Linear transformations · Matrices & linear transformations

The matrix of a transformation

A linear transformation $T : (x, y) \to (ax + by, cx + dy)$ has the matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ : read the coefficients of $x$ and $y$ row by row. Its columns are the images of $(1, 0)$ and $(0, 1)$ :

What the columns mean(1, 0) → (a, c) and (0, 1) → (b, d)

To find an image, multiply the matrix by the point written as a column. The area of any shape is multiplied by $|ad - bc|$ .

A matrix as a transformationSet the entries or pick a preset

(4, 5)image of P(1, 2)3area scale |ad − bc|

a = 2b = 1c = 1d = 2

(1, 0) goes to (2, 1) and (0, 1) goes to (1, 2): the columns of the matrix. P(1, 2) goes to (2×1 + 1×2, 1×1 + 2×2) = (4, 5). ad − bc = 3.

Rotation about the origin

Turn the plane anticlockwise through an angle $\theta$ about the origin. The unit vectors $\mathbf i = (1, 0)$ and $\mathbf j = (0, 1)$ turn with it: $\mathbf i$ goes to $(\cos\theta, \sin\theta)$ and $\mathbf j$ goes to $(-\sin\theta, \cos\theta)$ . Those images are the two columns, so the matrix of the rotation is

R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}

Rotation through θi′ = (cos θ, sin θ) is column 1, j′ = (−sin θ, cos θ) is column 2

Put in $\theta = 90^\circ$ , $180^\circ$ and $270^\circ$ to get the special cases. A clockwise turn through $\theta$ is an anticlockwise turn through $-\theta$ , so $270^\circ$ anticlockwise is $90^\circ$ clockwise.

R_{90^\circ} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad R_{180^\circ} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}, \quad R_{270^\circ} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}

They send $(x, y)$ to $(-y, x)$ , $(-x, -y)$ and $(y, -x)$ . To find the image of any point, multiply $R_\theta$ by the point written as a column.

Worked example · WAEC 2008

WAEC 2008 · Paper 2 · Q12 (a)

The point $P(3, -5)$ is rotated through an angle of $60^\circ$ anticlockwise about the origin. (i) Obtain the matrix for the rotation. (ii) Find the image $P_1$ of the point $P$ under the rotation.

The matrix
- Anticlockwise through ${60^\circ}$ , so use ${R_\theta}$ with ${\theta = 60^\circ}$ .
- ${\cos 60^\circ = \frac12}$ and ${\sin 60^\circ = \frac{\sqrt3}{2}}$ .
- So the matrix is ${\begin{pmatrix} \frac12 & -\frac{\sqrt3}{2} \\ \frac{\sqrt3}{2} & \frac12 \end{pmatrix}}$ .
Think first. Put θ = 60° into the rotation matrix.
The image
- Top row: ${\frac12(3) - \frac{\sqrt3}{2}(-5) = \frac{3 + 5\sqrt3}{2}}$ .
- Bottom row: ${\frac{\sqrt3}{2}(3) + \frac12(-5) = \frac{3\sqrt3 - 5}{2}}$ .
- So ${P_1\left(\frac{3 + 5\sqrt3}{2}, \frac{3\sqrt3 - 5}{2}\right)}$ .
- As decimals, ${P_1}$ is about ${(5.83, 0.10)}$ .
Think first. Multiply the matrix by (3, −5) written as a column.

One transformation followed by another

To apply $N$ first and then $M$ , multiply by $N$ and then by $M$ : the combined matrix is $MN$ , with the first transformation written on the right.

N followed by MThe combined matrix is MN

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q11

Given that $M : (x, y) \to (7y, 3x - y)$ and $N : (x, y) \to (2x - y, 5x + 3y)$ ,

write down the matrices $M$ and $N$ of the linear transformations;

find the image of $P(2, -3)$ under the linear transformation $N$ followed by $M$ ;

find the coordinates of the point $Q$ whose image is $Q'(2, 4)$ under the linear transformation $N$ .

The matrices
- ${M = \begin{pmatrix} 0 & 7 \\ 3 & -1 \end{pmatrix}}$ and ${N = \begin{pmatrix} 2 & -1 \\ 5 & 3 \end{pmatrix}}$ .
Think first. Read the coefficients of x and y in each output.
N followed by M
- ${MN = \begin{pmatrix} 0 + 35 & 0 + 21 \\ 6 - 5 & -3 - 3 \end{pmatrix} = \begin{pmatrix} 35 & 21 \\ 1 & -6 \end{pmatrix}}$ .
- ${MN\begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 70 - 63 \\ 2 + 18 \end{pmatrix} = \begin{pmatrix} 7 \\ 20 \end{pmatrix}}$ : the image is ${(7, 20)}$ .
Think first. The combined matrix is MN, not NM.
The point whose image is (2, 4)
- ${2x - y = 2}$ and ${5x + 3y = 4}$ .
- ${|N| = 6 + 5 = 11}$ , so ${x = \frac{3(2) + 1(4)}{11} = \frac{10}{11}}$ and ${y = \frac{-5(2) + 2(4)}{11} = -\frac{2}{11}}$ .
- So ${Q\left(\frac{10}{11}, -\frac{2}{11}\right)}$ .
Think first. Undo N: solve N(x, y) = (2, 4).

Finding a matrix from images

If you know the images of two points, write $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and multiply it by each point. That gives four equations: two in $a$ and $b$ , two in $c$ and $d$ .

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q11

The images of points $(2, 3)$ and $(4, 5)$ under a linear transformation $\mathbf A$ are $(3, 4)$ and $(5, 6)$ respectively. Find the:

matrix $\mathbf A$ ;

inverse of $\mathbf A$ ;

point whose image is $(-1, 1)$ .

Four equations
- First rows: ${2a + 3b = 3}$ and ${4a + 5b = 5}$ .
- Second rows: ${2c + 3d = 4}$ and ${4c + 5d = 6}$ .
Think first. Multiply (a b / c d) by (2, 3) and by (4, 5).
Solve each pair
- Double the first and subtract: ${b = 1}$ , so ${a = 0}$ .
- Likewise ${d = 2}$ , so ${c = -1}$ : ${\mathbf A = \begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix}}$ .
The inverse and the pre-image
- ${|\mathbf A| = 0(2) - 1(-1) = 1}$ , so ${\mathbf A^{-1} = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}}$ .
- ${\mathbf A^{-1}\begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} -3 \\ -1 \end{pmatrix}}$ : the point is ${(-3, -1)}$ .

More: linear transformations

Your turn

WAEC 2016 · Paper 2 · Q9 (b)

(b)
Two linear transformations $A$ and $B$ in the $Oxy$ plane are defined by $A : (x, y) \to (x + 2y, -x + y)$ and $B : (x, y) \to (2x + 3y, x + 2y)$ . (i) Write down the matrices $A$ and $B$ . (ii) Find the image of the point $P(-2, 2)$ under the linear transformation $A$ followed by $B$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(b)(i)

Read the coefficients off each rule: $A = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$ .

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