A-Level Further Maths 9231 / Topic 1.4 / AS & A Level

Read the transformation from the columns.

A \(2 \times 2\) matrix moves every point of the plane at once. Its first column is where \((1, 0)\) lands and its second column is where \((0, 1)\) lands, so you can build the matrix for any rotation, reflection, enlargement, stretch or shear, combine transformations by multiplying, undo them with the inverse and read the change in area from the determinant.

Topic 1.4All matrices lessons →

The key idea

The columns of \(\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) are the images of \((1, 0)\) and \((0, 1)\). The matrix \(\mathbf{AB}\) means 'apply \(\mathbf{B}\) first, then \(\mathbf{A}\)'. \(\mathbf{A}^{-1}\) undoes \(\mathbf{A}\). Areas are multiplied by \(|\det \mathbf{M}|\), and a negative determinant means the orientation of shapes is reversed.

Fig. 1: the unit square (dashed) and its image (shaded). Left: the shear with matrix columns (1, 0) and (2, 1); the x-axis is invariant and the area is unchanged. Right: a rotation of 30° anticlockwise about O, whose columns are (cos 30°, sin 30°) and (-sin 30°, cos 30°)

Section 01

The columns are the images.

Multiply \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) by \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\) and you get the first column \(\begin{pmatrix} a \\ c \end{pmatrix}\); multiply by \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\) and you get the second column \(\begin{pmatrix} b \\ d \end{pmatrix}\). So to find the matrix of a transformation, ask where the two unit vectors go and write their images as the columns. This is the unit square method: Fig. 1 shows the square with corners \(O\), \((1, 0)\), \((1, 1)\) and \((0, 1)\) carried to a parallelogram whose two sides from \(O\) are the columns. Every transformation here keeps the origin fixed.

Transformation	Matrix
Rotation through \(\theta\) anticlockwise about \(O\)	\(\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\)
Reflection in the line \(y = x\tan\theta\)	\(\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\)
Reflection in the \(x\)-axis; in the \(y\)-axis	\(\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\); \(\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\)
Reflection in \(y = x\); in \(y = -x\)	\(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\); \(\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\)
Enlargement, scale factor \(k\), centre \(O\)	\(\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}\)
Stretch parallel to the \(x\)-axis, factor \(k\); parallel to the \(y\)-axis	\(\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix}\); \(\begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}\)
Shear, \(x\)-axis invariant; \(y\)-axis invariant	\(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\); \(\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}\)

You do not need to memorise the whole table: each row comes from drawing the images of \((1, 0)\) and \((0, 1)\). A shear with the \(x\)-axis invariant keeps every point of the \(x\)-axis fixed and slides each point parallel to the \(x\)-axis by \(k\) times its \(y\)-coordinate.

Worked example A rotation from the unit vectors

Find the matrix that represents a rotation through \(120^{\circ}\) anticlockwise about the origin.

\((1, 0)\) goes to \((\cos 120^{\circ}, \sin 120^{\circ}) = \left(-\tfrac{1}{2}, \tfrac{\sqrt{3}}{2}\right)\): this is the first column.
\((0, 1)\) goes to \((-\sin 120^{\circ}, \cos 120^{\circ}) = \left(-\tfrac{\sqrt{3}}{2}, -\tfrac{1}{2}\right)\): this is the second column.
Check: the determinant is \(\tfrac{1}{4} + \tfrac{3}{4} = 1\), as it must be for a rotation.

Answer: \(\begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}\)

Worked example Reflection in a sloping line

Find the matrix that represents a reflection in the line \(y = \sqrt{3}\,x\).

\(\tan\theta = \sqrt{3}\), so the line makes \(\theta = 60^{\circ}\) with the \(x\)-axis and \(2\theta = 120^{\circ}\).
\(\cos 120^{\circ} = -\tfrac{1}{2}\) and \(\sin 120^{\circ} = \tfrac{\sqrt{3}}{2}\), so the matrix is \(\begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\).
Check with a point on the mirror line, \((1, \sqrt{3})\): its image is \(\left(-\tfrac{1}{2} + \tfrac{3}{2}, \tfrac{\sqrt{3}}{2} + \tfrac{\sqrt{3}}{2}\right) = (1, \sqrt{3})\), so it stays put, as it should.

Answer: \(\begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\)

Your turn

A shear with the \(x\)-axis invariant maps \((0, 1)\) to \((2, 1)\), as in Fig. 1. Write down its matrix and find the image of \((3, 2)\).

Examiner trap

The reflection matrix uses \(2\theta\), not \(\theta\): for the mirror line \(y = x\) (\(\theta = 45^{\circ}\)) it gives \(\cos 90^{\circ} = 0\) and \(\sin 90^{\circ} = 1\), the familiar \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\). Also, the images are COLUMNS, not rows: writing them as rows gives the transpose, which for a rotation is the rotation the other way.

Section 02

Order matters: B first, then A.

For a point with position vector \(\mathbf{x}\), \(\mathbf{AB}\mathbf{x} = \mathbf{A}(\mathbf{B}\mathbf{x})\): the matrix next to the point acts first. So the single matrix for 'transformation \(\mathbf{B}\) followed by transformation \(\mathbf{A}\)' is \(\mathbf{AB}\), written right to left. Matrix multiplication is not commutative, and neither are transformations: the two orders usually give different results.

Combined: \(\mathbf{P}\) then \(\mathbf{Q}\) then \(\mathbf{R}\) is \(\mathbf{RQP}\).
Inverse: \(\mathbf{A}^{-1}\) represents the transformation that undoes \(\mathbf{A}\), because \(\mathbf{A}^{-1}\mathbf{A} = \mathbf{I}\), the identity (nothing moves). The inverse of a rotation through \(\theta\) is a rotation through \(-\theta\); a reflection is its own inverse; an enlargement with factor \(k\) is undone by factor \(\tfrac{1}{k}\); a shear with factor \(k\) by the same shear with factor \(-k\).
Undoing a combination: \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\). To undo 'B then A', undo A first, then undo B, like taking off shoes and socks.
Finding an object: if \(\mathbf{A}\) maps \(P\) to \(P'\), then \(P = \mathbf{A}^{-1}P'\).

Worked example Two orders, two answers

\(\mathbf{P}\) is a reflection in the \(x\)-axis and \(\mathbf{Q}\) is a rotation through \(90^{\circ}\) anticlockwise about \(O\). Find the single matrix for \(\mathbf{P}\) followed by \(\mathbf{Q}\), and describe the transformation. Compare with \(\mathbf{Q}\) followed by \(\mathbf{P}\).

\(\mathbf{P} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\) and \(\mathbf{Q} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\).
\(\mathbf{P}\) first, then \(\mathbf{Q}\): \(\mathbf{QP} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\), a reflection in \(y = x\).
Check with \((1, 0)\): the reflection leaves it at \((1, 0)\), then the rotation takes it to \((0, 1)\), the first column. Correct.
The other order: \(\mathbf{PQ} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\), a reflection in \(y = -x\).

Answer: \(\mathbf{QP} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\), reflection in \(y = x\); the other order gives a reflection in \(y = -x\)

Worked example Working backwards with the inverse

The matrix \(\mathbf{M} = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\) maps the point \(P\) to the point \((5, 3)\). Find the coordinates of \(P\).

\(\det \mathbf{M} = 2 - 1 = 1\), so \(\mathbf{M}^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}\).
\(P = \mathbf{M}^{-1}\begin{pmatrix} 5 \\ 3 \end{pmatrix} = \begin{pmatrix} 5 - 3 \\ -5 + 6 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\).
Check: \(\mathbf{M}\begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}\).

Answer: \(P = (2, 1)\)

Your turn

Write down the matrix of the transformation that undoes the shear \(\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}\), and check that the product is \(\mathbf{I}\).

Examiner trap

'A followed by B' is \(\mathbf{BA}\), not \(\mathbf{AB}\). Reading the words left to right and writing the matrices in the same order is the most common error in this topic. Test your product on \((1, 0)\) by doing the two transformations one at a time.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Which matrix represents a reflection in the \(y\)-axis?

Quick check+10 XP

What is the image of \((0, 1)\) under \(\begin{pmatrix} 2 & 5 \\ -1 & 3 \end{pmatrix}\)?

Quick check+10 XP

Transformation \(\mathbf{T}\) is followed by transformation \(\mathbf{U}\). Which single matrix represents the combined transformation?

Quick check+10 XP

What is the area scale factor of the transformation \(\begin{pmatrix} 3 & 2 \\ 4 & 1 \end{pmatrix}\)?

Section 03

The determinant measures area.

The unit square has area 1, and its image is a parallelogram of area \(|ad - bc|\). Every shape is scaled in the same way, so the area scale factor of \(\mathbf{M}\) is \(|\det \mathbf{M}|\). The sign carries information too: if \(\det \mathbf{M} < 0\), the image is 'flipped over' (the order of the vertices is reversed, as in a mirror). If \(\det \mathbf{M} = 0\), the plane is squashed onto a line or a point, and the transformation cannot be undone, which is why \(\mathbf{M}^{-1}\) does not exist. In the next lesson you will find the points and lines that a matrix leaves unchanged.

Rotation: \(\det = \cos^{2}\theta + \sin^{2}\theta = 1\). Reflection: \(\det = -\cos^{2} 2\theta - \sin^{2} 2\theta = -1\).
Enlargement, factor \(k\): \(\det = k^{2}\). Stretch, factor \(k\): \(\det = k\). Shear: \(\det = 1\).
\(\det(\mathbf{AB}) = \det \mathbf{A} \times \det \mathbf{B}\): area factors multiply when transformations are combined.
To describe a matrix fully, look for a number \(k\) you can take out so that what is left is a rotation or reflection matrix: \(k\) is the enlargement factor (and \(k^{2}\) must equal \(|\det \mathbf{M}|\)).

Worked example Area of an image

A triangle of area \(6\) units\(^{2}\) is transformed by \(\begin{pmatrix} 3 & 1 \\ 2 & -4 \end{pmatrix}\). Find the area of its image and say what happens to its orientation.

\(\det = 3 \times (-4) - 1 \times 2 = -14\).
Area scale factor \(|-14| = 14\), so the image has area \(6 \times 14 = 84\).
The determinant is negative, so the image is reversed: vertices labelled anticlockwise in the object are clockwise in the image.

Answer: \(84\) units\(^{2}\); orientation reversed

Worked example Describe a matrix fully

Describe fully the transformation represented by \(\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}\).

\(\det = 3 + 1 = 4\), so if there is an enlargement its factor is \(\sqrt{4} = 2\).
Take out 2: \(2\begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix}\).
The matrix left over has the rotation pattern with \(\cos\theta = \tfrac{\sqrt{3}}{2}\) and \(\sin\theta = \tfrac{1}{2}\), so \(\theta = 30^{\circ}\).
An enlargement with centre \(O\) commutes with a rotation about \(O\), so the order does not matter here.

Answer: A rotation through \(30^{\circ}\) anticlockwise about \(O\) and an enlargement with scale factor \(2\), centre \(O\)

Your turn

Describe fully the transformation represented by \(\begin{pmatrix} 0 & 3 \\ 3 & 0 \end{pmatrix}\), and state its area scale factor.

Examiner trap

An enlargement with factor \(k\) multiplies areas by \(k^{2}\), not by \(k\). And the area scale factor is \(|\det \mathbf{M}|\): an image never has a negative area, so a determinant of \(-14\) means 'area times 14, flipped over'. 'Describe fully' needs every detail: the angle AND its direction, the mirror line, the scale factor AND the centre.

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Further Mathematics Paper 1 to 4 standard.

Finish the four checks above to unlock the exam questions

Exam style+20 XP

Find the matrix that represents a rotation through \(45^{\circ}\) clockwise about the origin.

Exam style+20 XP

A shear with the \(y\)-axis invariant maps the point \((2, 1)\) to \((2, 7)\). Find its matrix.

Exam style+20 XP

Describe fully the transformation represented by \(\begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}\).

Exam style+20 XP

A triangle of area 5 units\(^{2}\) is transformed by \(\mathbf{A} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\) and its image is then transformed by \(\mathbf{B} = \begin{pmatrix} 3 & 0 \\ 1 & -1 \end{pmatrix}\). Find the area of the final image.

Skill unlocked

Columns read, products ordered, areas scaled.

This skill is now recorded on your profile. Keep the chain going.

-Rank -Level -Score -Topics

Go deeper · practice

Six original Cambridge-style questions

Standard matrices from the unit square, a rotation followed by a reflection, finding an object with the inverse, the order of a stretch and a shear, describing a matrix fully, and a proof that two reflections make a rotation.