Matrices, order and notation

What a matrix is, how its order is written, how to name an element, and when two matrices count as equal.

Module contents

A matrix is a rectangular arrangement of numbers set inside square brackets. The numbers are called elements (or entries). Nothing more is claimed at this stage — a matrix is not a number and it does not have a value. It is a container that holds numbers in a fixed arrangement of rows and columns.

$A = [20 - 1 4 57]$

This matrix has $2$ horizontal lines (rows) and $3$ vertical lines (columns).

Order of a matrix

If a matrix has $m$ rows and $n$ columns, its order (or size, or dimension) is written

$m \times n read as “ m by n ”$

Rows first, always. The matrix above is of order $2 \times 3$ . A matrix of order $m \times n$ contains $mn$ elements.

Exam habit. Whenever a question hands you a matrix, write its order beside it before doing anything else. Half the marks lost in matrix questions come from operating on matrices whose orders do not permit that operation.

Naming an element

The element sitting in the $i$ -th row and $j$ -th column of $A$ is written $a_{ij}$ — row index first.

$A = [a_{11} a_{21} a_{12} a_{22} a_{13} a_{23}]$

So in the matrix at the top, $a_{11} = 2$ , $a_{13} = 5$ , $a_{22} = 4$ , and $a_{21} = 0$ .

A whole matrix can be written compactly as

$A = [a_{ij}]_{m \times n}$

which reads: “ $A$ is the matrix whose general element is $a_{ij}$ , of order $m$ by $n$ .”

Building a matrix from a rule

Questions often define a matrix by a formula rather than by listing it. To construct it, run $i$ over the rows and $j$ over the columns and evaluate.

Equality of matrices

Two matrices $A = [a_{ij}]$ and $B = [b_{ij}]$ are equal ( $A = B$ ) when both of the following hold:

They have the same order.
Every corresponding element matches: $a_{ij} = b_{ij}$ for all $i$ and $j$ .

Order alone is not enough, and matching elements are meaningless if the orders differ. This gives a standard exam device: equate two matrices and read off a system of equations, one per position.

Why computer science cares

A matrix is the natural shape of anything indexed by two things at once, which is most of the data a program handles:

A 2-D array in C or Java is stored exactly as a matrix; a[i][j] is $a_{ij}$ .
A greyscale image is a matrix of pixel intensities; a colour image is three of them.
A graph on $n$ vertices is stored as an $n \times n$ adjacency matrix.
A transformation in 2-D or 3-D graphics — rotate, scale, translate — is applied by multiplying by a matrix.
A system of linear equations collapses into the single statement $A X = B$ .

The rules in this module are not arbitrary conventions. They are the rules that make those applications work.

Worked examples

3 solved

Every step is shown, in the order you would write it in an answer book.

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).

Practice problems

5 with solutions

Work each one on paper first. The full solution — not just the answer — is one click away.

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