is then interpreted or just-in-time compiled, taking the portability of one and much of the speed of the other.

AnswerCompared on unit translated, output, error reporting, speed, memory, debugging and distribution. An interpreter is better for learning — immediate feedback, errors located at the failing line, and partly working programs still run.

Problem 4Exam level

Distinguish between packaged and custom software. Which should a small bookshop choose for its accounts, and why?

Show solution

	Packaged (general-purpose)	Custom (tailor-made)
Written for	Many users with similar needs	One organisation
Cost	Low — development is shared across all buyers	High — one buyer pays for all development
Availability	Immediate; buy and install	Weeks to months of development first
Fit to requirement	Approximate; the user adapts to the software	Exact; the software matches existing procedures
Reliability	High — tested by very many users	Depends on the developer; fewer users find the bugs
Documentation and training	Plentiful; books, courses, online help	Must be produced specially
Support	Vendor support, large user community	Only from the developer, who may move on
Modification	Limited to what the settings allow	Anything can be changed, at a price
Examples	Tally, MS Excel, MS Office	A university examination system, a bank's core banking application

Recommendation for a small bookshop — packaged software.

Cost. Custom development would cost many times the shop's likely annual software budget. A packaged accounts program is a small one-off purchase.
The requirement is ordinary. Sales, purchases, stock, GST and a ledger are needs common to every small retailer, and packaged accounting software already handles them all.
Available now. The shop can be running this week rather than after months of development.
Reliability and support. Software used by thousands of shops has had its bugs found already, and help is available from the vendor, from books and from other users.
Legal changes are handled for you. When tax rules change, the vendor issues an update — a custom program would need paid modification each time.

When a custom system would be justified instead: if the organisation's processes are genuinely unusual, if the volume is large enough that small inefficiencies matter, or if the software is itself a competitive advantage. None of those applies to a small bookshop.

AnswerPackaged software is written for many, cheap, immediate and approximate; custom software is written for one, costly, slow to arrive and exact. A small bookshop should buy packaged accounting software — its needs are ordinary, the cost is a fraction, and support and legal updates come with it.

Problem 5Challenge

Explain the terms freeware, shareware, proprietary software and open source software. Is all free software open source, and is all open source software free?

Show solution

Proprietary (commercial) software. Sold or licensed for use, with the source code withheld. The licence forbids modification, reverse engineering and redistribution. You buy permission to use it, not the program itself. Windows, MS Office, Adobe Photoshop.

Freeware. Distributed free of cost, for unlimited time, but the source code is not provided and modification is not permitted. The author retains full copyright. Adobe Reader, Google Chrome, VLC's binaries as distributed.

Shareware. Distributed free on a trial basis — for a limited period, or with some features disabled, or with reminders to pay. Continued or full use requires payment. Source is not provided. WinRAR is the classic example.

Open source software. The source code is published, and the licence explicitly permits anyone to study it, modify it and redistribute it (usually on condition that the same freedoms are passed on). Linux, Firefox, LibreOffice, MySQL, Python.

Is all free software open source? No. This is the commonest confusion, and the two examples above settle it. Google Chrome costs nothing — it is freeware — but Google does not publish the whole of its source, so you cannot inspect it, modify it or build your own version. Free of charge says nothing about whether the source is available.

Is all open source software free of cost? No, not necessarily. Open source licences guarantee the freedom to obtain and modify the source, not that no one may charge. Red Hat Enterprise Linux is open source and is sold, with the customer paying for support, certification and updates rather than for the code. Many open-source projects also sell commercial licences or hosted versions alongside a free one.

The two independent questions. It helps to see this as a grid:

	Source available	Source withheld
No charge	Linux, Firefox, LibreOffice	Chrome, Adobe Reader (freeware)
Paid	Red Hat Enterprise Linux	Windows, MS Office (proprietary)

All four boxes are occupied, which is the proof that price and openness are separate matters.

Why the distinction is worded so carefully. In the phrase free software, "free" refers to freedom, not price — the freedoms to run, study, modify and share the program. The term FOSS (Free and Open Source Software) is used when both properties hold at once, precisely because "free" alone is ambiguous in English.

AnswerProprietary = paid, source withheld; freeware = free, source withheld; shareware = free trial then paid; open source = source published and modifiable. Neither implication holds: Chrome is free but not open source, and Red Hat Enterprise Linux is open source but sold. Price and openness are independent.

BCA1B02T · M2

Operating systems

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Example 1

Choosing the right type of operating system

Question

Name the type of operating system needed for each, with a reason: (a) a car's air-bag deployment controller, (b) generating electricity bills for 200,000 consumers overnight, (c) a college laboratory where 40 students work simultaneously on one server, (d) a bank's ATM network appearing as one system.

(a) Air-bag controller
The bag must inflate within milliseconds of a collision being detected. A correct decision delivered late is worthless — indeed fatal.

$\Rightarrow$ Hard real-time operating system. The defining requirement is a guaranteed deadline, not speed on average.
(b) 200,000 electricity bills
Every bill is processed the same way, no human is present, and nobody needs any individual bill immediately. What matters is total throughput.

$\Rightarrow$ Batch operating system. All the jobs are collected and run one after another without interaction.
(c) Forty students on one server
Many interactive users at once, each of whom must feel that the machine is responding to them. Each needs a small share of CPU time, very frequently.

$\Rightarrow$ Time-sharing (multitasking) operating system, giving each user a time quantum in rotation.
(d) An ATM network as one system
Many separate machines in different places, sharing data, presenting themselves to the user as a single banking service, and required to keep working when one node fails.

$\Rightarrow$ Distributed operating system.
The question behind the question
Each answer follows from asking what is being optimised: a deadline (real-time), total throughput (batch), response time (time-sharing), or transparency across machines (distributed).

Answer

(a) hard real-time (b) batch (c) time-sharing (d) distributed.

NoteIdentify the quantity being optimised — deadline, throughput, response time or transparency — and the type follows.

Example 2

Why the CPU is not idle while a file is being read

Question

A program reads a record from disk, which takes $10$ ms, then computes for $1$ ms, repeatedly. Explain what a multiprogramming operating system does with the waiting time, and estimate how many such programs could share one CPU.

Establish how the time divides
Each cycle is $10$ ms of waiting on I/O and $1$ ms of computing, so out of every $11$ ms the CPU is needed for only $1$ ms.

$CPU utilisation with one program = \frac{1}{11} \approx 9%$

For $91%$ of the time the processor would have nothing to do.
What the operating system does instead
When the program issues its disk read, the OS marks it as blocked, saves its state, and immediately gives the CPU to another program that is ready to run. This switch is a context switch. When the disk finishes it raises an interrupt, and the OS marks the first program ready again.
Estimate the number of programs
Each program needs the CPU for $1$ ms out of every $11$ ms, so in principle

$\frac{11 ms}{1 ms} = 11 programs$

could keep the processor fully occupied.
State why the real number is lower
Two effects reduce it. Context switching itself costs time — saving and restoring state is real work that produces no output. And the programs' requests do not arrive politely spaced: two may become ready at the same moment, so one waits. In practice utilisation rises steeply with the first few programs and then flattens.
Note what has and has not happened
No program has been made faster. Each still takes at least $11$ ms per cycle. What has improved is the utilisation of the CPU and therefore the total work done by the system — which is precisely the purpose of multiprogramming.

Answer

The OS blocks the waiting program and switches the CPU to another that is ready. About $11$ such programs could in theory saturate the CPU; in practice fewer, because context switching costs time and requests are unevenly spaced.

Problem 1Basic

Define an operating system and list its main functions.

Show solution

An operating system is the system software that manages all the hardware and software resources of a computer and acts as an interface between the user and the machine. It is loaded during booting and remains in memory while the computer is on.

Main functions.

Process management — creating and terminating processes, scheduling the CPU among them, and handling deadlock.
Memory management — allocating and reclaiming main memory, protecting one process from another, and providing virtual memory.
File management — organising data into files and folders, tracking their location on disk, and controlling access.
Device management — controlling devices through drivers, queueing requests, and spooling output.
Security and protection — user accounts, passwords, permissions and encryption.
User interface — providing a command-line or graphical means of issuing instructions.
Error detection — reporting hardware and software faults.
Accounting — recording which resources were used and by whom.

Examples: Windows, Linux, UNIX, macOS, Android, iOS.

AnswerSystem software that manages resources and interfaces user to machine; functions are process, memory, file, device and security management, plus the user interface, error detection and accounting.

Problem 2Basic

What is booting? Distinguish between cold and warm booting.

Show solution

Booting is the process of starting a computer and loading the operating system from secondary storage into main memory so that the machine becomes usable.

The sequence.

Power is switched on and the CPU begins executing the BIOS/UEFI firmware in ROM — the only program available at that moment.
POST (Power-On Self Test) checks that RAM, keyboard, display and drives are present and working.
The firmware finds the boot device and reads its boot sector.
The bootstrap loader there loads the operating system kernel into RAM.
The OS initialises itself, loads drivers and services, and presents the login screen.

The name comes from the phrase "to pull oneself up by one's bootstraps" — the machine must load the program that loads programs.

Cold against warm booting.

	Cold booting	Warm booting
Starting state	Computer switched off	Computer already running
Started by	Pressing the power switch	Ctrl+Alt+Del, the Restart option, or the reset button
POST performed?	Yes, in full	Usually skipped or abbreviated
Time taken	Longer	Shorter
Also called	Hard boot	Soft boot / reboot
Used when	Beginning work	The system has hung, or after installing software

AnswerBooting loads the OS into memory at startup, via BIOS → POST → boot sector → bootstrap loader → kernel. Cold booting starts from power off with a full POST; warm booting restarts a running machine, skipping most of the POST.

Problem 3Exam level

Explain any five types of operating system with one application of each.

Show solution

1. Batch operating system. Jobs with similar requirements are grouped into a batch and executed one after another with no user interaction during execution. Optimised for throughput on bulk work.
Application: payroll processing, electricity and telephone billing, bank statement generation.

2. Time-sharing (multitasking) operating system. Each program receives a small slice of CPU time (a time quantum) in rotation, switching fast enough that every user believes the machine is theirs alone. Optimised for response time.
Application: a multi-user server, and every modern desktop or laptop.

3. Multiprocessing operating system. Manages a computer with two or more processors or cores, so instructions genuinely execute at the same time. Gives higher throughput and greater reliability, since one processor failing need not stop the system.
Application: database servers, scientific computation, all current multi-core PCs.

4. Real-time operating system. Guarantees a response within a strict time limit; a correct result delivered late counts as a failure. Hard real-time systems must never miss a deadline; soft real-time systems merely degrade.
Application (hard): air-bag controllers, pacemakers, aircraft control, nuclear plant monitoring. (Soft): video streaming, reservation systems.

5. Distributed operating system. Manages several independent computers connected by a network so that they appear to the user as one system, sharing resources and continuing to work when a node fails.
Application: ATM networks, cloud platforms, large search engines.

Others worth naming: network operating systems (a server sharing files and printers with clients), embedded operating systems (routers, washing machines, televisions) and mobile operating systems (Android, iOS).

AnswerBatch (billing), time-sharing (multi-user servers and PCs), multiprocessing (database and scientific servers), real-time (air-bag controllers), distributed (ATM networks and cloud).

Problem 4Exam level

Compare a command line interface with a graphical user interface. Give one situation in which each is clearly better.

Show solution

	Command Line Interface (CLI)	Graphical User Interface (GUI)
Input	Typed commands with exact syntax	Clicking icons, menus and buttons with a pointer
Memory load on the user	High — commands and options must be recalled	Low — the available options are visible
Ease of learning	Difficult; needs training	Easy; can be explored
Speed for an expert	Very fast, especially on repetitive work	Slower — many clicks per operation
System resources	Very small	Larger — needs memory and a graphics display
Automation	Excellent; commands can be saved as scripts and scheduled	Poor; each action needs a person
Error tolerance	Low — a typing mistake may do something unintended	Higher — dialogs confirm and constrain
Remote use	Works over a very slow link	Needs substantial bandwidth
Examples	MS-DOS, Linux shell, Command Prompt, PowerShell	Windows, macOS, GNOME, Android

A GUI is built on WIMP — Windows, Icons, Menus, Pointer.

Where a GUI is clearly better. A student editing a photograph. The task is visual and exploratory: the user needs to see the image, try an adjustment, judge the result and undo it. There is no way to express "make this shade slightly warmer" as a typed command, and the work is done once rather than repeated.

Where a CLI is clearly better. Renaming 5,000 files according to a pattern, or performing the same maintenance job on a server at 2 a.m. every night. One command does what would be 5,000 sets of clicks, it can be saved and rerun exactly, it can be scheduled without a person present, and it works over a slow remote connection.

The underlying principle: a GUI shows you the options, which helps when you do not know what you want; a CLI lets you state exactly what you want, which helps when you do — and lets you state it once for a thousand items.

AnswerCLI: typed commands, hard to learn, fast, scriptable, tiny resource needs. GUI: visual, easy, resource-hungry, hard to automate. GUI wins for exploratory visual work such as photo editing; CLI wins for repetitive or scheduled bulk work such as renaming thousands of files.

Problem 5Challenge

Distinguish clearly between multiprogramming, multitasking and multiprocessing.

Show solution

All three words describe more than one program making progress, but they answer different questions.

Multiprogramming — how is the CPU kept busy?

Several programs are loaded in main memory at once. When the running program stops to wait for input or output, the operating system switches the CPU to another program that is ready. There is one CPU, executing one instruction at a time; the gain is that it is rarely idle. Switching happens when a program blocks — it is event-driven, not timed.

Purpose: maximise CPU utilisation.

Multitasking — how are interactive users kept happy?

Multiprogramming with a timer added. Each program is given a fixed time quantum; when it expires the OS switches to the next program whether or not the current one has finished or blocked. Switching is fast enough that a user perceives several programs running at once. Still one CPU, still one instruction at a time.

Purpose: minimise response time. Time-sharing is the multi-user form of the same idea.

Multiprocessing — how is more work done at once?

The computer physically contains two or more processors or cores, and the OS distributes work across them. Now instructions really are executed simultaneously — two programs can each be mid-instruction at the same instant, which is impossible under either of the others.

Purpose: increase throughput and reliability (one processor failing need not stop the system).

The summary table.

	Multiprogramming	Multitasking	Multiprocessing
Number of CPUs	One	One	Two or more
True simultaneity	No	No	Yes
Switch triggered by	A program blocking on I/O	Expiry of a time quantum	Not applicable — work is shared
Optimises	CPU utilisation	Response time	Throughput and reliability
Extra hardware needed	No	No	Yes

The one distinction to state. Multiprogramming and multitasking create the appearance of simultaneity by switching a single processor quickly; multiprocessing achieves actual simultaneity with additional processors. A modern laptop does all three at once — several programs in memory (multiprogramming), time-sliced for responsiveness (multitasking), across several cores (multiprocessing).

AnswerMultiprogramming: several programs in memory on one CPU, switching when one blocks, to maximise CPU use. Multitasking: the same with timed slices, to minimise response time. Multiprocessing: two or more physical processors, giving genuine simultaneity. Only the last needs extra hardware.

BCA1B02T · M2

Working with Windows

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Problem 1Basic

Name and explain the parts of a window.

Show solution

A window is a rectangular frame in which a program or folder is displayed.

Title bar — the strip across the top showing the name of the document or folder. Dragging it moves the window; double-clicking it maximises or restores.
Minimise button — reduces the window to a taskbar button. The program keeps running and unsaved work is retained.
Maximise / Restore button — expands the window to fill the screen; when already maximised the same button restores the previous size.
Close button — closes the window, and ends the program if it was the last window. Prompts to save unsaved work.
Menu bar / ribbon — the commands available in the program, grouped into menus or tabs.
Toolbar — buttons for frequently used commands.
Work area (client area) — where the document or folder contents appear.
Scroll bars — vertical and horizontal, appearing only when the contents are larger than the window, and dragged to see the rest.
Status bar — information at the foot: page number, word count, number of items selected.
Borders and corners — dragged to resize the window.

The important distinction: minimise hides, close terminates. A minimised program still occupies memory.

AnswerTitle bar, minimise, maximise/restore and close buttons, menu bar/ribbon, toolbar, work area, scroll bars, status bar, and resizable borders.

Problem 2Basic

What is a file extension? Give six extensions with the file type each denotes.

Show solution

A file extension is the group of letters following the final dot in a file name — the .docx in notes.docx. It identifies the type of the file, and Windows uses it to decide which program should open the file and which icon to display.

Extension	File type	Usually opened by
`.docx`	Word document	MS Word
`.xlsx`	Excel workbook	MS Excel
`.pptx`	PowerPoint presentation	MS PowerPoint
`.pdf`	Portable Document Format	A PDF reader or browser
`.jpg`	Compressed image	Photos, any image viewer
`.exe`	Executable program	Run directly by Windows

Others: .txt plain text, .mp3 audio, .mp4 video, .zip compressed archive, .html web page, .c C source code.

A caution. Renaming photo.jpg to photo.docx does not convert it. The extension is only a label; the contents are unchanged, so Word will open it and report a damaged file. Conversion requires a program that actually rewrites the contents.

A security note. Because the extension decides what Windows does with a file, attackers disguise programs as documents — invoice.pdf.exe looks like a PDF when extensions are hidden but is an executable. This is why displaying file extensions should be left switched on.

AnswerThe letters after the final dot, identifying the file type and the program that opens it — e.g. .docx Word, .xlsx Excel, .pptx PowerPoint, .pdf document, .jpg image, .exe program.

Problem 3Exam level

Explain the difference between copying and moving a file, and between deleting to the Recycle Bin and permanent deletion.

Show solution

Copy against move.

	Copy	Move (cut)
Keys	Ctrl+C then Ctrl+V	Ctrl+X then Ctrl+V
Original	Remains in place	Removed from the original location
Number of copies afterwards	Two	One
Disk space used	Doubled	Unchanged
Purpose	Keep a second copy, e.g. a backup	Reorganise; put the file where it belongs

Note also that dragging a file with the mouse within the same drive moves it, but dragging it to a different drive copies it — which is why cut-and-paste is the safer habit when the intention matters.

Deleting to the Recycle Bin against permanent deletion.

	Delete (Del)	Permanent delete (Shift+Del)
Where the file goes	The Recycle Bin	Removed from the file system immediately
Recoverable?	Yes — right-click and Restore	Not by ordinary means
Disk space freed	No — the file still occupies space	Yes, at once
Confirmation	Usually none	A warning dialog

A subtlety worth stating. "Permanent" deletion removes the file's directory entry and marks its blocks as free, but does not immediately overwrite the data. Until those blocks are reused, specialist recovery software can often retrieve the file. Genuinely destroying data requires overwriting it, or encrypting the drive and discarding the key — which matters when disposing of an old computer.

When the Bin is not used at all: Shift+Delete; deletion from a pen drive, memory card or network drive; or a file too large for the space allotted to the Bin.

AnswerCopy leaves the original and duplicates the file; move removes it from the source. Delete sends the file to the Recycle Bin, recoverable but still using disk space; Shift+Delete removes it immediately, freeing space but leaving the data recoverable by specialist tools until overwritten.

Problem 4Exam level

What is a shortcut? How does it differ from the file it points to, and what happens if each is deleted?

Show solution

A shortcut is a small file that acts as a pointer to another item — a file, folder, program or device stored elsewhere. It is recognised by a small curved arrow on its icon. Double-clicking a shortcut opens the item it points to, exactly as if the item itself had been opened.

Differences from the actual file.

	Shortcut	Actual file
Contains	Only the path to the target	The real data
Size	A few kilobytes, whatever the target's size	Its full size
Icon	Carries a small arrow overlay	No arrow
How many can exist	Any number, in any folder	One, in one place
Purpose	Convenient access from several places	Storage of the data

What happens on deletion.

Delete the shortcut — only the pointer is removed. The actual file is untouched and still in its folder. Nothing is lost.
Delete the actual file — the data is gone. The shortcut survives but is now broken: double-clicking it produces an error saying the item cannot be found. Windows may offer to delete the dead shortcut.
Move or rename the actual file — Windows can often repair the shortcut automatically within the same drive, but a move to a different drive or a deletion usually breaks it.

Why shortcuts exist. A program installed once in C:\Program Files can be reached from the desktop, the Start menu and the taskbar simultaneously, without three copies of the program existing. Space is saved and, more importantly, there is only one copy to update.

Common exam trap. Deleting a desktop shortcut does not uninstall a program. Uninstalling is done through Settings → Apps or Control Panel → Programs and Features.

AnswerA small file holding only a path to a target, marked with an arrow. Deleting the shortcut leaves the file intact; deleting the file leaves the shortcut broken. Deleting a shortcut never uninstalls a program.

Problem 5Exam level

What is the clipboard? Explain its role in cut, copy and paste, and state two of its limitations.

Show solution

The clipboard is a temporary area of main memory, managed by the operating system, that holds data which has been cut or copied so that it can be pasted elsewhere.

Its role in the three operations.

Copy (Ctrl+C) — places a duplicate of the selection on the clipboard. The original is unchanged.
Cut (Ctrl+X) — places the selection on the clipboard and removes it from its source.
Paste (Ctrl+V) — inserts the clipboard contents at the current position. The clipboard is not emptied, so the same item can be pasted repeatedly, in as many places as needed.

Why it matters that the OS owns it. Because the clipboard belongs to the operating system rather than to any one program, it works between programs: a table copied in Excel can be pasted into Word, and text copied from a browser into Notepad. This is one of the main practical benefits of a common operating system beneath all applications.

Limitations.

It holds one item at a time. Each new cut or copy replaces the previous contents, which are then gone. Copying a second item before pasting the first loses the first — the commonest cause of accidental data loss in office work. (Windows offers Clipboard History, Win+V, to mitigate this, and Office keeps a multi-item clipboard of its own.)
It is volatile. The clipboard lives in RAM, so its contents are lost when the computer is shut down or restarted, and often when the source program is closed for large items. Data must never be "stored" on the clipboard even briefly across a restart.

Further points: pasting may lose formatting depending on the destination, which is why Paste Special offers choices such as keeping the source formatting, matching the destination, or pasting unformatted text; and copying very large items consumes real memory.

AnswerA temporary OS-managed area of RAM holding cut or copied data. Copy duplicates to it, cut moves to it, paste inserts from it without emptying it, and because the OS owns it, it works between programs. Limitations: it holds only one item, each copy overwriting the last, and it is volatile so contents are lost on shutdown.

BCA1B02T · M3

Documents, editing and formatting

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Problem 1Basic

What is a word processor? List six of its features.

Show solution

A word processor is application software used to create, edit, format, store and print text documents. Examples: MS Word, LibreOffice Writer, Google Docs, WPS Writer.

Features.

Editing — text can be inserted, deleted, moved and copied freely at any time, unlike a typewriter where a mistake means retyping the page.
Formatting at three levels — character (font, size, bold, italic, colour), paragraph (alignment, indentation, line spacing, bullets) and page (margins, orientation, headers and footers).
Find and Replace — locating text anywhere in the document and substituting it, optionally throughout in one action.
Spelling and grammar checking, with a thesaurus and AutoCorrect.
Tables, images and objects — inserting and formatting tables, pictures, charts, shapes and equations within the text.
Mail merge — producing many personalised copies of one letter from a data source.

Others: styles and templates for consistent formatting; automatic table of contents and page numbering; track changes and comments for review; multiple views and zoom; saving in several formats including PDF; and print preview before committing to paper.

AnswerSoftware for creating, editing, formatting, storing and printing documents — features include editing, three levels of formatting, find and replace, proofing tools, tables and images, and mail merge.

Problem 2Basic

Distinguish between Save and Save As. When must Save As be used?

Show solution

	Save (Ctrl+S)	Save As (F12)
Effect	Writes changes to the existing file	Creates a new file
Asks for a name?	Only the very first time	Every time
Original file	Overwritten and lost	Left untouched
Files afterwards	One	Two

On the first save of a new document the two behave identically, because there is no existing file to overwrite — Word must ask for a name and location. From then on they differ.

When Save As must be used.

To keep the original. Editing last year's report to make this year's: Save would destroy last year's version, Save As keeps both.
To change the file format. Saving a .docx as .pdf for submission, as .doc for an older version of Word, or as .txt for plain text.
To change the name or location of a document already saved — to move it into a project folder, or to give it a clearer name.
To create a working copy before a risky change, such as a large Replace All across a long document.

Practical rule: Save often while working; Save As whenever the earlier version still matters.

AnswerSave overwrites the current file; Save As creates a new one and leaves the original intact. Save As is needed to keep the original, to change format (e.g. to PDF), or to change the name or location.

Problem 3Exam level

Explain the three levels of formatting in a word processor, with four examples of each.

Show solution

Formatting operates at three levels, distinguished by what the command applies to.

1. Character (font) formatting — applies to the selected characters only, so text must be selected first.

Examples: font name and size; bold, italic, underline; font colour and highlighting; superscript and subscript.
Also: strikethrough, change case, character spacing, small caps.

2. Paragraph formatting — applies to the entire paragraph containing the cursor. No selection is needed, and selecting only part of a paragraph still formats the whole of it. A paragraph is all the text up to the next press of Enter.

Examples: alignment (left, centre, right, justified); indentation (left, right, first line, hanging); line spacing (single, 1.5, double); bullets and numbering.
Also: spacing before and after, borders and shading, tab stops, widow and orphan control.

3. Page formatting — applies to the whole document, or to a section of it.

Examples: margins; orientation (portrait or landscape); paper size (A4, Letter, Legal); headers and footers with page numbers.
Also: columns, page borders, watermark, page and section breaks.

Why the distinction is worth knowing. It tells you what to select before issuing a command. Trying to make one word bold without selecting it does nothing; trying to centre only half a paragraph is impossible, because centring is a paragraph property. And to change page settings for part of a document you must first insert a section break, since page formatting otherwise applies to everything.

AnswerCharacter formatting applies to selected text (font, size, bold, colour); paragraph formatting to the whole paragraph containing the cursor (alignment, indents, line spacing, bullets); page formatting to the document or section (margins, orientation, paper size, headers and footers).

Problem 4Exam level

What is a style? Why is using styles better than formatting text directly?

Show solution

A style is a named set of formatting instructions — font, size, weight, colour, alignment, spacing, indentation — that is applied to text in a single action. Word supplies built-in styles including Normal, Title, Quote and Heading 1 to Heading 9, and users can create their own.

Why styles beat direct formatting.

1. One change updates everywhere. In a report with forty section headings all set to Heading 1, deciding that headings should be dark blue and 14 pt means modifying the style once — all forty change. Formatted by hand, it would mean forty separate edits, and experience says one will be missed.

2. Guaranteed consistency. Applying a style gives exactly the same result every time. Formatting by hand, the tenth heading acquires 13.5 pt instead of 14, or bold but not the right colour — differences nobody notices until the document is printed.

3. Styles carry structure, not just appearance. This is the deeper reason. Applying Heading 1 tells Word "this is a top-level heading". Making text big and bold by hand tells Word nothing — it merely looks like a heading. Consequences of real structure:

Automatic table of contents. Word builds it from heading styles, with correct page numbers, and updates it when the document changes. Without styles, the contents page must be typed and maintained by hand.
Navigation pane — jump straight to any section.
Cross-references — "see Section 3.2" stays correct when sections are renumbered.
Outline view and the ability to collapse and reorder whole sections.
Accessibility — a screen reader announces headings, letting a blind reader navigate the document.

4. Speed. One click or shortcut against five or six separate commands, on every heading in the document.

5. Templates. A set of styles saved as a template gives every document in an organisation the same appearance, without anyone remembering the specification.

The summary. Direct formatting records what text should look like; a style records what the text is, and derives the appearance from that. Everything automatic in a long document depends on the second.

AnswerA named collection of formatting applied in one action. Styles guarantee consistency, let one edit update every instance, and — crucially — record document structure, which is what makes automatic tables of contents, navigation, cross-references and accessibility possible.

Problem 5Challenge

A student's 40-page project has its table of contents typed by hand, gaps made by pressing Enter repeatedly, and headings formatted individually. Explain the problems this causes and how each should have been done.

Show solution

Three habits, each of which fails as soon as the document changes.

Problem 1 — a hand-typed table of contents.

Every page number in it is a guess frozen at the moment it was typed. Inserting one paragraph on page 5 shifts every subsequent page, so all the numbers below become wrong at once, and the student must recheck forty entries by hand — before every reprint. Renaming a section means editing it in two places, and sooner or later the contents page and the heading disagree.

How it should be done: apply Heading 1/2/3 styles to the headings, then References → Table of Contents. Word generates it with correct page numbers, and Update Table (or F9) corrects the whole thing in one action.

Problem 2 — blank lines made by pressing Enter.

Each press creates an empty paragraph, not spacing. This causes:

Gaps that move. When text reflows, the empty paragraphs travel with it, so a gap intended before a heading can end up in the middle of a page or, worse, as blank lines at the top of the next page.
Uneven spacing. Some gaps are one Enter, some two, some three, so no two sections are spaced alike.
Wrong spacing when the font changes. An empty paragraph's height depends on its font size, so changing the body font changes every gap by a different amount.
Broken numbering and contents. Empty paragraphs are counted as paragraphs; they interrupt numbered lists and can appear as blank entries.

How it should be done: set Spacing Before and Spacing After in Paragraph settings — ideally as part of a style, so every heading gets identical space automatically and it stays attached to the heading wherever it moves. Also tick Keep with next so a heading can never be stranded at the foot of a page.

Problem 3 — headings formatted individually.

Inconsistency is certain: over forty headings some will differ in size, colour or spacing.
Any change means forty edits, and one will be missed.
No structure exists. Word does not know these are headings, so the navigation pane is empty, no table of contents can be generated, cross-references are impossible, and a screen reader cannot navigate the document.

How it should be done: apply heading styles, and modify the style when the appearance needs to change.

The common cause. All three mistakes come from formatting what the text should look like rather than declaring what it is. Once structure is declared — this is a heading, this paragraph needs space after it — appearance follows automatically, and stays correct when the document changes. The extra effort is a few minutes at the start; the saving is every reprint thereafter.

AnswerA typed contents page goes wrong the moment anything shifts; Enter-key gaps create empty paragraphs that move, vary and break numbering; hand-formatted headings are inconsistent, need forty edits to change, and carry no structure — so no contents page, navigation or accessibility is possible. Use heading styles, paragraph Spacing Before/After, and an automatic table of contents.

BCA1B02T · M3

Tables, graphics and page layout

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Problem 1Basic

Explain the terms merge cells, split cells and AutoFit in a table.

Show solution

Merge cells. Combines two or more adjacent cells into a single larger cell. The contents of the merged cells are kept, one below another. Its commonest use is a heading spanning the whole width of a table, or a label covering several rows.

How: select the cells → right-click → Merge Cells, or Layout → Merge Cells.

Split cells. The reverse: divides one cell into a specified number of rows and columns. Word asks how many of each. Used when a single column must be subdivided — for example splitting "Marks" into "Theory" and "Practical".

How: click the cell → Layout → Split Cells → enter the numbers.

Related: Split Table divides one table into two separate tables at the cursor's row.

AutoFit. Sets column widths automatically, in one of three ways:

AutoFit Contents — each column shrinks or grows to exactly fit the widest entry in it, so no space is wasted.
AutoFit Window — the table stretches to fill the full width between the margins, columns keeping their relative proportions.
Fixed Column Width — switches automatic adjustment off, so columns stay exactly as set even when longer text is typed.

How: Layout → AutoFit.

Also useful: Distribute Rows and Distribute Columns, which make all rows or columns exactly equal.

AnswerMerge cells combines adjacent cells into one (for spanning headings); split cells divides one cell into several; AutoFit sets column widths automatically to fit the contents, to fill the window, or fixes them.

Problem 2Basic

What is a header and footer? Give four items typically placed in each.

Show solution

A header is text or graphics that appears in the top margin of every page of a document (or of a section); a footer does the same in the bottom margin. They are entered once and repeated automatically, so they never have to be retyped.

Typically in a header: the document or chapter title; the organisation's name and logo; the author's name; the date; a confidentiality marking such as "Draft" or "Confidential".

Typically in a footer: the page number; "Page 3 of 40"; the file name and path; a copyright line; the date of printing.

Important options.

Different First Page — suppresses the header and footer on a title page.
Different Odd and Even Pages — for double-sided printing and binding, so the page number always falls on the outer edge.
Link to Previous — when switched off after a section break, each section can carry its own header, which is how a chapter name can appear in the header of that chapter only.

Why page numbers are inserted as fields. A page number typed by hand would be the same on every page. Word instead inserts a field, a placeholder it evaluates for each page, so numbering stays correct however much text is added or removed — and can be made to restart at 1 after the preliminary pages.

How to access: Insert → Header / Footer, or double-click in the top or bottom margin. Double-click in the body of the page to leave.

AnswerRepeating text in the top and bottom margins of every page. Headers typically hold the title, logo, author or date; footers the page number, "Page x of y", file name or copyright. Page numbers are fields, so they stay correct automatically.

Problem 3Exam level

What is text wrapping? Describe five wrapping options and say when each is appropriate.

Show solution

Text wrapping determines how the text of a document behaves around an inserted picture or shape — whether it flows round it, stops for it, or ignores it entirely. It is set from Picture Format → Wrap Text, or from the small layout-options button that appears beside a selected picture.

Five options.

1. In line with text. The picture is treated as a single very large character sitting in the line of text. It moves with the surrounding words as text is added or deleted, and cannot be dragged freely.
Appropriate for: a small diagram or equation image that must stay exactly with the sentence that refers to it — the safest choice for academic work, because the picture can never drift away from its text.

2. Square. Text flows around the rectangular boundary enclosing the picture, leaving a neat margin on all sides.
Appropriate for: a photograph placed beside a paragraph in a report or newsletter — the standard choice for a floating illustration.

3. Tight. Text follows the picture's actual outline rather than its rectangle, so it hugs an irregular shape.
Appropriate for: a cut-out image with a transparent background, such as a logo or a photograph with the background removed.

4. Top and bottom. Text stops above the picture and resumes below it; nothing appears to the left or right.
Appropriate for: a wide chart, table image or map that occupies the full width — trying to squeeze text alongside would leave unreadably narrow columns.

5. Behind text. The picture sits underneath the text, which runs straight over it.
Appropriate for: a watermark, a background pattern, or a faint logo behind the body of a certificate. (The opposite, In front of text, places the picture over the text and is used for a stamp or an overlaid label, but hides whatever is beneath.)

In line against floating. In line with text is the only option that keeps the picture in the text flow; the rest make it floating, anchored to a position on the page. A floating picture shows an anchor mark against the paragraph it is attached to — and if that paragraph is deleted, the picture disappears with it, which surprises many users.

AnswerText wrapping controls how text behaves around a picture. In line with text (keeps the picture with its sentence); Square (a floating illustration beside a paragraph); Tight (an irregular cut-out); Top and bottom (a full-width chart); Behind text (a watermark).

Problem 4Exam level

Distinguish between a page break and a section break, and give two situations requiring a section break.

Show solution

	Page break	Section break
Effect	Moves the following text to a new page	Starts a new section, which may have its own page settings
Inserted by	Ctrl+Enter, or Insert → Page Break	Layout → Breaks → (Next Page / Continuous / Even / Odd)
Allows different page settings?	No	Yes
Types	One	Four
Used for	Beginning a new chapter on a fresh page	Changing margins, orientation, columns or numbering part-way through

Both are far better than the alternative of pressing Enter repeatedly, which collapses as soon as text above is edited.

Situations requiring a section break.

1. One landscape page in a portrait document. Orientation is a property of a section, so a wide table needing landscape must be isolated between two Next Page section breaks; only then can that section be rotated while the rest stays portrait. The same applies to different margins for an appendix, or a different paper size.

2. Page numbering that restarts, or differs in style. A project report typically has preliminary pages (title, certificate, acknowledgement, contents) numbered in roman numerals i, ii, iii, and a main body numbered 1, 2, 3 starting from the introduction. This needs a section break between them, Link to Previous switched off for the header/footer, and the second section set to Start at 1.

Other cases: two-column text below a full-width headline (Word inserts Continuous section breaks for this automatically); a different header carrying each chapter's own name; and starting every chapter on an odd-numbered page for double-sided printing, using an Odd Page section break.

Practical tip. Turn on formatting marks with Ctrl+Shift+8 to see exactly where breaks lie. Invisible section breaks in the wrong place are a common cause of page settings that "refuse" to change.

AnswerA page break only starts a new page; a section break starts a new section that can have its own page settings. Section breaks are needed for a landscape page inside a portrait document, and for page numbering that restarts or changes style between the preliminary pages and the main body.

Problem 5Challenge

Describe the steps to create an automatic table of contents, and explain why it must be built from heading styles rather than from manually formatted headings.

Show solution

Steps.

Apply heading styles. Give each chapter title Heading 1, each section Heading 2, each sub-section Heading 3. Select the text and click the style on the Home tab, or use Ctrl+Alt+1, 2, 3.
Adjust the appearance if wanted — right-click a style and choose Modify to change its font, size, colour and spacing. Every heading using it changes at once.
Place the cursor where the contents should appear, normally a page of its own created with a page break.
Insert the table. References → Table of Contents → Automatic Table 1 (or Custom Table of Contents to choose how many levels to show, and whether to show page numbers and dot leaders).
Update after editing. Click the table and choose Update Table, or press F9, then choose Update page numbers only or Update entire table. Word also updates it when the document is printed if that option is set.

Why heading styles are essential.

A table of contents is generated by Word reading the document's outline. That outline exists only where heading styles have been applied. Applying Heading 1 records the fact "this text is a top-level heading" as a property of the paragraph. Making text 16 pt bold by hand records only "this text is 16 pt bold" — an appearance, carrying no meaning.

The consequence is direct: Word looks for paragraphs marked as headings. If none exist it finds nothing, and inserts an empty table or the message that no entries were found. It cannot infer intent from formatting, because plenty of text is bold and large without being a heading — a company name, an emphasised warning, the title of a figure.

What else the same structure provides, all of it lost with manual formatting:

Navigation pane — a clickable outline for jumping between sections
Collapsible sections in the body of the document
Cross-references — "see Section 3.2" that stays correct when sections are renumbered
Outline view — reordering an entire section, with its contents, by dragging one heading
Automatic numbering of chapters and sections, renumbering itself when one is inserted
Accessibility — a screen reader announces headings and their levels, letting a blind reader move through the document rather than listening to it end to end
Correct conversion — a document exported to PDF, HTML or EPUB keeps its bookmarks and structure

The general lesson. Manual formatting describes what text should look like; a style declares what text is. Every automatic feature in a long document is built on the second, and none of them can be recovered later from the first. Twenty minutes applying styles at the start saves the hour spent renumbering a contents page before every reprint — and produces a document that stays correct.

AnswerApply Heading 1/2/3 styles, place the cursor, then References → Table of Contents, and Update Table (F9) after edits. It must use styles because Word generates the table from the document outline, which exists only where heading styles are applied — manual bold text records appearance, not structure, so it also loses navigation, cross-references, outline editing and accessibility.

BCA1B02T · M3

Mail merge and printing

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Problem 1Basic

What is mail merge? Name and explain its three components.

Show solution

Mail merge is the process of combining a standard document with a list of data to produce many personalised copies of that document — one for each record in the list — in a single operation.

$Main document + Data source ⟶ one merged document per record$

The three components.

1. Main document. The document containing the text that is the same in every copy — the body of a letter, the layout of a label, the wording of a certificate — together with merge fields marking the places where varying data is to appear. It is written and proofread once.

2. Data source. A table holding the data that varies between copies. Its first row contains the field names (Name, Address, RollNo, Marks) and each following row is one record, that is one recipient. It may be a Word table, an Excel worksheet (the usual choice), an Access database, Outlook contacts, or a delimited text file.

3. Merge field. A placeholder inserted into the main document, displayed as «Name», which is replaced during the merge by the corresponding value from the current record. Fields must be inserted with Insert Merge Field; typing the angle brackets by hand produces ordinary text that never merges.

Applications. Circular letters to students or customers, address labels, envelopes, certificates, mark lists, salary slips, and personalised bulk email.

AnswerCombining a standard document with a data list to produce one personalised copy per record. Components: the main document (fixed text plus merge fields), the data source (field names in row 1, one record per row) and the merge fields (placeholders replaced at merge time).

Problem 2Basic

Explain collated and uncollated printing with an example, and state when each is wanted.

Show solution

Both settings concern the order in which pages emerge when more than one copy of a multi-page document is printed.

Take 3 copies of a 4-page document.

Collated — each complete copy is printed in turn:

$1, 2, 3, 4 ∣ 1, 2, 3, 4 ∣ 1, 2, 3, 4$

Three finished documents come out, each in the right order and ready to staple.

Uncollated — all copies of each page are printed together:

$1, 1, 1 ∣ 2, 2, 2 ∣ 3, 3, 3 ∣ 4, 4, 4$

Four stacks emerge, and the pages must be sorted by hand into three documents.

When collated is wanted — almost always for documents to be read or bound: reports, project submissions, question papers, handouts. The printer does the sorting.

When uncollated is wanted — when the pages are to be separated rather than kept together: printing 40 copies of a single-page notice for distribution, or a set of forms where all copies of each form go to a different office. Uncollated is also slightly faster on some printers, since the page image is sent once and printed repeatedly.

A note on single-page documents: the setting makes no difference at all, which is why it goes unnoticed until someone prints thirty copies of a booklet uncollated and spends an afternoon sorting.

AnswerCollated prints complete copies in order (1,2,3,4 | 1,2,3,4) and is wanted for reports and anything to be bound; uncollated prints all copies of each page together (1,1,1 | 2,2,2) and suits pages to be distributed separately.

Problem 3Exam level

Describe the steps in performing a mail merge to send a letter to 100 customers from an Excel list.

Show solution

Step 1 — Prepare the data source. In Excel create a worksheet with the field names in the first row — Name, Address1, City, PIN, Amount — and one customer per row for all 100. Requirements: no blank rows, no title line above the headings, no merged cells, and consistent data in each column. Save and close the workbook.

Step 2 — Create the main document. In Word, type the letter with the text common to every customer, leaving the personalised parts empty for now.

Step 3 — Start the merge. Mailings → Start Mail Merge → Letters.

Step 4 — Select the recipients. Mailings → Select Recipients → Use an Existing List. Choose the Excel file, then the worksheet, and confirm that the first row contains column headers.

Step 5 — Insert the merge fields. Place the cursor where each item belongs and use Insert Merge Field to place «Name», «Address1», «City», «PIN» and «Amount». Address Block and Greeting Line insert a conventionally formatted address and salutation in one step, and Address Block suppresses lines that are empty.

Step 6 — Refine the list if required. Edit Recipient List allows records to be filtered (only customers of one city), sorted (by PIN, for postal discount), or individually unticked.

Step 7 — Add rules if the wording must vary. Rules → If…Then…Else can insert "Sir" or "Madam" from a Gender field, or a different closing paragraph depending on the amount.

Step 8 — Preview. Preview Results and step through the records. Check the first, the last, and any with unusual data — a very long name, a missing address line, an amount that appears as 1500.000000 (which needs a numeric formatting switch).

Step 9 — Complete the merge. Finish & Merge, then either:

Edit Individual Documents — creates one document holding all 100 letters, which can be checked, corrected and saved as a record before printing
Print Documents — prints directly
Send E-mail Messages — emails each customer, using an email-address field

Step 10 — Save both source files. Keep the main document and the Excel list. The next letter reuses the same recipient list, and the merge can be run again at any time.

AnswerPrepare the Excel data with field names in row 1 → Start Mail Merge (Letters) → Select Recipients from the workbook → Insert Merge Fields → filter/sort the list → add rules if needed → Preview Results → Finish & Merge (Edit Individual Documents, Print, or Email) → save both files for reuse.

Problem 4Exam level

What is spooling? Explain the print queue and why printing does not lock up the computer.

Show solution

Spooling — from Simultaneous Peripheral Operation On-Line — is the technique of placing output intended for a slow device into a temporary file on disk, and then feeding it to the device at whatever rate the device can accept.

Why it is needed. A printer is enormously slower than a processor. Printing a page may take several seconds; in that time the CPU could execute billions of instructions. Without spooling the program would have to hand over each line of output and wait for the printer to accept it, so the application — and the user — would be blocked for the whole duration of the print.

What happens instead. The application writes the entire print job to a spool file on disk as fast as the disk will take it, which is a fraction of a second. It is then finished and the user regains control immediately. A separate part of the operating system, the print spooler service, reads that file and sends it to the printer in the background at the printer's own pace.

The print queue. Because jobs are held on disk, several can wait at once, and the queue is the list of them. From it a job may be:

paused and later resumed
restarted from the beginning
cancelled and removed
reordered by priority

The queue also allows one printer to serve many users on a network, each job being printed to completion in turn rather than the outputs becoming interleaved.

Why the computer does not lock up. Three things together: the transfer to disk is fast; the spooler runs as a background process so the CPU continues with foreground work; and the operating system's multitasking shares the processor between the spooler and the application. The user can close the document, or even the program, and printing continues — the job no longer depends on the application, because the output is already safely on disk.

A practical consequence. Cancelling a job from the queue may not stop printing at once, because pages already transferred are held in the printer's own memory and will finish. And if the spool folder's disk fills up, printing fails with an error that appears to be a printer fault but is really a disk-space problem.

AnswerSpooling writes print output to a temporary disk file which a background spooler service then feeds to the printer at the printer's pace. It creates the print queue (jobs can be paused, restarted, cancelled or reordered) and, because the transfer to disk is fast and the spooler runs as a background process under multitasking, the application is freed immediately instead of waiting on the printer.

Problem 5Challenge

A merge produces the letters, but every letter shows «Name» instead of a name, and the addresses have blank lines. Diagnose both faults and give the remedy.

Show solution

Two independent faults, with different causes.

Fault 1 — «Name» appears instead of the data.

There are three possible causes, and they are distinguished by what the document does when Preview Results is toggled.

(a) The field was typed rather than inserted. The characters « and » can be typed from the keyboard or pasted, producing text that looks exactly like a merge field but is ordinary text. Word has no reason to replace it, so it prints literally.
Test: switch Preview Results on. A genuine field changes to data; typed text does not.
Remedy: delete the typed text and insert the field properly with Mailings → Insert Merge Field.

(b) Preview Results is simply switched off. In this state Word deliberately shows the field names so you can see the structure — and if the merge is completed from here the output is still correct.
Test: look at the merged output rather than the main document.
Remedy: switch Preview Results on; nothing else is wrong.

(c) Field codes are being displayed rather than their results, which shows as { MERGEFIELD Name }.
Remedy: press Alt+F9 to toggle field codes off.

A fourth possibility if the data genuinely fails to arrive: the data source is not attached, or the field name in the document no longer matches a column heading in the source — for instance the Excel heading was changed from Name to Student Name after the field was inserted. Edit Recipient List shows whether the source is connected and what its columns are called; Match Fields repairs a mismatch.

Fault 2 — blank lines in the addresses.

The cause is empty fields in the data source. A merge field with no value produces nothing, but the paragraph mark or line break you typed after it in the main document remains — so a customer with no second address line still gets that line's worth of empty space, and the city and PIN are pushed down.

$Address1 \to Address2 (empty) \to City ⟹ a gap where Address2 should be$

Three remedies, best first.

Use Address Block instead of inserting the address fields one by one. It is designed for exactly this: it assembles the address and suppresses any line whose fields are all empty. This is the correct solution and needs no per-record work.
Use an If…Then…Else rule where a specific line must be conditional: Rules → If…Then…Else — if Address2 is not blank, insert Address2 and a new line, otherwise insert nothing.
Fill the gaps in the data, if the values genuinely exist and were simply omitted. Never "fix" it by putting a space in the empty cells — Word then treats the field as non-empty and the blank line returns.

The general lesson. A merge fault is almost always in one of two places: the link between document and data (fields not inserted, source not attached, names not matching), or the shape of the data (empty cells, blank rows, headings not in row 1). Preview Results, stepping through the first and last records and one deliberately incomplete record, catches nearly all of them before anything is printed.

Answer«Name» printing literally means the field was typed rather than inserted (or Preview Results is off, or field codes are displayed with Alt+F9, or the source/column name no longer matches). Blank address lines come from empty fields whose typed line break remains — use Address Block, which suppresses empty lines, or an If…Then…Else rule.

BCA1B02T · M4

Workbooks, cells and references

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Example 1

Predicting what a formula becomes when copied

Question

Cell D2 contains =B2*$F$1+C$1-$A2. Write the formula as it appears after copying to F5.

1
Find the displacement
From D2 to F5:
- columns: D → F, so 2 columns right
- rows: 2 → 5, so 3 rows down
Every relative part shifts by exactly this amount; every locked part does not.
Take each reference in turn — B2
B2 is fully relative. Both parts move: column B + 2 = D, row 2 + 3 = 5.

$B2 ⟶ D5$
$F$1 — both parts locked
Both parts are locked, so nothing changes.

$$F$1 ⟶ $F$1$
C$1 — mixed, row locked
The column has no dollar, so it moves: C + 2 = E. The row is locked at 1.

$C$1 ⟶ E$1$
$A2 — mixed, column locked
The column is locked at A. The row has no dollar, so it moves: 2 + 3 = 5.

$$A2 ⟶ $A5$
Assemble
$=D5*$F$1+E$1-$A5$

Answer

=D5*$F$1+E$1-$A5

NoteWork reference by reference, not formula by formula. Ask of each part: is there a dollar in front of it? If not, add the displacement.

Example 2

Why one formula needs an absolute reference

Question

Column B holds prices in B2:B6 and cell E1 holds the GST rate $0.18$ . A student enters =B2*E1 in C2 and fills down to C6, and gets the right answer in C2 and zeros below. Explain and correct it.

Work out what the fill actually produced
E1 is relative, so it moved down with the formula:

Cell Formula after filling What E-column cell it reads
C2 =B2*E1 E1 — the rate ✓
C3 =B3*E2 E2 — empty
C4 =B4*E3 E3 — empty
C5 =B5*E4 E4 — empty
C6 =B6*E5 E5 — empty
Explain the zeros
An empty cell used in arithmetic is treated as $0$ . So every row below the first computes

$price \times 0 = 0$

which is why the results are zeros rather than error values — the formula is perfectly valid, it is simply multiplying by nothing.
Identify what should have been locked
The price must change with each row; the rate must not. So the price reference stays relative and the rate reference becomes absolute:

$=B2*$E$1$
Apply and verify
Enter =B2*$E$1 in C2 and fill down. Now:

Cell Formula
C2 =B2*$E$1
C3 =B3*$E$1
C6 =B6*$E$1

Every row multiplies its own price by the one rate.
Note the further advantage
Because the rate lives in one cell that every formula consults, changing E1 from $0.18$ to $0.12$ updates all five results at once. Typing $0.18$ into each formula would give the same answers today and five edits to make tomorrow — and one of them would be missed.

Cell	Formula after filling	What E-column cell it reads
C2	`=B2*E1`	E1 — the rate ✓
C3	`=B3*E2`	E2 — empty
C4	`=B4*E3`	E3 — empty
C5	`=B5*E4`	E4 — empty
C6	`=B6*E5`	E5 — empty

Cell	Formula
C2	`=B2*$E$1`
C3	`=B3*$E$1`
C6	`=B6*$E$1`

Answer

E1 was relative and drifted to empty cells, which count as zero. Use =B2*$E$1, filled down.

NoteThe rule of thumb: a value that appears once on the sheet but is used by many formulas needs an absolute reference.

Example 3

Filling a multiplication table with mixed references

Question

Row 1 holds $1$ to $10$ in B1:K1, and column A holds $1$ to $10$ in A2:A11. Write a single formula for B2 that, filled across and down, produces the whole multiplication table.

1
State what must stay fixed in each direction
B2 must multiply the number at the top of its column by the number at the left of its row.
- The top number is always in row 1, whichever column we are in. So the row must be locked and the column free.
- The left number is always in column A, whichever row we are in. So the column must be locked and the row free.
Write the two references
Top of the column: B$1 — column free, row locked.

Left of the row: $A2 — column locked, row free.
Give the formula
$=B$1*$A2$

Enter this in B2, then select B2 and fill across to K2, then select B2:K2 and fill down to row 11.
Check a cell away from the corner — D5
From B2 to D5 is 2 columns right and 3 rows down.

B$1: column B + 2 = D, row locked → D$1 — the top of column D, which holds $3$ .

$A2: column locked, row 2 + 3 = 5 → $A5 — the left of row 5, which holds $4$ .

So D5 computes $3 \times 4 = 12$ ✓
See why neither alternative works
Fully relative =B1*A2 would drift both references diagonally and give nonsense away from the first cell.

Fully absolute =$B$1*$A$2 would put $1 \times 1 = 1$ in all 100 cells.

Only the mixed form varies in one direction per reference, which is exactly what a two-dimensional table needs.

Answer

=B$1*$A2 in B2, filled across and down.

NoteAny grid computed from a header row and a header column uses this pattern: lock the row for the header row, lock the column for the header column.

Problem 1Basic

Define workbook, worksheet, cell, cell address and range, and state how many rows and columns a worksheet has.

Show solution

Workbook — the spreadsheet file, saved with the extension .xlsx. One workbook contains one or more worksheets.
Worksheet — a single grid of rows and columns within the workbook, reached by its tab at the foot of the window. Sheets can be renamed, reordered, copied, coloured, hidden and deleted.
Cell — the intersection of one row and one column; the smallest unit that holds data.
Cell address (reference) — the column letter followed by the row number, such as C5 or AB120. A cell on another sheet is written Sheet2!C5.
Range — a rectangular block of cells, written with a colon. A1:C10 is every cell from A1 to C10 — $3$ columns $\times$ $10$ rows $= 30$ cells.

Size of a worksheet.

$Rows: 1, 048, 576 (= 2^{20}) Columns: 16, 384 (= 2^{14})$

Columns are lettered A to Z, then AA to AZ, and so on up to XFD.

Also worth knowing: the active cell is the currently selected one, shown in the Name Box; and the formula bar shows what a cell actually contains, which may differ from what it displays.

AnswerWorkbook = the .xlsx file; worksheet = one grid within it; cell = a row–column intersection; cell address = column letter + row number (C5); range = a rectangular block (A1:C10). A worksheet has

1, 048, 576

rows and

16, 384

columns (A to XFD).

Problem 2Basic

A cell shows #####. What does this mean and how is it corrected? How does it differ from an error value?

Show solution

What it means. The column is simply too narrow to display the number in its current format. It is not an error of any kind: the value stored in the cell is complete and correct, and any formula referring to the cell gets the right answer. Excel shows hashes rather than a truncated number, because a number cut short would be worse than no number — 1234567 displayed as 123 would be actively misleading.

How to correct it. Any of:

Widen the column — drag the boundary in the column heading, or double-click it to autofit to the widest entry.
Reduce the decimal places shown, or choose a more compact number format.
Reduce the font size, or switch on Wrap Text / Shrink to Fit in Format Cells.

How it differs from an error value. Hashes filling the cell are a display problem. An error value is a computation problem, and the cell genuinely holds no usable result:

Shown	Meaning
`#####`	Column too narrow — value is fine
`#DIV/0!`	Division by zero or by an empty cell
`#VALUE!`	Wrong type of argument — arithmetic on text
`#REF!`	The referenced cell no longer exists (usually deleted)
`#NAME?`	A misspelled function name or an undefined name
`#N/A`	A lookup found no match
`#NULL!`	An invalid range operator
`#NUM!`	A numeric value that is impossible, e.g. the square root of a negative

The practical difference: widening the column fixes hashes and nothing else; an error value needs the formula or the data corrected, and it propagates — every formula depending on an error cell shows an error too.

AnswerIt means the column is too narrow to display the number; the stored value is unaffected. Widen the column, reduce decimals, or shrink the font. Unlike error values such as #DIV/0! or #VALUE!, it is a display issue and does not propagate to dependent formulas.

Problem 3Exam level

Explain relative, absolute and mixed cell references with an example of when each is needed.

Show solution

The three kinds differ in what happens to a reference when the formula is copied, and they are distinguished by the dollar signs. The dollar locks whatever follows it.

1. Relative reference — A1

No dollar signs. Both the column and the row adjust by the displacement of the copy. Moved one column right, A1 becomes B1; moved one row down, A2.

Needed when: every row must use its own data. Calculating =B2*C2 for the amount of an order in row 2, then filling down so that row 3 computes =B3*C3. This is the default, and correct most of the time.

2. Absolute reference — $A$1

Both parts locked. The reference is identical in every copy of the formula.

Needed when: one value stored in one cell must be consulted by every formula. A GST rate in E1, used as =B2*$E$1 and filled down; a currency conversion factor; a grand total used to compute each row's percentage share, =B2/$B$20.

An extra benefit: the constant exists in exactly one place, so changing it updates every dependent formula at once. Typing the number into each formula instead would mean editing every one.

3. Mixed reference — $A1 or A$1

One part locked, the other free.

$A1 — column fixed, row free. Copying across keeps pointing at column A; copying down moves through its rows.
A$1 — row fixed, column free. Copying down keeps pointing at row 1; copying across moves through its columns.

Needed when: a formula must be filled both across and down, taking one factor from a header row and the other from a header column. In a multiplication table with the multipliers in row 1 and the multiplicands in column A, the single formula

$=B$1*$A2$

fills the entire grid correctly. Neither a fully relative nor a fully absolute version can do this — the first drifts diagonally into nonsense, the second repeats one product in every cell.

Shortcut. Pressing F4 while the cursor is on a reference cycles A1 → $A$1 → A$1 → $A1 → A1.

AnswerRelative (A1) shifts with the copy — used so each row calculates from its own data. Absolute ($A$1) never changes — used for one constant consulted by many formulas, such as a tax rate. Mixed ($A1, A$1) locks one part — needed when a formula is filled both across and down from a header row and a header column.

Problem 4Exam level

Cell C3 contains =$B3+C$2-A1. State the formula after copying to (a) E3 (b) C6 (c) F8.

Show solution

First fix the rule for each reference:

$B3 — column locked at B, row free
C$2 — column free, row locked at 2
A1 — both free

(a) C3 → E3. Displacement: 2 columns right, 0 rows.

Reference	Column	Row	Result
`$B3`	locked	3 + 0 = 3	`$B3`
`C$2`	C + 2 = E	locked	`E$2`
`A1`	A + 2 = C	1 + 0 = 1	`C1`

$=$B3+E$2-C1$

(b) C3 → C6. Displacement: 0 columns, 3 rows down.

Reference	Column	Row	Result
`$B3`	locked	3 + 3 = 6	`$B6`
`C$2`	C + 0 = C	locked	`C$2`
`A1`	A + 0 = A	1 + 3 = 4	`A4`

$=$B6+C$2-A4$

(c) C3 → F8. Displacement: 3 columns right, 5 rows down.

Reference	Column	Row	Result
`$B3`	locked	3 + 5 = 8	`$B8`
`C$2`	C + 3 = F	locked	`F$2`
`A1`	A + 3 = D	1 + 5 = 6	`D6`

$=$B8+F$2-D6$

Method to use in the exam: write the displacement as (columns, rows) first, then treat each reference independently — add the column displacement unless a dollar precedes the letter, and the row displacement unless a dollar precedes the number.

Answer(a) =$B3+E$2-C1 (b) =$B6+C$2-A4 (c) =$B8+F$2-D6

Problem 5Challenge

A column of numbers is formatted to show no decimal places. Each cell displays a whole number, but SUM of the column shows a total one greater than adding the displayed numbers by hand. Explain, and say how it should be handled.

Show solution

The cause: formatting changes the display, not the value.

Reducing the decimal places shown is a format. The cell continues to hold its full value, and every formula uses that full value. So the sheet is doing arithmetic on numbers that are not the ones on the screen.

A concrete illustration. Suppose four cells hold

$10.4, 10.4, 10.4, 10.4$

Displayed with no decimals, each rounds to $10$ , so the column reads

$10, 10, 10, 10$

Adding what is displayed gives $40$ . But SUM works on the stored values:

$10.4 + 10.4 + 10.4 + 10.4 = 41.6$

which is then displayed as $42$ . The reader sees four tens totalling forty-two, and quite reasonably concludes the spreadsheet is wrong. It is not — the displayed figures are each rounded down by $0.4$ , and four such losses come to $1.6$ , enough to move the total by two.

$apparent error = \sum (rounding of each cell)$

The discrepancy therefore grows with the number of rows, which is why it is invisible in a short test and obvious in a real report.

How it should be handled. Which remedy is right depends on what is actually required.

1. If the underlying precision matters — show it. Display enough decimal places that the figures shown do add up. Honest and simplest, and correct for engineering or scientific data.

2. If the values themselves should be whole — round them, do not merely format them. Use the ROUND function to change the stored value:

=ROUND(A2,0)

Now the cell genuinely holds $10$ , and SUM over the rounded column gives exactly $40$ . Use ROUND(x,2) for two decimals. This is the correct approach for currency, where the paisa either exists or it does not, and a printed invoice must add up.

3. Round the total rather than the parts, if the parts are estimates and only the total is reported. Perfectly valid, but it must be documented, because the total will not equal the sum of the printed parts.

What not to do. Excel offers a setting called Set precision as displayed, which permanently discards the hidden decimals across the whole workbook. It appears to solve the problem, and it does — by destroying data irreversibly. There is no undo and no way to recover the original values.

The general lesson. Formatting is for the reader; functions are for the data. Any requirement expressed as "the values must be whole" or "amounts must be to two decimals" is a requirement about values, and must be met with ROUND — never with a display format.

AnswerFormatting only rounds the display; SUM uses the full stored values, so hidden decimals accumulate — four cells of

10.4

display as four tens but total

41.6

, shown as

42

. Fix it by showing enough decimals, or by rounding the values themselves with =ROUND(A2,0). Never use Set precision as displayed, which discards data irreversibly.

BCA1B02T · M4

Formulas and functions

Read the topic →

Example 1

Evaluating a formula by precedence

Question

Given A1 $= 4$ , A2 $= 6$ , A3 $= 2$ , evaluate (a) =A1+A2*A3 (b) =(A1+A2)*A3 (c) =A1+A2/A3^2 (d) =A1-A2-A3

Recall the order
$brackets \to power \land \to \times, \div \to +, -$

with equal precedence evaluated left to right.
(a) =A1+A2*A3
Multiplication comes before addition:

$4 + (6 \times 2) = 4 + 12 = 16$

Not $(4 + 6) \times 2 = 20$ .
(b) =(A1+A2)*A3
Brackets first:

$(4 + 6) \times 2 = 10 \times 2 = 20$
(c) =A1+A2/A3^2
Power first, then division, then addition:

$A 3^{2} = 2^{2} = 4$
$6 \div 4 = 1.5$
$4 + 1.5 = 5.5$
(d) =A1-A2-A3
Two subtractions have equal precedence, so left to right:

$(4 - 6) - 2 = - 2 - 2 = - 4$

Right to left would give $4 - (6 - 2) = 0$ , which is wrong. This is why left-to-right matters for $-$ and $\div$ , though not for $+$ and $\times$ .

Answer

(a) $16$ (b) $20$ (c) $5.5$ (d) $- 4$

NoteWhen in doubt, add brackets. They cost nothing and make the formula readable to whoever inherits the sheet.

Example 2

Grading with a nested IF

Question

Marks are in B2. Write one formula giving A for $\geq 90$ , B for $\geq 80$ , C for $\geq 70$ , D for $\geq 60$ and F below that.

Understand how a nested IF is read
IF(test, value_if_true, value_if_false). To test more than one condition, the false slot holds another IF. Excel stops at the first test that is true, so the order of the tests does the work.
Test from the highest boundary downwards
$=IF(B2>=90,"A",IF(B2>=80,"B",IF(B2>=70,"C",IF(B2>=60,"D","F"))))$

Note there is no need to write AND(B2>=80,B2<90). If the formula reaches the second test at all, the first has already failed, so B2 is known to be below $90$ .
3
Trace a value — B2 = 85
- B2>=90? $85 \geq 90$ is FALSE → go to the false slot
- B2>=80? $85 \geq 80$ is TRUE → return "B" and stop
Nothing further is evaluated.
Trace a boundary value — B2 = 90
B2>=90? $90 \geq 90$ is TRUE → "A".

Boundaries are where grading formulas go wrong. Using > instead of >= would give a student with exactly $90$ a B. Always test the exact boundary values.
See why the order cannot be reversed
Written from the bottom up:

$=IF(B2>=60,"D",IF(B2>=70,"C",...))$

a mark of $95$ satisfies the very first test, so it returns "D" and stops. Every mark above $60$ would be graded D. Descending order is not a style preference; it is required.
Note the alternatives
=IFS(B2>=90,"A",B2>=80,"B",B2>=70,"C",B2>=60,"D",TRUE,"F") is easier to read.

For many bands, a lookup table with VLOOKUP and approximate match is better still, because the boundaries then live in cells that can be changed without rewriting a formula.

Answer

=IF(B2>=90,"A",IF(B2>=80,"B",IF(B2>=70,"C",IF(B2>=60,"D","F"))))

NoteTest boundaries from the top down, and check each boundary value itself — 90, 80, 70, 60 — before trusting the formula.

Example 3

Counting and totalling under a condition

Question

A sheet holds Department in column B, Marks in column C and Fee in column D, for rows 2 to 101. Write formulas for: (a) how many students are in BCA, (b) how many scored above 400, (c) the total fee from BCA students, (d) the average mark of BCA students, (e) how many BCA students scored above 400.

(a) Counting on a text condition
$=COUNTIF(B2:B101,"BCA")$

The criterion is quoted because it is text. It could equally be a cell reference — =COUNTIF(B2:B101,F1) — which is better practice, since the department can then be changed without editing the formula.
(b) Counting on a numeric comparison
$=COUNTIF(C2:C101,">400")$

The comparison is written inside quotation marks — the whole expression ">400" is the criterion. =COUNTIF(C2:C101,>400) is a syntax error.
(c) Totalling a different column from the one tested
$=SUMIF(B2:B101,"BCA",D2:D101)$

Read as: examine B, and where it says BCA, add the corresponding cell from D. The three arguments are where to look, what to look for, what to add — and the first and third ranges must be the same height, or the results silently misalign.
(d) Averaging under a condition
$=AVERAGEIF(B2:B101,"BCA",C2:C101)$

Note that AVERAGEIF ignores blank cells in the average range, so a missing mark reduces the divisor rather than counting as zero.
(e) Two conditions at once
One condition needs COUNTIF; two need COUNTIFS, in which the ranges and criteria alternate:

$=COUNTIFS(B2:B101,"BCA",C2:C101,">400")$

Only rows satisfying both are counted. SUMIFS works the same way, except that the range to be added comes first rather than last — the one inconsistency in this family, and worth remembering.
A caution on matching text
COUNTIF is not case sensitive, so "BCA" and "bca" both match — usually convenient. But a trailing space in the data ("BCA ") will not match "BCA", and nothing on screen reveals the difference. Cleaning imported data with TRIM prevents an entire class of undercounting.

Answer

(a) =COUNTIF(B2:B101,"BCA") (b) =COUNTIF(C2:C101,">400") (c) =SUMIF(B2:B101,"BCA",D2:D101) (d) =AVERAGEIF(B2:B101,"BCA",C2:C101) (e) =COUNTIFS(B2:B101,"BCA",C2:C101,">400")

Example 4

Looking up a price from a rate list

Question

A rate list occupies H2:J20 — item code, description, price. Cell B5 holds an item code. Write a formula that returns its price, and shows "Not listed" if the code is absent.

Check the leftmost-column requirement
VLOOKUP searches only the first column of the table it is given. The item code is in column H, which is the first column of H2:J20, so VLOOKUP can be used.
Count the column to return
Columns are numbered within the table, not on the sheet:

$H = 1, I = 2, J = 3$

The price is in J, so the column index is $3$ .
Write the lookup with an exact match
$=VLOOKUP(B5,$H$2:$J$20,3,FALSE)$

FALSE demands an exact match. Omitting it defaults to TRUE, which performs an approximate match and can return the price of a completely different item without any warning — the most dangerous default in the whole application.
Note why the table reference is absolute
$H$2:$J$20 is locked so the formula can be filled down a column of item codes while the rate list stays put. Left relative, the table would drift downward and rows would fall off the bottom of it.
Handle the missing code
An unmatched code returns #N/A, which then propagates into any total. Wrap the lookup:

$=IFERROR(VLOOKUP(B5,$H$2:$J$20,3,FALSE),"Not listed")$
6
If the lookup still fails on a code that is visibly present
Two usual causes, neither visible on screen:
1. Trailing spaces — "A101 " is not "A101". Fix with TRIM.
2. Number stored as text — a code that is a number in one place and text in the other never matches. The left-aligned entry in a right-aligned column is the giveaway.

Answer

=IFERROR(VLOOKUP(B5,$H$2:$J$20,3,FALSE),"Not listed")

NoteAlways pass FALSE explicitly. An approximate match on unsorted data returns wrong answers rather than errors, which is far harder to notice.

Problem 1Basic

Distinguish between a formula and a function. Write one formula and one function that both total A1 to A10.

Show solution

	Formula	Function
What it is	Any calculation the user writes	A ready-made calculation supplied by Excel, called by name
Begins with	`=`	`=` then the function name
Written by	The user, from operators and references	The user, but the logic is built in
Length	Grows with the task	Fixed, however large the range
Example	`=A1+A2+A3`	`=SUM(A1:A3)`

A function is used inside a formula, so the two are not alternatives: =SUM(A1:A10)*0.18 is a formula that contains a function.

Totalling A1 to A10.

As a formula:

=A1+A2+A3+A4+A5+A6+A7+A8+A9+A10

As a function:

=SUM(A1:A10)

Why the function is better. It is shorter and readable; it adjusts automatically when rows are inserted inside the range, whereas the long formula silently omits the new row; it ignores text and blanks instead of failing; and it scales — the same nine characters total ten cells or ten thousand.

AnswerA formula is any user-written calculation beginning with =; a function is a built-in named calculation used within a formula. =A1+A2+…+A10 against =SUM(A1:A10) — the function is shorter, adjusts when rows are inserted, and scales.

Problem 2Basic

State the difference between COUNT, COUNTA and COUNTBLANK. For the range A1:A6 holding $10$ , "Pass", empty, $25$ , empty, "Fail", give each result.

Show solution

COUNT counts cells containing numbers only. Text, blanks and logical values are ignored.
COUNTA counts cells that are not empty — numbers, text, dates, logical values, even a formula returning an empty string.
COUNTBLANK counts cells that are empty.

Applying them to the range.

Cell	Contents	COUNT	COUNTA	COUNTBLANK
A1	$10$	✓	✓
A2	"Pass"		✓
A3	(empty)			✓
A4	$25$	✓	✓
A5	(empty)			✓
A6	"Fail"		✓

$=COUNT(A1:A6) = 2 =COUNTA(A1:A6) = 4 =COUNTBLANK(A1:A6) = 2$

Note that COUNTA and COUNTBLANK together account for every cell in the range:

$4 + 2 = 6 ✓$

The examinable trap. To count students in a column of names, COUNT returns $0$ , because names are not numbers. COUNTA is the correct function. Conversely, to count how many students actually sat an examination from a column of marks with blanks for absentees, COUNT is correct — it excludes the blanks automatically.

AnswerCOUNT counts numbers (

= 2

); COUNTA counts non-empty cells of any type (

= 4

); COUNTBLANK counts empty cells (

= 2

Problem 3Exam level

Explain the IF function. Write formulas for: (a) "Pass" if B2 is at least $40$ , else "Fail"; (b) "Pass" only if both B2 and C2 are at least $40$ ; (c) a grade of A, B, C or F using bands of $80$ , $60$ and $40$ .

Show solution

The IF function tests a condition and returns one of two values.

$=IF(logical_test, value_if_true, value_if_false)$

logical_test — any expression giving TRUE or FALSE, such as B2>=40
value_if_true — returned when the test succeeds
value_if_false — returned when it fails

Text results must be in quotation marks; numbers must not.

(a) One condition.

=IF(B2>=40,"Pass","Fail")

Note >=, not > — a student with exactly $40$ passes.

(b) Two conditions together. Combine with AND, which is TRUE only when every condition is TRUE:

=IF(AND(B2>=40,C2>=40),"Pass","Fail")

Use OR instead if passing either subject would suffice.

(c) Four bands — a nested IF. The false slot holds another IF, and the tests must run from the highest boundary downwards:

=IF(B2>=80,"A",IF(B2>=60,"B",IF(B2>=40,"C","F")))

Tracing $B 2 = 65$ : $65 \geq 80$ FALSE → $65 \geq 60$ TRUE → returns "B" and stops.

Why the order matters. Written from the bottom up — =IF(B2>=40,"C",IF(B2>=80,"A",…)) — a mark of $95$ satisfies the very first test and returns "C". Every mark above $40$ would be graded C. Descending order is required, not preferred.

Why no upper bound is needed. Once the first test has failed, the value is known to be below $80$ , so AND(B2>=60,B2<80) adds nothing.

A cleaner alternative for many bands is IFS:

=IFS(B2>=80,"A",B2>=60,"B",B2>=40,"C",TRUE,"F")

Answer(a) =IF(B2>=40,"Pass","Fail") (b) =IF(AND(B2>=40,C2>=40),"Pass","Fail") (c) =IF(B2>=80,"A",IF(B2>=60,"B",IF(B2>=40,"C","F"))) — nested IFs must test from the highest boundary down.

Problem 4Exam level

Explain VLOOKUP and its four arguments. What is the difference between TRUE and FALSE as the last argument, and why is FALSE usually safer?

Show solution

VLOOKUP — vertical lookup — searches for a value down the first column of a table and returns a value from the same row of another column. It is how one sheet fetches data held in another, replacing manual copying.

$=VLOOKUP(lookup_value, table_array, col_index_num, [range_lookup])$

1. lookup_value — what to search for. A value, a cell reference, or text in quotation marks.

2. table_array — the range to search. Its first column must be the one containing the lookup value; VLOOKUP cannot look leftward. Normally made absolute ($H$2:$J$20) so the formula can be filled down without the table drifting.

3. col_index_num — which column of the table to return, counted within the table starting at $1$ for its first column. If the table is H2:J20, then H is $1$ , I is $2$ , J is $3$ — not $8$ , $9$ , $10$ .

4. range_lookup — the kind of match, and the argument that matters most.

TRUE against FALSE.

	`FALSE` (or $0$ )	`TRUE` (or omitted)
Match	Exact	Approximate — the largest value not exceeding the lookup value
Data must be sorted?	No	Yes, ascending, by the first column
If not found	Returns `#N/A`	Returns the nearest lower entry
Used for	Codes, names, IDs, item numbers	Banded lookups: grade boundaries, tax slabs, discount tiers

Why FALSE is usually safer. Two reasons, and the second is the serious one.

Most lookups are of identifiers, where an "approximate" item code or roll number is meaningless. There is no such thing as nearly the right student.
An approximate match on unsorted data fails silently. It does not report an error; it returns a value — just the wrong one. VLOOKUP with TRUE assumes ascending order and effectively stops as soon as it passes the target, so on unsorted data it returns whatever happened to be there. A plausible-looking wrong price is far more dangerous than a visible #N/A, because nobody investigates it.

Since TRUE is the default when the argument is omitted, the fourth argument should always be written explicitly.

Handling a genuine miss:

=IFERROR(VLOOKUP(B5,$H$2:$J$20,3,FALSE),"Not found")

When a lookup fails on a value that is visibly present, the cause is almost always invisible: a trailing space (fix with TRIM) or a number stored as text in one of the two places.

AnswerVLOOKUP searches the first column of a table and returns a value from column

n

of the matching row; arguments are the lookup value, the table (usually absolute), the column index counted within the table, and the match type. FALSE demands an exact match; TRUE gives an approximate match needing ascending sorted data. FALSE is safer because an approximate match on unsorted data returns a plausible wrong value instead of an error.

Problem 5Exam level

Explain any six error values in Excel, with a cause and a remedy for each.

Show solution

1. #DIV/0! — division by zero, or by a cell that is empty (an empty cell counts as zero in arithmetic).
Common case: =B2/C2 where C2 has not yet been filled in.
Remedy: guard the formula — =IF(C2=0,"",B2/C2) or =IFERROR(B2/C2,"").

2. #VALUE! — an argument of the wrong type, typically arithmetic on text.
Common case: =A1*2 where A1 contains "10 kg", or a number imported as text.
Remedy: find the text entry — it will be left-aligned in a column of right-aligned numbers — and correct it, or clean it with VALUE or TRIM.

3. #REF! — the reference points to a cell that no longer exists, because its row, column or worksheet was deleted.
Common case: deleting a column that a formula elsewhere depended on.
Remedy: the original address is unrecoverable, so the reference must be rewritten. Undoing the deletion immediately restores it.

4. #NAME? — Excel does not recognise a name in the formula: a misspelled function (=SUME(A1:A5)), text left unquoted (=IF(A1>5,Pass,Fail)), or an undefined range name.
Remedy: correct the spelling, and put quotation marks round text. Choosing the function from the drop-down list as you type prevents this entirely.

5. #N/A — "not available": a lookup found no match.
Common case: VLOOKUP for a code absent from the table, or present but with a trailing space, or stored as text on one side and as a number on the other.
Remedy: verify the value exists, clean it with TRIM, and wrap the lookup in IFERROR for the genuinely missing cases.

6. #NUM! — a numeric value that is impossible or out of range.
Common case: =SQRT(-4), or an iterative function that fails to converge.
Remedy: check the argument; guard with IF(A1>=0,SQRT(A1),"").

Also: #NULL! — an invalid range operator, usually a space typed instead of a colon (=SUM(A1 A5)); and #####, which is not an error at all but merely a column too narrow to display its number.

Two general points. Errors propagate — any formula referring to an error cell shows an error too, so one bad cell can spoil an entire report; and Formulas → Error Checking, with Trace Precedents and Evaluate Formula, is the quickest way to find the cell where the trouble started.

Answer#DIV/0! (divide by zero or empty cell); #VALUE! (wrong type, arithmetic on text); #REF! (referenced cell deleted); #NAME? (misspelled function or unquoted text); #N/A (lookup found no match); #NUM! (impossible number). Errors propagate to dependent formulas; #####, by contrast, is only a narrow column.

Problem 6Challenge

A column of ten marks has two blank cells for absent students. AVERAGE gives $62$ but the teacher expects $49.6$ . Explain the discrepancy and give the formula for each interpretation.

Show solution

The cause: AVERAGE ignores blank cells entirely — it does not treat them as zero.

Let the eight marks that were entered total $496$ .

What AVERAGE computes. It sees eight numeric cells and divides by eight:

$\frac{496}{8} = 62$

What the teacher expected. That an absentee scores zero, so the divisor should be the full ten students:

$\frac{496 + 0 + 0}{10} = \frac{496}{10} = 49.6$

Both figures are arithmetically correct. They answer different questions, and the spreadsheet cannot know which was meant:

$62$ is the average mark of those who sat the examination
$49.6$ is the average over the whole class, counting an absentee as zero

The formula for each.

1. Average of those who appeared — the default behaviour, and correct if a blank means "no data":

=AVERAGE(B2:B11)

2. Average over all ten students, absentees counting as zero. Divide the total by the full count rather than by the count of numbers:

=SUM(B2:B11)/COUNTA(A2:A11)

using the names column, which has ten entries, for the divisor. Or, if the class size is fixed:

=SUM(B2:B11)/10

Note that =SUM(B2:B11)/COUNT(B2:B11) would not work — COUNT also ignores blanks and returns eight, reproducing the $62$ .

Which is right, and how to make the sheet say so. For most academic purposes the two are genuinely different statistics and both are worth reporting. Ambiguity is the real fault here, and it should be removed from the data rather than patched in the formula:

If an absentee's mark is zero, enter $0$ . Then AVERAGE gives $49.6$ by itself, and every other function agrees.
If the mark is genuinely unknown — the student will sit a supplementary paper — leave it blank, and label the column heading "Average (present only)" so no reader is misled.
Use a distinct marker such as "AB" if absence must be visible on the sheet. Being text, it is ignored by AVERAGE exactly as a blank is, but a reader can see why.

The general lesson. A blank cell is an absence of data, not a zero, and Excel's functions treat it that way consistently. Whenever a blank actually means something — zero, not applicable, not yet known, absent — that meaning must be recorded explicitly, because no formula can recover an intention that was never written down.

AnswerAVERAGE ignores blanks, so it divided the total

496

by the

8

marks present, giving

62

; the teacher wanted division by all

10

, giving

49.6

. Use =AVERAGE(B2:B11) for the present-only mean, or =SUM(B2:B11)/COUNTA(A2:A11) to count absentees as zero — and better, enter an explicit

0

when zero is what is meant.

BCA1B02T · M4

Charts, sorting and data tools

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Problem 1Basic

Name and explain six parts of a chart.

Show solution

Chart area — the entire chart object, including titles and legend. Selecting it moves or resizes the whole chart.
Plot area — the rectangle inside which the data is actually drawn, bounded by the axes.
Data series — one complete set of related values, usually a single row or column of the worksheet. A chart may hold several series, each in its own colour.
Data point — one individual value within a series: one column, one slice, one marker on a line.
Category (X) axis — the horizontal axis, carrying the labels of the things being compared (subjects, months, regions).
Value (Y) axis — the vertical axis, carrying the numeric scale against which the data points are measured.

Others worth naming: the legend, which says which colour belongs to which series; the chart title and axis titles, which state what is shown and in what units; data labels, printing the value on each point; gridlines, helping the eye read a value off the scale; and a trendline, a fitted line showing the overall direction.

On single-series charts, the legend should be removed — it merely repeats the title.

AnswerChart area, plot area, data series, data point, category (X) axis and value (Y) axis; also legend, titles, data labels, gridlines and trendline.

Problem 2Basic

Differentiate between sorting and filtering. Why must the whole table be selected before sorting?

Show solution

	Sorting	Filtering
Effect	Rearranges the rows into order	Hides rows that do not meet a condition
All rows still present?	Yes, in a new order	Yes, but some are hidden from view
Changes the data?	The arrangement, permanently	Nothing at all
Reversible?	Only by sorting again or undoing	Yes — Clear Filter restores every row
Based on	Values, colour, icon, or a custom list	A condition on one or more columns
Command	Data → Sort	Data → Filter (Ctrl+Shift+L)

The two are often used together: filter to the rows of interest, then sort those into a useful order.

Why the whole table must be selected before sorting.

Sorting rearranges rows. For the data to stay meaningful, every column must be rearranged together, so that each row's values remain attached to the same record.

If only the Marks column is selected and sorted, Excel reorders only that column. The names, roll numbers and fees stay exactly where they were, so the highest mark ends up beside whichever name happened to be in the first row. Every student now has someone else's mark.

The damage is undetectable afterwards: the sheet looks perfectly normal, every name is present and every mark is present, and nothing indicates that they have been paired wrongly. There is no way to reconstruct the original pairing from the file.

In practice: click one cell inside the list and let Excel detect the whole block, or select the entire table including headings. Excel usually offers to expand the selection — that warning must be read rather than dismissed. Before a large sort, copy the sheet.

AnswerSorting rearranges rows into order; filtering merely hides rows failing a condition and is fully reversible. The whole table must be selected because sorting one column alone reorders that column only, permanently attaching each value to the wrong record — damage that is undetectable afterwards.

Problem 3Exam level

Explain any five types of chart and state a situation suited to each. Why is a pie chart unsuitable for showing marks in five subjects?

Show solution

1. Column chart. Vertical bars whose heights represent values, for comparing separate categories.
Suited to: marks by subject, sales by branch, population by district — anything where distinct items are compared.

2. Line chart. Points joined by a line, showing how a value changes along a continuous scale.
Suited to: monthly sales over a year, daily temperature, share price — the category axis must be something genuinely ordered, normally time.

3. Pie chart. A circle divided into slices, each proportional to its share of the total.
Suited to: the composition of one whole — how a monthly budget divides between rent, food, travel and savings; the share of votes each party received.

4. XY (scatter) chart. Points plotted against two numeric axes, showing whether two quantities are related.
Suited to: hours studied against marks obtained; height against weight; advertising spend against sales.

5. Stacked column chart. Columns divided into segments, showing both the total for each category and its composition.
Suited to: total marks per student split by subject; monthly revenue split by product line.

Also: bar charts (a column chart turned horizontal, for long category names); area charts; combo charts (columns and a line together, for two different scales); doughnut and radar.

Why a pie chart is wrong for marks in five subjects.

A pie chart makes one claim about the data: these values are parts of a single whole, and their sum is meaningful. That claim is false here.

The total means nothing. Marks in Mathematics, English and Computer Science do not combine into any real quantity. A slice reading "Mathematics 28%" invites the reader to think mathematics is 28% of something — but 28% of an arbitrary sum is not a fact about the student.
The question being asked is comparison, not composition. A reader wants to know which subject is strongest and by how much. Judging that from the angles of slices is much harder than from the lengths of columns — an established finding about how people read charts, and the reason a column chart is right.
A pie hides the scale entirely. Marks of $80, 70, 60, 50, 40$ and marks of $40, 35, 30, 25, 20$ produce identical pie charts, because the proportions are the same. A pie chart cannot distinguish an excellent student from a failing one — it discards exactly the information that matters.

The correct choice is a column chart, with subjects on the category axis, a value axis running from $0$ to the maximum mark, and data labels. A pie chart would only be right if the question were genuinely about composition — for instance how a student's 168 hours in a week divide between classes, study, sleep and travel.

AnswerColumn (comparing categories), line (change over time), pie (composition of one whole), scatter (relation between two numeric variables), stacked column (total plus composition). A pie is wrong for marks because the total is meaningless, angles are harder to compare than lengths, and identical proportions at different mark levels produce identical pies — hiding the very information that matters.

Problem 4Exam level

What is conditional formatting? Describe four of its types and one use of each.

Show solution

Conditional formatting applies formatting to a cell only when a stated condition is true, and removes it when the value changes. It makes a sheet indicate its own exceptions, so a reader is not obliged to check every number.

It differs from ordinary formatting in being dynamic: colouring a failing mark red by hand is wrong the moment the mark is corrected, whereas a conditional rule re-evaluates itself.

1. Highlight Cells Rules. Formats cells meeting a comparison — greater than, less than, between, equal to, text containing, a date occurring, or duplicate values.
Use: colour every mark below $40$ red, so failures are visible at a glance in a list of 200 students. Or highlight duplicate roll numbers, which finds data-entry errors that no formula would reveal.

2. Data Bars. Draws a bar inside each cell, its length proportional to the value — a small chart within the column.
Use: a sales column where relative sizes must be seen while the exact figures remain readable. It gives the comparative power of a chart without leaving the table.

3. Colour Scales. Applies a two- or three-colour gradient across a range, so each cell's colour indicates its position between the lowest and highest values — a heat map.
Use: a grid of monthly sales by region, where the eye immediately finds the strong and weak combinations without reading a single number.

4. Icon Sets. Places a small symbol in each cell — traffic lights, arrows, flags — according to which band the value falls in.
Use: a project tracker where green, amber and red show progress against target; or up, level and down arrows against last month's figures.

5. Formula-based rules deserve mention as the most powerful form. Choosing Use a formula to determine which cells to format allows any condition, including one that formats a whole row based on one column:

=$D2<40

applied to the range A2:E101 colours the entire row of any student failing. The mixed reference $D2 is essential — the column is locked so every cell in the row consults column D, while the row is free so each row tests its own value.

Managing rules. Conditional Formatting → Manage Rules lists them in priority order, since several may apply to one cell, and Stop If True prevents lower rules being applied once one has matched.

AnswerFormatting applied only while a condition holds, and re-evaluated automatically. Highlight Cells Rules (mark failures below 40, or find duplicate roll numbers); Data Bars (in-cell bars comparing sales); Colour Scales (a heat map of sales by region and month); Icon Sets (traffic lights on a progress tracker); and formula rules such as =$D2<40 to colour a whole row.

Problem 5Challenge

What is a pivot table? Explain its four areas and give an example of what it produces from a sales list. Why is it better than writing SUMIF formulas?

Show solution

A pivot table is an interactive summary that cross-tabulates a large list, grouping records and aggregating a numeric field — without writing a single formula and without altering the source data.

Insert → PivotTable, then drag field names into four areas.

Area	Effect
Rows	One row of the summary for each distinct value of the field
Columns	One column for each distinct value
Values	The field to be aggregated — Sum, Count, Average, Max, Min, or a percentage of the total
Filters	Restricts the whole table to selected values

An example. Take $5, 000$ sales records with the fields Date, Region, Salesperson, Product, Quantity, Amount.

Drag Region to Rows, Month to Columns and Amount to Values as Sum. The result is a complete region-by-month sales table:

Region	Jan	Feb	Mar	Total
North	45,000	52,000	48,000	145,000
South	38,000	41,000	55,000	134,000
East	29,000	33,000	31,000	93,000
Total	112,000	126,000	134,000	372,000

Now drag Product into Rows in place of Region, and the same 5,000 records are re-summarised by product instantly. Add Salesperson beneath Region and the table nests one inside the other. That rearrangement is the pivot.

Why it beats SUMIF formulas.

1. No formulas to write, and none to get wrong. The table above would need $3 \times 3$ SUMIFS formulas plus row and column totals — around sixteen formulas, each with absolute references to get right. The pivot table needs three drags.

2. It discovers the categories itself. SUMIFS requires you to know in advance that the regions are North, South and East, and to have typed those labels as row headings. A pivot table reads the distinct values out of the data. If a fourth region appears next month, the pivot table shows it on refresh; the SUMIFS grid silently omits it, and its sales vanish from a total that still looks complete.

3. Rearranging costs nothing. Summarising by product instead of by region means one drag. With formulas it means rebuilding the entire grid.

4. It scales. The same three drags handle 5,000 rows or 500,000. Sixteen formulas over half a million rows recalculate slowly, and a hundred such grids make a workbook unusable.

5. Extras come free: row and column totals, percentage of total, grouping dates into months, quarters and years, expanding a cell to see the underlying records (drill down), slicers, and pivot charts.

The one caution. A pivot table does not update itself when the source data changes — it must be refreshed (right-click → Refresh). SUMIF formulas recalculate at once. This is the single genuine advantage of formulas, and the reason a live dashboard cell is often better as a formula while analysis is better as a pivot table.

When formulas are still right: a handful of specific figures embedded in a report, values that must recalculate instantly, or a result that feeds further calculations. When a pivot table is right: exploring a large list, or any summary whose shape may need to change.

AnswerAn interactive cross-tabulation of a list. Rows and Columns give the two grouping fields, Values the aggregated number, Filters restricts the whole table — e.g. Region × Month summing Amount from 5,000 records. Better than SUMIF because it needs no formulas, discovers the categories itself (so a new region is never silently omitted), rearranges with one drag, and scales — but it must be refreshed, whereas formulas recalculate at once.

BCA1B02T · M5

Presentations with PowerPoint

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Problem 1Basic

Distinguish between animation and transition, with an example of each.

Show solution

$Transition = between slides Animation = within a slide$

	Transition	Animation
Applies to	A whole slide, as it replaces the previous one	An individual object on a slide — text, picture, chart, shape
When it happens	At the change of slide	While the slide is displayed
How many per slide	One	Any number, in a set order
Ribbon tab	Transitions	Animations
Types	Fade, Push, Wipe, Morph, Cut	Entrance, Emphasis, Exit, Motion Path
Example	Slide 3 fades into slide 4	The four bullets on slide 3 appear one at a time as the speaker reaches each

The four kinds of animation. Entrance — how an object arrives (Appear, Fade, Fly In). Emphasis — drawing attention to an object already visible (Pulse, Grow). Exit — how it leaves. Motion Path — moving it along a drawn route.

Controlling them. The Animation Pane lists a slide's animations in order, each set to start On Click, With Previous or After Previous. Transitions have a duration and advance either On Mouse Click or After a set time, and Apply To All gives the whole deck the same transition.

The one legitimate use worth naming: revealing a bulleted list one point at a time, so the audience attends to the point being spoken rather than reading ahead.

AnswerA transition is the effect between slides (slide 3 fading into slide 4); an animation is the effect applied to an object within a slide (bullets appearing one at a time). Animations come in four kinds — entrance, emphasis, exit and motion path.

Problem 2Basic

Name and explain any five views in PowerPoint.

Show solution

1. Normal view. The default working view: the current slide shown large for editing, a thumbnail strip of all slides at the side, and a notes pane beneath. Nearly all editing happens here.

2. Outline view. Shows only the text of the presentation, as a hierarchy of titles and bullets, with no design. The fastest way to draft content, reorder whole slides and promote or demote points, because attention is on what will be said rather than on how it looks. Typing a title and pressing Tab creates its bullets; Shift+Tab starts a new slide.

3. Slide Sorter view. Every slide as a thumbnail on one screen. Used for reordering by dragging, duplicating, deleting, hiding slides and applying transitions to several at once. It is also where the overall balance of the deck becomes visible — three slides of dense text in a row show up immediately.

4. Notes Page view. Each slide printed small at the top of a page with the speaker's notes below. Used for preparing and printing what the speaker will hold, as opposed to what the audience will see.

5. Slide Show view (F5). The actual presentation, full screen, with transitions and animations running. Shift+F5 starts from the current slide rather than the first.

Also: Presenter View. On a machine with two displays, the audience sees only the slide while the speaker sees the current slide, the next slide, the notes, a timer and the annotation tools. This is what makes the notes pane worth filling in.

Reading View runs the presentation full screen but inside a window, for checking it without leaving the application.

AnswerNormal (editing), Outline (text hierarchy, for drafting and reordering), Slide Sorter (thumbnails, for reordering and transitions), Notes Page (slide plus speaker's notes) and Slide Show / Presenter View (the live presentation).

Problem 3Exam level

What is a slide master? Why is it better than formatting each slide individually?

Show solution

A slide master is the top-level slide that stores the design of the entire presentation — background, theme colours, heading and body fonts and their sizes, the position and size of every placeholder, bullet styles, and anything that should appear on every slide such as a logo, footer, date or slide number.

Reached by View → Slide Master. Beneath the master sit layout masters, one for each layout (Title Slide, Title and Content, Two Content), so a change can be made to every slide at once or to every slide of one layout.

Why it beats formatting slides individually.

1. One change updates the whole deck. Deciding that headings should be dark blue at $36$ pt means one edit on the master. Done slide by slide across forty slides it is forty edits, and one will be missed — leaving a single slide that visibly does not match.

2. Guaranteed consistency. Because every slide inherits its placeholder positions from the master, headings sit at exactly the same height on every slide. Text boxes drawn by hand differ by a few millimetres each time, so headings appear to jump as the slides advance. An audience notices this even without knowing why the deck looks unprofessional.

3. Elements appear automatically. A logo, footer or slide-number field placed on the master appears on every slide, including slides added later. Placed by hand it must be copied to each new slide, and will be forgotten.

4. Speed. New slides arrive already formatted. There is nothing to style, so the work goes into content.

5. Reuse. A master saved as a template (.potx) gives every presentation in a college or company the same appearance without anyone remembering the specification.

6. A safe route to changing the design. Applying a different theme restyles the whole presentation, because the content sits in placeholders that the theme controls. Content in hand-drawn text boxes ignores the theme and has to be reformatted by hand.

The underlying principle, the same one as styles in a word processor: put the content in the structure the software provides, and control the appearance in one place. Formatting each slide separately records how forty slides happen to look; a master records what the presentation is — and stays correct when it changes.

AnswerThe slide that stores the design of the whole presentation — background, fonts, colours, placeholder positions and repeating elements such as a logo and slide number. It gives one-edit updates across the deck, guaranteed consistency of placeholder positions, automatic repeating elements, faster new slides, reuse as a template, and safe theme changes.

Problem 4Exam level

List and explain the guidelines for designing an effective presentation.

Show solution

1. A slide is a signpost, not a document. Roughly six lines of six words. The slide carries the headings of what is being said; the speaker supplies the substance. Anything needing a paragraph belongs in the speaker's notes or a handout.

2. Font size at least $28$ pt for body text, larger for headings. This is not a style preference but a limit: smaller text cannot be read from the back of a room. If text must be shrunk to fit, the answer is less text or another slide.

3. Contrast decides legibility. Dark text on a light background, or light on dark. Mid-tone on mid-tone is unreadable when projected, however acceptable it looks on a laptop.

4. Prefer sans-serif faces. They hold their shape better than serif faces when projected at a distance.

5. One idea per slide. Two ideas need two slides. Slides cost nothing; a confused audience costs a great deal.

6. Prefer a picture to a table, and a chart to a paragraph. A column chart is understood in a second; the same figures in a table take half a minute; described in prose, they are not understood at all.

7. Be consistent, using the slide master. Headings in the same position, the same fonts and the same colours throughout. Inconsistency is the most visible mark of a hurried deck.

8. Use animation only when it has a reason. Revealing bullets in step with the speaker is a reason. Decoration is not. Excess animation lengthens the delivery and distracts from the content.

9. One transition, applied to all. A different effect on each slide looks accidental.

10. Number the slides, so a question can refer to one.

11. Never read the slides aloud. If the slide says everything the speaker is redundant; if the speaker says everything the slide should be shorter. The slide and the speech should complement one another.

12. Structure the deck. Title slide, then an agenda, then the content in sections, then a summary, then an invitation for questions. An audience that knows where it is follows better.

13. Rehearse with timings, and know how long the talk actually runs.

14. Check it on the real projector. Colours, contrast and aspect ratio all differ from a laptop screen. A presentation is judged on the wall.

15. Have a fallback. A PDF on a pen drive and a printed handout, in case the machine, the file or the projector fails.

The principle behind all of them: the presentation is the speaker, supported by slides. Every rule above follows from asking whether a choice helps the audience follow the person talking.

AnswerSix-by-six lines of text; at least 28 pt; strong contrast and sans-serif faces; one idea per slide; charts rather than tables or prose; consistency through the slide master; animation only with a purpose and one transition throughout; numbered slides; never read the slides aloud; a clear structure; rehearsed timings; a check on the real projector; and a fallback copy.

BCA1B02T · M5

The Internet, the Web and email

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Example 1

Reading a URL

Question

Identify every part of https://www.uoc.ac.in/academics/bca/syllabus.pdf and say what each tells you.

The protocol — https://
The rules by which the page will be fetched. HTTPS is HTTP with SSL/TLS encryption, so the traffic cannot be read in transit.

Its presence is a security fact, not a formality: on plain http:// anything typed into the page travels readable across every network on the route.
The subdomain — www
A name for a particular host within the domain. Conventionally www for a web server, but any label may be used — mail., exam., library. — and each may point to a different machine.
3
The domain name — uoc.ac.in
The registered name of the organisation, read right to left in order of generality:
- .in — the country code top-level domain, India
- .ac — the second-level domain for academic institutions in India
- uoc — the organisation itself
So the address is an Indian academic institution before a single character of the page has loaded. That is a genuine reason to trust a syllabus found here over one on an anonymous .com.

DNS translates this name into the server's numeric IP address; the network itself routes only by number.
The path — /academics/bca/
The directory structure on the server leading to the file, exactly like a folder path on a disk. Paths are often case sensitive on the server, unlike Windows file names.
The file — syllabus.pdf
The resource requested, and its extension says what will arrive: a PDF, which the browser will display or download rather than render as a page.
Putting it together
$protocol https:// host www. domain uoc.ac.in path /academics/bca/ resource syllabus.pdf$

A URL may also carry a query string after a ? (parameters sent to the server) and a fragment after a # (a position within the page).

Answer

Protocol https://; host www; domain uoc.ac.in (.in India, .ac academic, uoc the institution); path /academics/bca/; resource syllabus.pdf.

NoteRead a domain right to left. It is the quickest way to judge who is really publishing a page — and the trick phishing links rely on you not doing.

Example 2

Spotting a phishing message

Question

An email reads: "Dear Customer, Your account will be blocked in 24 hours. Click here to verify: http://sbi-verify-login.com/secure. Reply with your OTP to confirm." Identify every warning sign.

A generic greeting
"Dear Customer" — a bank that holds your account holds your name and uses it. Bulk fraud cannot, because the same message is being sent to thousands of addresses harvested indiscriminately.
Manufactured urgency and threat
"blocked in 24 hours" exists to stop you thinking. A deadline discourages the one action that would defeat the fraud — pausing to check with the bank through its own channel. Genuine notices give reasonable notice and do not threaten.
The domain is wrong
Read sbi-verify-login.com right to left: the registered domain is sbi-verify-login.com, which belongs to whoever bought it. The letters "sbi" are merely part of a name anyone can register.

India's State Bank is at onlinesbi.sbi. Words before the final domain prove nothing — sbi.attacker.com is a site belonging entirely to attacker.com.
No HTTPS
The link begins http://, not https://. No bank operates a login page without encryption. Anything typed would cross the network readable.

Note the converse is not reassurance: a padlock proves the connection is encrypted, not that the site is honest. Fraudulent sites obtain certificates easily.
It asks for an OTP — decisive on its own
This single feature settles it. A One Time Password is the second factor protecting the account; its whole purpose is that only the account holder ever sees it.

No bank, and no genuine support desk, ever asks for an OTP, PIN or password, by email, message or telephone. Any request for one is fraud, with no exception worth entertaining.
"Click here" concealing the destination
Link text can say anything at all while pointing elsewhere. Hovering over a link shows its true destination in the status bar, and the two differing is a reliable sign of fraud.
7
What to do
1. Do not click, and do not reply.
2. If there is any doubt whether the account really needs attention, reach the bank independently — type its address yourself, use its official app, or telephone the number printed on the card. Never use contact details supplied by the suspicious message.
3. Report it to the provider as phishing and delete it.
4. If a credential has already been entered, change that password at once from a different device, and inform the bank immediately.

Answer

Generic greeting; manufactured urgency; a look-alike domain that is not the bank's; plain HTTP; a "click here" link hiding its destination; and — decisively — a request for an OTP, which no genuine organisation ever makes.

NoteThe OTP request alone is proof. Everything else is corroboration.

Problem 1Basic

Differentiate between the Internet and the World Wide Web.

Show solution

	Internet	World Wide Web
What it is	A global network of networks — physical infrastructure	A service that runs on the Internet: linked documents
Consists of	Cables, fibre, routers, satellites, connected computers	Web pages, websites and the hyperlinks between them
Protocol	TCP/IP	HTTP / HTTPS
Reached with	Any Internet application	A web browser
Began	ARPANET, $1969$	$1989$ , Tim Berners-Lee at CERN
Relationship	The Web is one service on it	It is one of many services

The essential point. The Internet is the roads; the Web is one kind of traffic on them. Email, file transfer (FTP), video calling, VoIP, online games and software updates all use the Internet without being part of the Web.

This is why "the Internet is down" and "this website is down" are different complaints, and why the Web could be replaced tomorrow while the Internet continued unchanged.

A useful test: opening Gmail in a browser uses the Web to reach email; collecting the same mail with an application over POP3 or IMAP uses the Internet but not the Web at all.

AnswerThe Internet is the global network of networks — the infrastructure, using TCP/IP. The Web is one service running on it, a collection of linked documents reached over HTTP with a browser. Email, FTP, VoIP and video calls use the Internet without being part of the Web.

Problem 2Basic

What is a URL? Identify the parts of https://www.example.org/notes/unit1.html

Show solution

A URL — Uniform Resource Locator — is the complete address of a resource on the Internet, stating both how to fetch it and where it is.

The parts.

$protocol https:// host www. domain example.org path /notes/ file unit1.html$

Protocol — https://. The rules for the transfer. HTTPS is HTTP secured with SSL/TLS, so the traffic is encrypted. Others: http://, ftp://, mailto:.
Host (subdomain) — www. Names a particular server within the domain. Conventional but not compulsory; mail. or library. may name other servers.
Domain name — example.org. The registered name, read right to left: .org is the top-level domain (organisation) and example the organisation. DNS translates this into the server's numeric IP address.
Path — /notes/. The directory on the server, like a folder path. Often case sensitive.
File — unit1.html. The resource requested, its extension indicating the type.

A URL may also carry a query string after ? (parameters, such as ?page=2) and a fragment after # (a position within the page).

AnswerA URL is the full address of a resource. Here: protocol https://, host www, domain example.org (.org being the top-level domain), path /notes/, file unit1.html.

Problem 3Exam level

Explain LAN, MAN and WAN with one example and one distinguishing feature each.

Show solution

1. LAN — Local Area Network. Connects computers within a single building or campus, typically up to a few kilometres. Owned and administered privately by the organisation that uses it, so it needs no permission from any authority.
Example: a college computer laboratory, or an office floor.
Distinguishing feature: highest data rate and lowest error rate, because the distances are short and the cabling is under the owner's control. Speeds of 100 Mbps to 10 Gbps are ordinary.

2. MAN — Metropolitan Area Network. Spans a city or a large town, up to a few tens of kilometres, usually by joining several LANs. Often built and run by a service provider or a municipal body, using leased lines or fibre laid along public routes.
Example: a cable television network across a city; a university linking its several campuses within one city; a city-wide corporation network.
Distinguishing feature: intermediate in every respect — larger than a LAN, faster and cheaper per kilometre than a WAN — and it typically requires public rights of way, so it cannot be purely private.

3. WAN — Wide Area Network. Spans a country, a continent or the world, with no distance limit. Built from links leased from telecommunication carriers, and usually owned by no single organisation.
Example: the Internet — the largest WAN in existence; also a bank's network joining branches across India.
Distinguishing feature: greatest coverage but lowest speed per link and highest error rate and cost, because signals travel very long distances over infrastructure the user does not control.

The pattern. As the area grows, coverage rises while speed, reliability and degree of private control all fall — the trade-off that makes all three necessary rather than one being simply better.

Also worth naming: PAN, a Personal Area Network of a few metres, such as a phone paired with earphones over Bluetooth.

	LAN	MAN	WAN
Area	Building, campus	City	Country, world
Speed	Highest	Medium	Lowest per link
Cost of setup	Low	Medium	High
Ownership	Private	Provider or public body	Multiple, usually leased
Error rate	Lowest	Medium	Highest

AnswerLAN covers a building or campus, privately owned, fastest and most reliable (a college laboratory). MAN covers a city, usually via a provider (a cable television network). WAN covers a country or the world, leased and multi-owned, greatest reach but slowest and least reliable (the Internet).

Problem 4Exam level

Explain the parts of an email message, and the difference between To, Cc and Bcc. When should Bcc be used?

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Parts of a message.

Field	Purpose
To	The main recipient or recipients — those the message is addressed to
Cc	Carbon copy — others who should see it for information
Bcc	Blind carbon copy — recipients hidden from all the others
Subject	A short line stating what the message concerns
Body	The message itself
Attachment	A file sent along with the message
Signature	A block of sender details appended automatically

An address has the form username@domain, the @ separating the user's name from the domain of the mail server, and containing no spaces.

To, Cc and Bcc compared.

	To	Cc	Bcc
Who it is for	Those expected to act or reply	Those who need to know	Those who need to know, privately
Visible to other recipients?	Yes	Yes	No
Sees the other Bcc addresses?	—	—	No — each Bcc recipient sees only their own
Reply All includes them?	Yes	Yes	No

When Bcc should be used.

1. Writing to a large group of people who do not know one another. This is the important case. Sending a notice to sixty students with all sixty addresses in To discloses every student's address to every recipient — a breach of their privacy, and a gift to anyone collecting addresses for spam. The same notice with the addresses in Bcc reaches everyone while disclosing nothing.

2. Copying someone discreetly — keeping a supervisor informed without making that visible to the main recipient.

3. Preventing a Reply All storm. Because Bcc recipients are not on the visible list, no one can reply to the whole group. On a mailing of hundreds this alone saves a great deal of unnecessary mail.

4. Keeping a copy for a record, by Bcc'ing an archive address.

A caution. Bcc should not be used to conceal a recipient from someone who has a legitimate right to know they are involved — that is a misuse of a privacy feature rather than an application of it. And the ordinary care applies to Reply All itself: it goes to everyone in To and Cc, so the recipient list is worth reading before sending.

AnswerFields are To, Cc, Bcc, Subject, Body, Attachment and Signature; the address is username@domain. To and Cc are visible to all recipients, Bcc is hidden from everyone. Use Bcc when mailing a large group who do not know each other — it protects their addresses and prevents Reply All storms.

Problem 5Exam level

Distinguish between SMTP, POP3 and IMAP, and explain why IMAP is preferred when mail is read on several devices.

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All three are email protocols, but they do different jobs. SMTP is for sending; POP3 and IMAP are alternatives for receiving.

SMTP — Simple Mail Transfer Protocol. Carries an outgoing message from the sender's program to the sender's mail server, and onward from server to server until it reaches the recipient's mail server. It is a push protocol and is used for sending only — it cannot retrieve mail.

POP3 — Post Office Protocol version 3. Downloads messages from the server to one device, and by default deletes them from the server afterwards. The mail then exists only on that machine.

IMAP — Internet Message Access Protocol. Synchronises with the server. Messages remain on the server, and the client displays them along with their folder structure and read/unread status. Actions taken on the device are carried out on the server.

	POP3	IMAP
Mail stored on	The device	The server
Left on the server?	No, by default	Yes
Multiple devices	Poorly supported	Fully supported
Folders synchronised?	No	Yes
Read/unread status shared?	No	Yes
Works offline	Yes, fully	Only for what has been cached
Server storage used	Little	More

Why IMAP is preferred for several devices.

Suppose the same account is used on a phone and a laptop.

With POP3, whichever device connects first downloads the messages and removes them from the server. Open the laptop later and the inbox is empty — the mail is on the phone. Read a message on the phone and the laptop has no way of knowing. Reply from the laptop and the sent copy exists only there. Folders created on one device do not exist on the other. The single account behaves as two unrelated mailboxes, each holding part of the mail.

With IMAP, the server holds the definitive mailbox and every device is a view of it. Reading a message on the phone marks it read on the laptop; filing it into a folder files it everywhere; deleting it deletes it once. Every device shows the same inbox, because there is only one inbox.

$POP3: mail lives on the device IMAP: mail lives on the server$

Two further advantages of IMAP: the mail survives the loss or failure of any device, since the server holds it; and only headers need be fetched until a message is opened, so a large mailbox is usable over a slow connection.

When POP3 is still reasonable: a single device, a wish to keep everything locally rather than on a provider's server, or very limited server storage.

AnswerSMTP sends mail; POP3 downloads it to one device and removes it from the server; IMAP synchronises, leaving the mail on the server. IMAP suits several devices because the server holds the one authoritative mailbox and every device is a view of it — read status, folders and deletions all agree, whereas POP3 splits the mail across whichever device downloaded it first.

Problem 6Challenge

What is phishing? Describe how to recognise it and the precautions to be taken online.

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Phishing is a fraud in which a message impersonating a trusted organisation — a bank, a service provider, an employer, a university — persuades the reader to disclose a credential such as a password, PIN, card number or OTP, or to install malicious software.

The word plays on "fishing": bait is put out to a very large number of people in the expectation that a few will take it. It is an attack on the person, not on the computer, which is why it defeats technical defences: no software fault is exploited when the victim types their own password into a convincing form.

Variants: spear phishing (aimed at one named individual using real details about them), vishing (by telephone), smishing (by SMS), and pharming (corrupting DNS so a correctly typed address reaches a fake site).

How to recognise it.

A generic greeting — "Dear Customer", "Dear User". An organisation holding your account holds your name.
Urgency and threat — "within 24 hours", "your account will be suspended", "immediate action required". The purpose is to prevent you pausing to check, which is the one action that defeats the fraud.
A domain that is close but wrong. Read the address right to left: in sbi-verify-login.com the registered domain is sbi-verify-login.com, which anyone may buy. Words appearing earlier prove nothing — sbi.attacker.com belongs entirely to attacker.com. Watch also for misspellings and swapped characters: paypa1, arnazon.
A link whose destination differs from its text. Hovering over it reveals the true target in the status bar.
Missing HTTPS on a page asking for credentials. Note the converse is not reassurance — a padlock proves only that the connection is encrypted, and fraudulent sites obtain certificates easily.
Poor spelling and grammar, or an unusual tone for the supposed sender.
Any request for a password, PIN or OTP — decisive on its own. No genuine bank or support desk ever asks for one. This single feature identifies fraud with no exception worth entertaining.
An unexpected attachment, particularly .exe, .zip or a document urging you to enable macros.

Precautions.

Never disclose an OTP, PIN or password to anyone, by any channel.
Verify independently. If a message might be genuine, reach the organisation through its own published address, its official app, or the number printed on your card — never through details supplied by the suspicious message.
Type addresses, or use your own bookmarks, for banking and other sensitive sites rather than following links.
Two-factor authentication everywhere it is offered, so a stolen password alone is insufficient.
Strong, different passwords per service, held in a password manager. Reuse means one breach compromises every account.
Keep the operating system, browser and antivirus updated — most successful attacks exploit a fault already patched.
Do not open unexpected attachments or links, even from a known sender, whose account may itself be compromised.
Avoid entering credentials over public Wi-Fi; use a VPN if unavoidable.
Back up important data — the only reliable defence against ransomware, since it removes the attacker's leverage.
Log out of shared computers, and clear the session.
Report and delete suspected phishing; providers use the reports to protect other users.

If a credential has already been given: change that password immediately from a different device, change it anywhere else it was reused, inform the bank or provider at once, and watch the account for unauthorised activity.

The principle. Phishing works on urgency and trust rather than on any technical weakness. The habit that defeats nearly all of it is simply to slow down: no legitimate organisation is harmed by your taking two minutes to verify a request through a channel you chose yourself.

AnswerA fraud in which a message impersonating a trusted organisation tricks the reader into surrendering a credential or installing malware — an attack on the person, not the software. Recognise it by generic greetings, manufactured urgency, look-alike domains read right to left, links whose destination differs from their text, missing HTTPS, and above all any request for a password or OTP. Precautions: never disclose an OTP, verify independently through your own channel, use two-factor authentication and unique passwords, keep software patched, avoid unexpected attachments, and back up data.