Matrices and Determinants of Order 2

Lessons

Senior 4 · Mathematics

Matrices and Determinants of Order 2

5 lessons

Lesson 3 of 5

Multiplication of matrices

25 min

Rows meet columns

Go back to the coffee roaster. In week 1 the two shops sold $P = \begin{pmatrix} 40 & 25 \\ 32 & 18 \end{pmatrix}$ bags (rows: shops; columns: $250\,\mathrm{g}$ and $500\,\mathrm{g}$ bags). Suppose a $250\,\mathrm{g}$ bag sells for 3000 Frw and a $500\,\mathrm{g}$ bag for 5500 Frw (illustrative prices). Shop 1 took in

40 \times 3000 + 25 \times 5500 = 257\,500 \text{ Frw},

a row of quantities multiplied, entry by entry, with a column of prices and added up. Do the same for Shop 2 and you have

\begin{pmatrix} 40 & 25 \\ 32 & 18 \end{pmatrix}\begin{pmatrix} 3000 \\ 5500 \end{pmatrix} = \begin{pmatrix} 257\,500 \\ 195\,000 \end{pmatrix}.

Add a second column of wholesale prices, 2500 Frw and 4800 Frw, and each shop gets a second total in the same way:

\begin{pmatrix} 40 & 25 \\ 32 & 18 \end{pmatrix}\begin{pmatrix} 3000 & 2500 \\ 5500 & 4800 \end{pmatrix} = \begin{pmatrix} 257\,500 & 220\,000 \\ 195\,000 & 166\,400 \end{pmatrix}.

This "row times column" rule is matrix multiplication. It is the same rule you already use for $M\vec{v}$ , applied to each column in turn.

Definition

Matrix product: if $A$ is $m \times n$ and $B$ is $n \times p$ , the product $A B$ is the $m \times p$ matrix whose entry in row $i$ , column $j$ is row $i$ of $A$ times column $j$ of $B$ :

(AB)_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \cdots + a_{in}b_{nj}.

For order 2:

\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{pmatrix}.

The product exists only when the number of columns of $A$ equals the number of rows of $B$ .

Example: $\begin{pmatrix} 2 & -3 \end{pmatrix}\begin{pmatrix} 4 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \end{pmatrix}$ (a $1 \times 1$ matrix), while $\begin{pmatrix} 4 \\ 1 \end{pmatrix}\begin{pmatrix} 2 & -3 \end{pmatrix} = \begin{pmatrix} 8 & -12 \\ 2 & -3 \end{pmatrix}$ .

Worked example 1: $A B$ and $B A$

Let $A = \begin{pmatrix} 2 & 1 \\ -1 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 4 \\ 1 & -2 \end{pmatrix}$ . Find $A B$ and $B A$ .

Solution.

AB = \begin{pmatrix} 2(0) + 1(1) & 2(4) + 1(-2) \\ -1(0) + 3(1) & -1(4) + 3(-2) \end{pmatrix} = \begin{pmatrix} 1 & 6 \\ 3 & -10 \end{pmatrix},

BA = \begin{pmatrix} 0(2) + 4(-1) & 0(1) + 4(3) \\ 1(2) + (-2)(-1) & 1(1) + (-2)(3) \end{pmatrix} = \begin{pmatrix} -4 & 12 \\ 4 & -5 \end{pmatrix}.

$AB \ne BA$ : matrix multiplication is not commutative.

Why this rule? Composition

Let $g$ and $f$ be linear with matrices $B$ and $A$ . The composite $f \circ g$ means "first $g$ , then $f$ ": $(f \circ g)(\vec{v}) = f(g(\vec{v})) = A(B\vec{v})$ . Writing out the components shows that $A(B\vec{v}) = (AB)\vec{v}$ for every $\vec{v}$ . Hence

\text{the matrix of } f \circ g \text{ is } AB \quad (\text{the matrix of the transformation done first is on the right}).

This explains the rule for the product, and also why order matters. Let $S = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ (reflection in the $x$ -axis) and $R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ (rotation of $90^\circ$ anticlockwise about $O$ ). Then

RS = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad SR = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}.

"Reflect, then rotate" is the reflection in $y = x$ ; "rotate, then reflect" is the reflection in $y = - x$ . The diagram shows the two different images of the same triangle.

Important

Properties of the product (whenever the products are defined). With the identity matrix $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ , the matrix of the transformation that changes nothing:

Associative: $(A B) C = A (B C)$ , because doing $h$ , then $g$ , then $f$ does not depend on how you group the steps.
Distributive: $A (B + C) = A B + A C$ and $(A + B) C = A C + B C$ .
Scalars: $(k A) B = k (A B) = A (k B)$ .
Identity: $A I = I A = A$ .
Not commutative: in general $AB \ne BA$ .
No zero-product rule: $A B = O$ can happen with $A \ne O$ and $B \ne O$ .

Powers of a square matrix are $A^{2} = A A$ , $A^{3} = A^{2} A$ , and so on; associativity makes them unambiguous.

Worked example 2: a zero product and powers

(a) Show that $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ satisfy $A B = O$ . (b) Find $R^{2}$ and $R^{4}$ for the rotation $R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ .

Solution.

(a) $AB = \begin{pmatrix} 1(0) + 0(0) & 1(0) + 0(1) \\ 0 & 0 \end{pmatrix} = O$ . Geometrically, $B$ projects onto the $y$ -axis and $A$ then projects onto the $x$ -axis, which sends everything on the $y$ -axis to $\vec{0}$ .

(b) $R^2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I$ , the rotation of $180^\circ$ . Then $R^{4} = (R^{2})^{2} = (- I)^{2} = I$ : four quarter-turns bring every point back.

Worked example 3: a matrix equation

Show that $A = \begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}$ satisfies $A^{2} - 5 A + 4 I = O$ .

Solution. $A^2 = \begin{pmatrix} 9 + 2 & 3 + 2 \\ 6 + 4 & 2 + 4 \end{pmatrix} = \begin{pmatrix} 11 & 5 \\ 10 & 6 \end{pmatrix}$ . Then

A^2 - 5A + 4I = \begin{pmatrix} 11 - 15 + 4 & 5 - 5 + 0 \\ 10 - 10 + 0 & 6 - 10 + 4 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}.

(The numbers 5 and 4 are not chance: $5 = 3 + 2$ is the sum of the diagonal entries and $4 = 3 (2) - 1 (2)$ is the determinant, the subject of the next lesson.)

Warning

Habits from numbers that fail for matrices. $(A + B)^{2} = A^{2} + A B + B A + B^{2}$ , which is not $A^{2} + 2 A B + B^{2}$ unless $A B = B A$ . From $A B = A C$ you may not cancel $A$ to get $B = C$ . And a product is never computed entry by entry: $\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} e & f \\ g & h \end{pmatrix} \ne \begin{pmatrix} ae & bf \\ cg & dh \end{pmatrix}$ .

Tip

Run your left index finger along a row of $A$ and your right index finger down a column of $B$ , multiplying as you go; the total lands where that row and that column cross in $A B$ . Before you start, check "inner orders match": $(m \times \mathbf{n})(\mathbf{n} \times p)$ .

3Blue1Brown: Matrix multiplication as composition, Chapter 4 of Essence of Linear Algebra (about 10 min). Watch for why the product is read from right to left, and why the order of two transformations matters.

✦

Practice

Work each one, then open the answers.

$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ , $B = \begin{pmatrix} 2 & 0 \\ 1 & -1 \end{pmatrix}$ . Find $A B$ and $B A$ .
$A$ is $2 \times 3$ , $B$ is $3 \times 2$ and $C$ is $2 \times 2$ . Which of $A B$ , $B A$ , $A C$ , $C A$ , $B C$ exist, and what are their orders?
Compute (a) $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 5 \\ -1 \end{pmatrix}$ (b) $\begin{pmatrix} 2 & -3 \end{pmatrix}\begin{pmatrix} 4 \\ 1 \end{pmatrix}$ .
For $S = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ , find $S^{2}$ and $S^{3}$ , and guess $S^{n}$ .
Find a matrix $N \ne O$ with $N^{2} = O$ .
$f$ is the shear with matrix $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $g$ the reflection in the $x$ -axis. Find the matrices of $f \circ g$ and $g \circ f$ , and the image of $(1, 1)$ under each.
Find $x$ and $y$ if $\begin{pmatrix} x & 1 \\ 2 & y \end{pmatrix}\begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 5 \\ -4 \end{pmatrix}$ .
Show that every matrix of the form $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ commutes with $R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ .

Show answersHide answers

$AB = \begin{pmatrix} 4 & -2 \\ 10 & -4 \end{pmatrix}$ , $BA = \begin{pmatrix} 2 & 4 \\ -2 & -2 \end{pmatrix}$ .
$A B$ : $2 \times 2$ . $B A$ : $3 \times 3$ . $A C$ : not defined (3 columns, 2 rows). $C A$ : $2 \times 3$ . $B C$ : $3 \times 2$ .
(a) $\begin{pmatrix} 3 \\ 11 \end{pmatrix}$ (b) $\begin{pmatrix} 5 \end{pmatrix}$ .
$S^2 = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix}$ , $S^3 = \begin{pmatrix} 1 & 6 \\ 0 & 1 \end{pmatrix}$ , $S^n = \begin{pmatrix} 1 & 2n \\ 0 & 1 \end{pmatrix}$ : $n$ shears of factor 2 make one shear of factor $2 n$ .
For example $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ : $N^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ .
$f \circ g$ : $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 0 & -1 \end{pmatrix}$ , image $(0, - 1)$ . $g \circ f$ : $\begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}$ , image $(2, - 1)$ .
$x + 3 = 5$ , so $x = 2$ ; $2 + 3 y = - 4$ , so $y = - 2$ .
$\begin{pmatrix} a & -b \\ b & a \end{pmatrix}R = \begin{pmatrix} -b & -a \\ a & -b \end{pmatrix}$ and $R\begin{pmatrix} a & -b \\ b & a \end{pmatrix} = \begin{pmatrix} -b & -a \\ a & -b \end{pmatrix}$ : equal.

Quick check

Key points

$(AB)_{ij}$ = row $i$ of $A$ times column $j$ of $B$ ; $A B$ exists when columns of $A$ = rows of $B$ .
The matrix of $f \circ g$ (first $g$ , then $f$ ) is $A B$ .
The product is associative and distributive, $A I = I A = A$ , but $AB \ne BA$ in general, and $A B = O$ is possible with $A, B \ne O$ .
Powers: $A^{2} = A A$ ; for the quarter-turn $R$ , $R^{2} = - I$ and $R^{4} = I$ .

Sources and credits

Written by Ganzaa: the explanations, the diagram, the worked examples (the prices are illustrative), the practice items and the questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
History: Matrices and determinants, J. J. O'Connor and E. F. Robertson, MacTutor History of Mathematics, University of St Andrews.
Engraving: Arthur Cayley, Richard Taylor after Herbert Rose Barraud, Wikimedia Commons, public domain.
Video: Matrix multiplication as composition, Chapter 4, Essence of linear algebra, 3Blue1Brown, embedded from YouTube.

1Rows meet columns

2Why this rule? Composition

Rows meet columns

Why this rule? Composition