Matrix Transformations

Every time a video game turns a character or a design app flips an image, a computer is multiplying coordinates by a matrix. Writing a point as a column vector $\begin{pmatrix} x \\ y \end{pmatrix}$, a $2 \times 2$ matrix can reflect it, stretch it, enlarge it or rotate it, and adding a vector translates it. This page builds the standard matrices, shows how to combine them (order matters!), and explains what the determinant tells you about area.

Key ideas

Points, images and the transformation rule

A transformation sends each point (the object) to a new point (its image). In this course, transformations have the form

$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} e \\ f \end{pmatrix}$$

The matrix does the reflecting, stretching, enlarging or rotating, and the vector $\begin{pmatrix} e \\ f \end{pmatrix}$ does a translation. With no translation, the origin always maps to itself.

To transform a whole shape, put the coordinates of its vertices in the columns of one matrix and multiply once.

Finding a matrix: follow the unit vectors

The columns of a matrix are the images of $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$:

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} a \\ c \end{pmatrix}, \qquad \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} b \\ d \end{pmatrix}$$

So to find the matrix of a transformation, work out where it sends $(1, 0)$ and $(0, 1)$, and write those images as the columns. That’s where every matrix in the table below comes from.

The standard matrices

Transformation	Matrix
Reflection in the $x$-axis	$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
Reflection in the $y$-axis	$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$
Reflection in $y = x$	$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
Reflection in $y = (\tan\theta)x$	$\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}$
Horizontal stretch, scale factor $k$	$\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}$
Vertical stretch, scale factor $k$	$\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}$
Enlargement, scale factor $k$, centre the origin	$\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$
Rotation by $\theta$ anticlockwise about the origin	$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$
Rotation by $\theta$ clockwise about the origin	$\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$
Translation by $\begin{pmatrix} e \\ f \end{pmatrix}$	add $\begin{pmatrix} e \\ f \end{pmatrix}$

In the reflection in $y = (\tan\theta)x$, $\theta$ is the angle the mirror line makes with the positive $x$-axis. With $\theta = 45^\circ$ it gives the reflection in $y = x$, and with $\theta = 0^\circ$ the reflection in the $x$-axis. Angles are usually in degrees here; just make sure your GDC is in the mode you mean.

The triangle from Example 1 (blue) and its images (orange) under a reflection in $y = x$ and a rotation of $90^\circ$ anticlockwise.

Compositions: the order matters

To do transformation $A$ first and then transformation $B$, multiply $\vec{x}$ by $A$ first, then by $B$:

$$\vec{x}' = B(A\vec{x}) = (BA)\vec{x}$$

So the single matrix for ”$A$ then $B$” is $BA$: the first transformation goes on the right, next to the vector. Because matrix multiplication isn’t commutative, doing them in the other order usually gives a different result. Repeating the same transformation $n$ times uses $A^n$; applying a few transformations over and over like this is how some fractals are generated.

When a translation is involved, work step by step: apply each matrix or vector to the result of the step before.

The determinant and area

Under the transformation with matrix $A$, every area is multiplied by the same amount:

$$\text{area of image} = |\det A| \times \text{area of object}$$

• $|\det A|$ is the area scale factor. Translations don’t change area at all.
• If $\det A \gt 0$, the image has the same orientation as the object (going round the vertices A, B, C anticlockwise stays anticlockwise).
• If $\det A \lt 0$, the orientation is reversed, as it is in a mirror. Every reflection has determinant $-1$.
• Rotations have determinant $\cos^2\theta + \sin^2\theta = 1$: they don’t change area or orientation.
• If $\det A = 0$, the whole plane is squashed onto a line (or a point), and the transformation can’t be undone.

The matrix with rows $(2, 1)$ and $(1, 3)$ has determinant $2 \cdot 3 - 1 \cdot 1 = 5$, so it maps the unit square to a parallelogram of area $5$.

Worked examples

Example 1: Reflecting and rotating a triangle

Triangle $T$ has vertices $A(1, 1)$, $B(4, 1)$ and $C(1, 3)$. Find the image of $T$ under:

• (a) a reflection in the line $y = x$
• (b) a rotation of $90^\circ$ anticlockwise about the origin

Solution. Put the vertices in the columns of one matrix.

(a)

$$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 4 & 1 \\ 1 & 1 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 3 \\ 1 & 4 & 1 \end{pmatrix}$$

The image has vertices $A'(1, 1)$, $B'(1, 4)$, $C'(3, 1)$. Each point’s coordinates are swapped, as you’d expect for a reflection in $y = x$. $A$ is on the mirror line, so it doesn’t move.

(b) With $\theta = 90^\circ$: $\cos 90^\circ = 0$ and $\sin 90^\circ = 1$, so the matrix is $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.

$$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 4 & 1 \\ 1 & 1 & 3 \end{pmatrix} = \begin{pmatrix} -1 & -1 & -3 \\ 1 & 4 & 1 \end{pmatrix}$$

The image has vertices $A'(-1, 1)$, $B'(-1, 4)$, $C'(-3, 1)$. The figure above shows both images.

Example 2: Reflection in a sloping line

Find the matrix for a reflection in the line $y = \dfrac{1}{\sqrt{3}}x$, and find the images of $P(2, 0)$ and $Q(0, 4)$.

Solution. The slope is $\tan\theta = \dfrac{1}{\sqrt{3}}$, so $\theta = 30^\circ$ and $2\theta = 60^\circ$:

$$\begin{pmatrix} \cos 60^\circ & \sin 60^\circ \\ \sin 60^\circ & -\cos 60^\circ \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}$$

Images:

$$\begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}\begin{pmatrix} 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix}, \qquad \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}\begin{pmatrix} 0 \\ 4 \end{pmatrix} = \begin{pmatrix} 2\sqrt{3} \\ -2 \end{pmatrix}$$

So $P'(1, \sqrt{3}) \approx (1, 1.73)$ and $Q'(2\sqrt{3}, -2) \approx (3.46, -2)$.

Check: a reflection keeps distances from the origin. $|OP'| = \sqrt{1 + 3} = 2 = |OP|$ ✓ and $|OQ'| = \sqrt{12 + 4} = 4 = |OQ|$ ✓.

Example 3: Order matters in a composition

Transformation $X$ is a reflection in the $x$-axis and $R$ is a rotation of $90^\circ$ anticlockwise about the origin.

• (a) Find the single matrix for ”$X$ followed by $R$”, and describe it.
• (b) Find the single matrix for ”$R$ followed by $X$”, and describe it.

Solution. $X = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ and $R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.

(a) $X$ first, so it goes on the right:

$$RX = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$

This is a reflection in the line $y = x$.

(b) $R$ first:

$$XR = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$$

This sends $(1, 0)$ to $(0, -1)$ and $(0, 1)$ to $(-1, 0)$: a reflection in the line $y = -x$.

Same two transformations, different order, different result. (Both have determinant $-1$: a reflection combined with a rotation is a reflection.)

Example 4: A matrix plus a translation

Triangle $T$ with vertices $(1, 1)$, $(4, 1)$ and $(1, 3)$ is transformed by

$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} 1 \\ -2 \end{pmatrix}$$

• (a) Find the vertices of the image.
• (b) Find the area of the image, and say whether the orientation is preserved.

Solution.

(a)

$$\begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 1 & 4 & 1 \\ 1 & 1 & 3 \end{pmatrix} + \begin{pmatrix} 1 & 1 & 1 \\ -2 & -2 & -2 \end{pmatrix} = \begin{pmatrix} 4 & 13 & 6 \\ 3 & 6 & 7 \end{pmatrix} + \begin{pmatrix} 1 & 1 & 1 \\ -2 & -2 & -2 \end{pmatrix} = \begin{pmatrix} 5 & 14 & 7 \\ 1 & 4 & 5 \end{pmatrix}$$

The image has vertices $(5, 1)$, $(14, 4)$ and $(7, 5)$.

(b) $T$ is a right triangle with legs $3$ and $2$, so its area is $\dfrac{1}{2}(3)(2) = 3$. The determinant is $3(2) - 1(1) = 5$, and the translation doesn’t change area, so

$$\text{area of image} = |5| \times 3 = 15$$

Since $\det = 5 \gt 0$, the orientation is preserved.

Common mistakes

Multiplying in the wrong order for a composition. ”$A$ then $B$” is $BA$, not $AB$. The first transformation must sit next to the vector, on the right.

Mixing up the clockwise and anticlockwise rotation matrices. The standard rotation matrix with $-\sin\theta$ in the top right turns points anticlockwise. Check with $(1, 0)$: rotating it $90^\circ$ anticlockwise should give $(0, 1)$.

Using the line’s angle instead of double the angle. The reflection in $y = (\tan\theta)x$ uses $2\theta$ in its matrix. For $y = x$, $\theta = 45^\circ$ and $2\theta = 90^\circ$, which correctly gives $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$.

Forgetting the absolute value for area. A reflection has $\det = -1$, but areas can’t be negative. Area scale factor $= |\det A|$; the sign only tells you about orientation.

Confusing a stretch with an enlargement. A stretch scales only one direction ($\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}$ multiplies areas by $k$). An enlargement scales both ($\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$ multiplies areas by $k^2$).

Trying to translate with a 2 × 2 matrix. No $2 \times 2$ matrix can move the origin, so a translation must be added as a vector.

Practice

1. (Warm-up) Find the image of the point $(3, -2)$ under:

• (a) a reflection in the $y$-axis
• (b) a rotation of $180^\circ$ about the origin
• (c) an enlargement with scale factor $2$, centre the origin

Solution

(a) $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} -3 \\ -2 \end{pmatrix}$

(b) $\cos 180^\circ = -1$, $\sin 180^\circ = 0$: $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}$

(c) $\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}$

2. (Warm-up) Write down the matrix for each transformation.

• (a) A vertical stretch with scale factor $3$
• (b) A rotation of $90^\circ$ clockwise about the origin
• (c) A reflection in the line $y = -x$

Solution

(a) $\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$

(b) $\cos 90^\circ = 0$, $\sin 90^\circ = 1$: $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$. Check: $(1, 0) \to (0, -1)$, a quarter turn clockwise ✓.

(c) The line $y = -x$ makes an angle $\theta = -45^\circ$ (or $135^\circ$) with the $x$-axis, so $2\theta = -90^\circ$: $\begin{pmatrix} \cos(-90^\circ) & \sin(-90^\circ) \\ \sin(-90^\circ) & -\cos(-90^\circ) \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$. Check: $(1, 0) \to (0, -1)$ ✓.

3. (Warm-up) A shape with area $6$ cm² is transformed by the matrix $\begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix}$. Find the area of the image.

Solution

$\det = 3(4) - (-1)(2) = 12 + 2 = 14$. Area of image $= 14 \times 6 = 84$ cm².

4. (Core) Describe fully the transformation given by each matrix.

• (a) $\begin{pmatrix} 0.6 & -0.8 \\ 0.8 & 0.6 \end{pmatrix}$
• (b) $\begin{pmatrix} 1 & 0 \\ 0 & 0.5 \end{pmatrix}$
• (c) $\begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix}$

Solution

(a) This matches $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ with $\cos\theta = 0.6$ and $\sin\theta = 0.8$. Both are positive, so $\theta$ is acute: $\theta = \cos^{-1} 0.6 = 53.1^\circ$ (3 s.f.). It’s a rotation of $53.1^\circ$ anticlockwise about the origin. (Its determinant is $0.36 + 0.64 = 1$, as for every rotation.)

(b) A vertical stretch with scale factor $0.5$ (it halves every $y$-coordinate).

(c) An enlargement with scale factor $-3$, centre the origin. (Equivalently, an enlargement with scale factor $3$ combined with a rotation of $180^\circ$ about the origin.)

5. (Core) A horizontal stretch with scale factor $2$ is followed by a rotation of $90^\circ$ anticlockwise about the origin.

• (a) Find the single matrix for this composition.
• (b) Find the image of $(3, 1)$.
• (c) Find the image of $(3, 1)$ if the transformations are done in the other order.

Solution

(a) Stretch $S = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$ first, so it goes on the right:

$$RS = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 2 & 0 \end{pmatrix}$$

(b) $\begin{pmatrix} 0 & -1 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 6 \end{pmatrix}$. Check step by step: the stretch gives $(6, 1)$, then the rotation gives $(-1, 6)$ ✓.

(c) $SR = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -2 \\ 1 & 0 \end{pmatrix}$, and $\begin{pmatrix} 0 & -2 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}$. A different image: the order matters.

6. (Core) Find the matrix for a reflection in the line $y = 2x$, and the image of $(5, 0)$.

Solution

$\tan\theta = 2$. Find $\cos 2\theta$ and $\sin 2\theta$ from $\theta = \tan^{-1} 2 = 63.43\ldots^\circ$, so $2\theta = 126.86\ldots^\circ$:

$$\cos 2\theta = -0.6, \qquad \sin 2\theta = 0.8$$

(These are exact: $\cos 2\theta = \dfrac{1 - \tan^2\theta}{1 + \tan^2\theta} = \dfrac{1 - 4}{5}$ and $\sin 2\theta = \dfrac{2\tan\theta}{1 + \tan^2\theta} = \dfrac{4}{5}$.) The matrix is

$$\begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix}, \qquad \begin{pmatrix} -0.6 & 0.8 \\ 0.8 & 0.6 \end{pmatrix}\begin{pmatrix} 5 \\ 0 \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$$

Check: the midpoint of $(5, 0)$ and $(-3, 4)$ is $(1, 2)$, which is on $y = 2x$ ✓, and both points are $5$ units from the origin ✓.

7. (Core) A transformation is given by

$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} 3 \\ -1 \end{pmatrix}$$

• (a) Find the image of $(1, 2)$.
• (b) The image of a point $P$ is $(4, 4)$. Find $P$.

Solution

(a) $\begin{pmatrix} 2 - 2 \\ 1 + 2 \end{pmatrix} + \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$. The image is $(3, 2)$.

(b) Undo the steps in reverse order: subtract the translation, then multiply by the inverse matrix.

$$\begin{pmatrix} 4 \\ 4 \end{pmatrix} - \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}$$

$\det = 2 + 1 = 3$, so the inverse is $\dfrac{1}{3}\begin{pmatrix} 1 & 1 \\ -1 & 2 \end{pmatrix}$, and

$$P = \frac{1}{3}\begin{pmatrix} 1 & 1 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 5 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 6 \\ 9 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$$

So $P = (2, 3)$. Check: $\begin{pmatrix} 4 - 3 \\ 2 + 3 \end{pmatrix} + \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \end{pmatrix}$ ✓.

8. (Challenge) A rotation of $90^\circ$ anticlockwise about the point $C(2, 1)$ can be done in three steps: translate by $\begin{pmatrix} -2 \\ -1 \end{pmatrix}$ (moving $C$ to the origin), rotate $90^\circ$ anticlockwise about the origin, then translate by $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$.

• (a) Show that the rotation can be written as $\vec{x}' = R\vec{x} + \begin{pmatrix} 3 \\ -1 \end{pmatrix}$, where $R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.
• (b) Find the image of $(5, 1)$, and explain why your answer makes sense.

Solution

(a) Following the three steps:

$$\vec{x}' = R\left(\vec{x} - \begin{pmatrix} 2 \\ 1 \end{pmatrix}\right) + \begin{pmatrix} 2 \\ 1 \end{pmatrix} = R\vec{x} - R\begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ 1 \end{pmatrix}$$

$R\begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}$, so the vector is $\begin{pmatrix} 2 \\ 1 \end{pmatrix} - \begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$. ✓

(b)

$$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 5 \\ 1 \end{pmatrix} + \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} -1 \\ 5 \end{pmatrix} + \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$$

The image is $(2, 4)$. This makes sense: $(5, 1)$ is $3$ units to the right of $C(2, 1)$, and a quarter turn anticlockwise about $C$ puts it $3$ units directly above $C$, at $(2, 4)$. (Also, $C$ itself maps to $R\begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$: the centre stays fixed ✓.)

9. (Challenge) The matrix $M = \begin{pmatrix} k & 2 \\ 1 & k \end{pmatrix}$ maps a shape of area $4$ to an image of area $20$, and preserves orientation. Find the possible values of $k$.

Solution

The area scale factor is $\dfrac{20}{4} = 5$, so $|\det M| = 5$. Orientation is preserved, so $\det M \gt 0$, which means $\det M = 5$:

$$k^2 - 2 = 5 \quad\Rightarrow\quad k^2 = 7 \quad\Rightarrow\quad k = \pm\sqrt{7}$$

(If the orientation had been reversed, we’d need $k^2 - 2 = -5$, so $k^2 = -3$, which has no real solutions.)

So $k = \sqrt{7}$ or $k = -\sqrt{7}$ ($\approx \pm 2.65$).