Matrix operations

Addition, subtraction, scalar multiplication and the row-by-column product — with the conformability rules, the algebraic laws that survive and the ones that fail.

Module contents

Four operations, three of them easy and one of them the reason matrices exist.

Addition and subtraction

Conformable for addition means: same order. Then add element by element.

$A + B = [a_{ij} + b_{ij}] A - B = [a_{ij} - b_{ij}]$

$[1324] + [50 - 1 2] = [6316]$

If the orders differ, the sum is simply not defined — that is the whole answer, and it earns marks.

Laws of addition

Law	Statement
Closure	$A + B$ is a matrix of the same order
Commutative	$A + B = B + A$
Associative	$(A + B) + C = A + (B + C)$
Additive identity	$A + O = O + A = A$
Additive inverse	$A + (- A) = O$ , where $- A = [- a_{ij}]$
Cancellation	$A + C = B + C \Rightarrow A = B$

Scalar multiplication

Multiply every element by the scalar $k$ :

$k A = [k a_{ij}]$

$3 [10 - 2 4] = [30 - 6 12]$

Compare with determinants. $k A$ multiplies every one of the $n^{2}$ elements, but $∣ k A ∣ = k^{n} ∣ A ∣$ , because a determinant only takes one factor of $k$ out of each row. Mixing these two up is a classic exam loss.

Properties: $k (A + B) = k A + k B$ , $(k + l) A = k A + l A$ , $k (l A) = (k l) A$ , $1 A = A$ , $0 A = O$ .

Matrix multiplication

This is the operation that does real work, and its rule is not element-by-element.

Conformability

$A B$ is defined only when the number of columns of $A$ equals the number of rows of $B$ .

$A_{m \times n} B_{n \times p} = (A B)_{m \times p}$

The inner dimensions must match and they vanish; the outer dimensions survive as the order of the product.

The rule

$(A B)_{ij} = k = 1 \sum n a_{ik} b_{k j}$

In words: the element in row $i$ , column $j$ of the product is row $i$ of $A$ dotted with column $j$ of $B$ .

$[a_{11} a_{21} a_{12} a_{22}] [b_{11} b_{21} b_{12} b_{22}] = [a_{11} b_{11} + a_{12} b_{21} a_{21} b_{11} + a_{22} b_{21} a_{11} b_{12} + a_{12} b_{22} a_{21} b_{12} + a_{22} b_{22}]$

A reliable procedure

Write both orders. Check inner dimensions match. Write down the order of the answer and draw an empty frame of that size.
Fill the frame one position at a time. For position $(i, j)$ : cover everything except row $i$ of $A$ and column $j$ of $B$ .
Multiply the pairs in order and add.

Working position by position, in order, is what stops the errors.

Laws of multiplication — and the three that fail

Law	Holds?
Associative $(A B) C = A (B C)$	✔
Distributive $A (B + C) = A B + A C$	✔
Distributive $(B + C) A = B A + C A$	✔
Identity $A I = I A = A$	✔
Scalar $k (A B) = (k A) B = A (k B)$	✔
Commutative $A B = B A$	✘
Zero product $A B = O \Rightarrow A = O$ or $B = O$	✘
Cancellation $A B = A C \Rightarrow B = C$	✘

These three failures are where exam questions live.

Non-commutativity. Usually $A B \neq = B A$ ; often one is defined and the other is not. Matrices for which $A B = B A$ are said to commute.

Zero divisors. With $A = [0010]$ and $B = [0020]$ , $A B = O$ though neither is $O$ .

No cancellation. $A B = A C$ does not give $B = C$ unless $A$ is non-singular (then multiply through by $A^{- 1}$ ).

The consequence for algebra

Familiar identities have to be re-derived. For instance

$(A + B)^{2} = (A + B) (A + B) = A^{2} + A B + B A + B^{2}$

which equals $A^{2} + 2 A B + B^{2}$ only if $A$ and $B$ commute. Likewise

$(A + B) (A - B) = A^{2} - A B + B A - B^{2}$

which is $A^{2} - B^{2}$ only when $A B = B A$ .

Transpose and its laws

$(A^{T})^{T} = A (A + B)^{T} = A^{T} + B^{T} (k A)^{T} = k A^{T}$

$(A B)^{T} = B^{T} A^{T} — the reversal law$

The order reverses. Extending: $(A B C)^{T} = C^{T} B^{T} A^{T}$ .

Powers of a square matrix

$A^{1} = A, A^{n} = A^{n - 1} A, A^{0} = I$

Only square matrices can be raised to powers, since $A^{2} = AA$ needs the column count to equal the row count.

Cost, and why it matters in computing

Computing an $n \times n$ product by the definition performs $n^{3}$ multiplications and about $n^{3}$ additions. Doubling $n$ makes the work eight times larger. That single fact drives a large part of numerical computing — blocked algorithms, GPU kernels and Strassen-type methods all exist to fight the $n^{3}$ .

Worked examples

4 solved

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

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.

Practice problems

6 with solutions

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

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