Matrix algebra and inverses · Matrices & linear transformations

This lesson builds on matrices and determinants.

Multiplying matrices

Each entry of a product is a row of the first matrix times a column of the second:

Row times columnMultiply across the row and down the column, then add

Order matters: in general $AB \ne BA$ . The identity $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ leaves a matrix unchanged: $AI = IA = A$ .

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q3

Given that $B = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ and $B^2 + 3B + 2I = 3N$ , where $I$ is the $2 \times 2$ unit matrix, find the matrix $N$ .

B squared
- ${B^2 = \begin{pmatrix} 4 + 3 & 6 + 12 \\ 2 + 4 & 3 + 16 \end{pmatrix} = \begin{pmatrix} 7 & 18 \\ 6 & 19 \end{pmatrix}}$ .
Think first. Multiply B by itself, row by column.
Add the three matrices
- ${3B = \begin{pmatrix} 6 & 9 \\ 3 & 12 \end{pmatrix}}$ and ${2I = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}}$ .
- ${3N = \begin{pmatrix} 7 + 6 + 2 & 18 + 9 + 0 \\ 6 + 3 + 0 & 19 + 12 + 2 \end{pmatrix} = \begin{pmatrix} 15 & 27 \\ 9 & 33 \end{pmatrix}}$ .
Divide by 3
- ${N = \begin{pmatrix} 5 & 9 \\ 3 & 11 \end{pmatrix}}$ .

More: matrix equations

The inverse of a 2 × 2 matrix

The inverse $A^{-1}$ undoes $A$ : $AA^{-1} = A^{-1}A = I$ . It exists only when the determinant $ad - bc$ is not zero.

The inverseA⁻¹ = (1 ÷ (ad − bc)) × (d, −b / −c, a)

Worked example · WAEC 2020

WAEC 2020 · Paper 2 · Q2

Given that $P = \begin{pmatrix} 3 & 4 \\ 5 & 6 \end{pmatrix}$ and $Q = \begin{pmatrix} -2 & 5 \\ -3 & 1 \end{pmatrix}$ , find $PQ^{-1}$ , where $Q^{-1}$ is the inverse of $Q$ .

The inverse of Q
- ${|Q| = (-2)(1) - (5)(-3) = -2 + 15 = 13}$ .
- ${Q^{-1} = \frac{1}{13}\begin{pmatrix} 1 & -5 \\ 3 & -2 \end{pmatrix}}$ .
Think first. ad − bc for Q?
Multiply
- ${PQ^{-1} = \frac{1}{13}\begin{pmatrix} 3 + 12 & -15 - 8 \\ 5 + 18 & -25 - 12 \end{pmatrix}}$ .
- ${= \frac{1}{13}\begin{pmatrix} 15 & -23 \\ 23 & -37 \end{pmatrix}}$ .

More: the inverse

WAEC 2020 · Paper 1 · Q2Find the inverse of $\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}$ .

Solving simultaneous equations

Two equations $ax + by = p$ and $cx + dy = q$ are the matrix equation $A\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} p \\ q \end{pmatrix}$ . Multiply both sides on the left by $A^{-1}$ :

\begin{pmatrix} x \\ y \end{pmatrix} = A^{-1}\begin{pmatrix} p \\ q \end{pmatrix}

Solving with an inverse matrixSet the coefficients and the right-hand sides

5determinant ad − bc(2, 3)solution (x, y)

a = 2b = 1c = 3d = 4p = 7q = 18

|A| = (2)(4) − (1)(3) = 5. A⁻¹ = (1/5) × (4, −1 / −3, 2). Then x = (4 × 7 − 1 × 18) ÷ 5 = 2 and y = ((−3) × 7 + 2 × 18) ÷ 5 = 3. Check: 2 × 2 + 1 × 3 = 7 ✓.

Worked example · NECO 2023

NECO 2023 · Paper 1 · Q13

If $\begin{pmatrix} 3 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 13 \\ 7 \end{pmatrix}$ , find the value of $y$ .

The inverse
- ${|A| = 3(1) - (-1)(1) = 4}$ , so ${A^{-1} = \frac14\begin{pmatrix} 1 & 1 \\ -1 & 3 \end{pmatrix}}$ .
Multiply
- ${\begin{pmatrix} x \\ y \end{pmatrix} = \frac14\begin{pmatrix} 13 + 7 \\ -13 + 21 \end{pmatrix} = \frac14\begin{pmatrix} 20 \\ 8 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix}}$ .
- So ${y = 2}$ : option B.

Your turn

WAEC 2018 · Paper 2 · Q11 (a)

(a)
Given the matrix $A = \begin{pmatrix} 3 & 8 \\ 5 & -2 \end{pmatrix}$ , find its inverse.
Show the answer
$A^{-1} = \begin{pmatrix} \frac{1}{23} & \frac{4}{23} \\ \frac{5}{46} & -\frac{3}{46} \end{pmatrix}$

Worked solution (try it first)

(a)

The determinant is $|A| = 3(-2) - 8(5) = -6 - 40 = -46$ .
Swap the leading diagonal and change the signs of the other two entries: $\begin{pmatrix} -2 & -8 \\ -5 & 3 \end{pmatrix}$ .
Divide by the determinant: $A^{-1} = -\dfrac{1}{46}\begin{pmatrix} -2 & -8 \\ -5 & 3 \end{pmatrix}$
$= \begin{pmatrix} \frac{1}{23} & \frac{4}{23} \\ \frac{5}{46} & -\frac{3}{46} \end{pmatrix}$ .

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