3.9 Matrix transformations

Learning objectives

• Find the image of a point or a shape under a transformation given by a matrix, or by a matrix and a translation vector.

• Write down the matrix of a reflection, a horizontal or vertical stretch, an enlargement or a rotation about the origin, and find a matrix from the images of two points.

• Describe fully the transformation represented by a given matrix.

• Find the single matrix that represents a composition of transformations, multiplying in the correct order, and use inverse matrices to undo a transformation.

• Find invariant points of a transformation, and write down transformations about centres other than the origin in the form $\boldsymbol{r}' = P(\boldsymbol{r} - \boldsymbol{c}) + \boldsymbol{c}$.

• Calculate the area of an image using the determinant, and interpret a negative or zero determinant.

• Find the equation of the image of a curve under a transformation.

• Use repeated transformations to generate sequences of points and fractals, and calculate areas and lengths in an iterative pattern.

Introduction

Every time a phone rotates a photo, a game shrinks a character as it walks into the distance or an engineer programs a robot arm to turn and reach, the coordinates of thousands of points are moved by the same rule. Writing that rule as a matrix turns a geometric move into a single multiplication. Two moves done one after the other become a product of matrices, undoing a move becomes an inverse matrix, and the determinant measures at once how much every area grows or shrinks.

Repeating a transformation many times produces patterns of a different kind. Shapes such as the Sierpinski triangle and Barnsley's fern are built by applying a few transformations over and over, and their areas and lengths follow geometric sequences.

Points, objects and transformation matrices

A point $(x, y)$ in the plane can be written as a column vector $\begin{pmatrix} x \\ y \end{pmatrix}$, its position vector from the origin. A transformation moves every point of the plane to a new position. The point or shape before the transformation is the object, and the result is its image.

Images are labelled with a prime: the image of $A$ is $A'$, read "A dashed" or "A prime", and the image of a general point $(x, y)$ is $(x', y')$

Multiplying a point by a matrix

Many of the transformations met in earlier years (reflections, rotations, enlargements and stretches with the origin fixed) share one property. The new coordinates are each a fixed combination of the old ones: $x' = ax + by$ and $y' = cx + dy$

Because $A\begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$ for every matrix, a transformation given by a matrix alone always leaves the origin where it is. It also maps straight lines to straight lines and keeps parallel lines parallel, so a square becomes a parallelogram (or, in the special case of a zero determinant, collapses onto a line or a point).

Transforming several points at once

Matrix multiplication works column by column, so the coordinates of all the vertices of a shape can be written as the columns of one matrix and transformed together. If the vertices are $(x_1, y_1)$, $(x_2, y_2)$

and each column of the answer is the image of the matching vertex. On a GDC the vertex matrix is entered as a $2 \times n$ matrix and multiplied by $A$ in one step. On a TI-84 Plus CE matrices are entered through 2nd x⁻¹ (the MATRIX menu, EDIT); on a TI-Nspire the matrix template is on the templates key; on a Casio fx-CG50 matrices are in the Run-Matrix application.

The columns are the images of (1, 0) and (0, 1)

Multiplying any matrix by $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ picks out its first column, and multiplying by $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$

So the first column of a transformation matrix is the image of $(1, 0)$ and the second column is the image of $(0, 1)$. Every point $(x, y)$ is $x$ steps of $(1, 0)$

The unit square and its image under $M = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}$ show the column rule at work: $(1, 0)$

The column rule works in both directions. To write down a matrix from a description, find where $(1, 0)$ and $(0, 1)$ go and write those images as the columns. To find a matrix from the images of other points, set up equations or use an inverse.

The inverse $X^{-1}$ is multiplied on the right because $X$ is on the right of $M$: from $MX = Y$, multiplying both sides on the right by $X^{-1}$

The standard transformation matrices

Five kinds of transformation about the origin have standard matrices. Each can be found from the column rule by asking where $(1, 0)$ and $(0, 1)$ go. The four panels of the flag diagram show a stretch, an enlargement, a rotation and a reflection applied to the same object: the stretch and the enlargement change size, the rotation and the reflection do not, and only the reflection turns the flag over.

Horizontal and vertical stretches

A horizontal stretch with scale factor $k$ multiplies every $x$-coordinate by $k$ and leaves $y$ alone, so $(1, 0) \mapsto (k, 0)$

Points on the $y$-axis have $x = 0$, so a horizontal stretch does not move them: the $y$-axis is an invariant line of the stretch, and every point on it is an invariant point. In the same way the $x$-axis is invariant under a vertical stretch. A full description of a stretch therefore names three things: that it is a stretch, its scale factor, and its invariant axis (or, equivalently, its direction). A scale factor between 0 and 1 brings points closer to the invariant axis, and a negative scale factor also reflects them across it.

A diagonal matrix $\begin{pmatrix} p & 0 \\ 0 & q \end{pmatrix}$ with $p \neq q$

Enlargements

An enlargement multiplies the distance of every point from a fixed centre by the same scale factor, so shapes keep their angles and proportions.

When the centre is the origin, $(1, 0) \mapsto (k, 0)$ and $(0, 1) \mapsto (0, k)$, so the matrix is $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} = kI$

The scale factor need not be greater than 1. With $0 < k < 1$ the image is smaller, and it is still called an enlargement. With $k < 0$ the image lies on the opposite side of the centre: an enlargement with scale factor $-2$ is the same as an enlargement with scale factor 2 followed by a half-turn about the centre, because $-2I = (2I)(-I)$

Rotations about the origin

The rotation matrix comes from the unit circle (subtopic 3.8). A rotation through angle $\theta$ anticlockwise about the origin moves $(1, 0)$ to the point on the unit circle at angle $\theta$, which is $(\cos\theta, \sin\theta)$. It moves $(0, 1)$

A clockwise rotation through $\theta$ is an anticlockwise rotation through $-\theta$. Replacing $\theta$ by $-\theta$ and using $\cos(-\theta) = \cos\theta$

The angle may be given in degrees or in radians, and the GDC must be in the matching mode. The most used rotations have simple matrices worth knowing on sight: 90° anticlockwise is $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, 180° is $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$

Reflections in lines through the origin

A reflection maps each point to its mirror image in a fixed line.

For a mirror line through the origin making angle $\theta$ with the positive $x$-axis, the line is $y = (\tan\theta)x$. Reflecting $(1, 0)$, which is at angle 0, in this line takes it to the point at angle $2\theta$

The angle that appears in the entries is $2\theta$, twice the angle of the mirror line. The table gives the four special cases, all found by substituting into this one matrix.

Recognising a matrix

Describing a matrix is the reverse task: given the matrix, name the transformation and all its details. The pattern of entries in each standard matrix identifies its type.

For a rotation, the first column $(p, q) = (\cos\theta, \sin\theta)$ gives the angle; the signs of $p$ and $q$ fix the quadrant, so $\theta$

Translations and affine transformations

A translation moves every point by the same translation vector $\begin{pmatrix} e \\ f \end{pmatrix}$. It moves the origin, so no $2 \times 2$ matrix can represent it on its own. Instead the vector is added after the multiplication.

The order inside the formula matters: the point is multiplied by $A$ first and the vector is added second. Translating first and then applying $A$ gives $A(\boldsymbol{r} + \boldsymbol{b}) = A\boldsymbol{r} + A\boldsymbol{b}$, which has translation part $A\boldsymbol{b}$

Finding the matrix and vector from images

Every part of an affine transformation can be read from three images. The origin goes to $A\boldsymbol{0} + \boldsymbol{b} = \boldsymbol{b}$, so the image of the origin is the translation vector. Then the image of $(1, 0)$ is (first column of $A$) $+ \boldsymbol{b}$

Invariant points of an affine transformation

A point that a transformation leaves in place satisfies $\boldsymbol{r}' = \boldsymbol{r}$. For an affine transformation this gives an equation for the invariant point.

When $I - P$ has an inverse there is exactly one invariant point. When $\det(I - P) = 0$ (for example, when $P = I$ and $T$

Transformations about other centres

The standard matrices all have the origin as centre or on the mirror line. A rotation, enlargement or reflection about another point $C$ with position vector $\boldsymbol{c}$ is built in three steps: translate so that $C$ moves to the origin, apply the matrix, then translate back.

Substituting $\boldsymbol{r} = \boldsymbol{c}$ gives $\boldsymbol{r}' = P\boldsymbol{0} + \boldsymbol{c} = \boldsymbol{c}$

Compositions of transformations

A composite transformation applies one transformation and then another. If the first has matrix $A$ and the second has matrix $B$, a point $\boldsymbol{r}$ first becomes $A\boldsymbol{r}$, and then $B$ acts on that result.

The step from $B(A\boldsymbol{r})$ to $(BA)\boldsymbol{r}$ uses the associativity of matrix multiplication. So "$A$ followed by $B$" has matrix $BA$

Because $BA$ and $AB$ are usually different, the order of the transformations usually matters. A reflection in the $x$-axis $F = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

Reflecting first and then rotating ($RF$) is a reflection in $y = x$, while rotating first and then reflecting ($FR$) is a reflection in $y = -x$. Some pairs do commute: two rotations about the origin (the angles add), two stretches in perpendicular directions, and an enlargement centred at the origin with anything, because $kI$

A transformation with $P^2 = I$ undoes itself: every reflection has this property, and so does a rotation of 180°. Such a transformation is its own inverse, like a self-inverse function in subtopic 2.7.

Two reflections and the rotation they make

Multiplying two reflection matrices always gives a rotation. Reflecting in a line at angle $\alpha$ and then in a line at angle $\beta$ (both through the origin) gives a rotation about the origin through $2(\beta - \alpha)$, anticlockwise if this is positive. The determinant gives a quick reason why the result cannot be another reflection: each reflection has determinant $-1$, and $(-1) \times (-1) = 1$

Compositions that include translations

When translations are involved, each step is applied in turn. If $T_1: \boldsymbol{r} \mapsto A\boldsymbol{r} + \boldsymbol{p}$ is followed by $T_2: \boldsymbol{r} \mapsto B\boldsymbol{r} + \boldsymbol{q}$

The matrix part is $BA$, as before, but the first translation vector is also transformed by $B$. Applying the same affine transformation twice, for example, gives $P^2\boldsymbol{r} + (P\boldsymbol{q} + \boldsymbol{q})$

Inverse transformations

The inverse matrix $A^{-1}$ undoes the transformation $A$: if $\boldsymbol{r}' = A\boldsymbol{r}$ then $\boldsymbol{r} = A^{-1}\boldsymbol{r}'$

The inverses of the standard transformations are the expected ones: a rotation through $\theta$ anticlockwise is undone by $\theta$ clockwise, a reflection by itself, and a stretch or enlargement with scale factor $k$ by scale factor $\frac{1}{k}$. A singular matrix

An unknown matrix in a composition

When a composition is known but one of its parts is not, the matrix equation is solved by multiplying by an inverse on the correct side. If "$X$ followed by $A$" equals $B$, then $AX = B$, and multiplying on the left by $A^{-1}$

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The determinant and area

The determinant of $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $\det A = ad - bc$

Why area is multiplied by the determinant

The unit square has area 1. Its image is the parallelogram with vertices $O$, $(a, c)$, $(a + b, c + d)$ and $(b, d)$

So the unit square, with area 1, becomes a parallelogram with area $|ad - bc|$ (the modulus covers the cases where the determinant is negative). Any other shape can be filled, as closely as needed, with tiny squares whose sides are parallel to the axes. Each tiny square is a scaled copy of the unit square, so each one's area is multiplied by the same number $|\det A|$, and so is their total.

A translation moves a shape without changing it, so the vector $\boldsymbol{b}$ plays no part in the area: only the matrix matters. The area scale factors of the standard transformations agree with what is expected from their geometry.

An enlargement with scale factor 3 multiplies every length by 3 and every area by 9, which matches the familiar rule for similar shapes.

The sign of the determinant and orientation

The modulus in the area rule discards the sign of the determinant, and that sign has a meaning of its own. Labelling the vertices of the flag-shaped object in order round its edge, say anticlockwise, and then following the same labels round the image gives an anticlockwise order for the stretch, enlargement and rotation, but a clockwise order for the reflection. A negative determinant always signals this flip.

Every reflection has determinant $-1$ and every rotation $+1$, so a composition containing an odd number of reflections reverses orientation and one with an even number does not. This gives a quick classification for a matrix whose area scale factor is 1: if $\det A = 1$ and the matrix has the rotation form, it is a rotation; if $\det A = -1$ and it has the reflection form, it is a reflection.

Because $\det(BA) = \det B \times \det A$, the area scale factor of a composition is the product of the separate area scale factors.

A zero determinant

If $\det A = 0$, every area is multiplied by 0, so the whole plane is squashed onto a line through the origin (or, for the zero matrix, onto the origin itself). For $A = \begin{pmatrix} 2 & 4 \\ 1 & 2 \end{pmatrix}$

so every image is a multiple of $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and lies on the line $y = \frac{1}{2}x$

Transforming lines and curves

A transformation moves every point of a curve, so the image of a curve is another curve. The equation of the image is found by writing the old coordinates $x$ and $y$ in terms of the new ones $x'$ and $y'$ and substituting into the old equation.

1. Write the transformation as equations: $x' = \ldots$ and $y' = \ldots$.

2. Rearrange (or use the inverse matrix) to get $x$ and $y$

The substitution goes "backwards" because the image point $(x', y')$ is on the new curve exactly when the original point $(x, y)$ it came from is on the old curve. This is the same reason why, in subtopic 2.8, a horizontal stretch with scale factor 2 replaces $x$

The two parabolas show the transformation acting on the whole curve: the vertex moves from the origin to $(1, -4)$, and the image is the object stretched by scale factor 2 horizontally and by scale factor 3 vertically, so overall it is wider, with $a = \frac{3}{2^2} = 0.75$

Lines are transformed in the same way. Under the rotation of 90° anticlockwise, $x' = -y$ and $y' = x$, so $x = y'$

Iteration and fractals

Applying the same transformation again and again to its own output is called iteration. Starting from a point $\boldsymbol{r}_0$ and a matrix $A$, the points $\boldsymbol{r}_1 = A\boldsymbol{r}_0$

Iteration links transformations to sequences. If $A$ is a rotation combined with an enlargement with scale factor $s$, each step multiplies the distance from the origin by $s$, so the distances form a geometric sequence with common ratio $s$ (subtopic 1.3). Areas multiply by $|\det A|$ each step, which is another geometric sequence, and when $|\det A| < 1$

Fractals and self-similarity

When several transformations are iterated together, each acting on the whole of the previous shape, the result can be a fractal: a shape made of smaller copies of itself.

In the fractals built from transformations, the parts that repeat the whole are exact copies, each one the image of the whole shape under one of the transformations.

The Sierpinski triangle

Start with a solid equilateral triangle with vertices at $(0, 0)$, $(1, 0)$ and $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$

The three copies together fill the triangle except for its middle quarter, an upside-down triangle. Applying all three transformations to the new shape, and again to the result, removes the middle quarter of every remaining triangle at each stage.

Each enlargement has determinant $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$

Iterated function systems and the chaos game

The same fractal can be drawn without tracking whole shapes. Start from any point, choose one of the transformations at random, apply it, plot the result, and repeat thousands of times. After the first few steps the plotted points all lie on, or extremely close to, the fractal. For the Sierpinski triangle this is the "chaos game": from the current point, jump halfway towards a randomly chosen vertex of the triangle, which is exactly applying $T_1$, $T_2$ or $T_3$

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The random method works because each transformation shrinks distances, so the effect of the starting point fades with each step. Whichever sequence of transformations is chosen, the point is soon drawn onto the one shape that the system maps onto itself.

Case study: the Barnsley fern

The mathematician Michael Barnsley showed in his 1988 book Fractals Everywhere how an iterated function system of just four affine transformations draws a realistic fern, modelled on the black spleenwort. Each map has the form $\boldsymbol{r}' = A\boldsymbol{r} + \boldsymbol{b}$, and at each step one map is chosen at random with the probability given.

Source: Barnsley, Fractals Everywhere (1988), as reproduced in Wikipedia, "Barnsley fern", 2025. Determinants calculated.

Plotting about 9,000 iterations of these four maps, starting at the origin, gives a fern in which every leaflet is itself a small fern.

Matrix transformations account for every feature of the picture.

• $f_2$ is close to $0.85$ times a rotation matrix: $0.85^2 + 0.04^2 = 0.7241$

The probabilities control how evenly the points fill the shape and leave the shape itself unchanged. The map that makes the largest copy, $f_2$, is chosen most often, so that the large upper part of the fern receives enough points. Changing the coefficients slightly gives ferns of other shapes, which is the sense in which small "mutations" of the numbers produce new forms. The same matrix operations, carried out in bulk by graphics processors, rotate, resize and mirror images and models in computer games and photo-editing software.

Exam focus

Matrix transformations appear on HL Papers 1 and 2, usually as one question of 6–18 marks, and the iterative and fractal ideas suit Paper 3. Questions write down standard matrices, find images, matrices from images, compositions and areas, describe fully a given matrix, and show that a matrix has a given form. They combine with matrix algebra (subtopic 1.14), the unit circle (subtopic 3.8), transformations of graphs (subtopic 2.8) and geometric series (subtopics 1.3 and 1.11).

The question shapes that recur are: writing the matrices of a list of moves and multiplying them in the correct order (the most common error); recognising the product as a single rotation or reflection; finding $\boldsymbol{q}$ and $P$ from the images of $(0, 0)$, $(1, 0)$ and $(0, 1)$

In "show that" parts, the working must move towards the given matrix, starting from the information in the question. Substituting the given matrix and checking that it works only verifies the result, and earns no marks.

Practice questions

1. Write down the matrix that represents (a) a reflection in the $y$-axis; (b) an enlargement with scale factor 3, centre the origin. [2]

2. Find the image of the point $(-2, 5)$ under a rotation of 90° clockwise about the origin. [2]

3. (a) Find the matrix of the reflection in the line $y = 3x$. (b) Hence find the image of $(10, 0)$

Summary

• A transformation matrix acts by $\boldsymbol{r}' = A\boldsymbol{r}$, with the matrix on the left; it fixes the origin. Its first column is the image of $(1, 0)$ and its second the image of $(0, 1)$.

Self-check

• To do:

  I can find the image of a point or a set of vertices by multiplying by a matrix, with the matrix on the left.

• To do:

  I can write down a matrix from the images of $(1, 0)$ and $(0, 1)$, and find a matrix from the images of any two points.

• To do:

  I can write down the matrices of stretches, enlargements, rotations and reflections, including the special reflections in the axes and in $y = \pm x$