AS & A Level · Practice questions · Matrices and transformations

Build it, combine it, undo it.

Six original Cambridge-style questions on matrices and transformations: matrices for rotations, reflections, stretches and shears, combined transformations in the right order, inverses, area scale factors and describing a matrix fully.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Columns first, order right to left, determinant for area.

Build a matrix from the images of \((1, 0)\) and \((0, 1)\): they are the columns.
Combine right to left: 'B followed by A' is \(\mathbf{AB}\). Test the product on \((1, 0)\).
Undo with the inverse: the object is \(\mathbf{M}^{-1}\) times the image.
Area: multiply by \(|\det \mathbf{M}|\); a negative determinant reverses orientation.
Describe fully: take out \(k = \sqrt{|\det \mathbf{M}|}\), match what is left to a rotation or reflection, and give the angle and direction, or the mirror line, plus the centre.

Recall

[2 marks]

Write down the matrix that represents a reflection in the line \(y = -x\), and use it to find the image of the point \((3, 1)\).

\((1, 0)\) goes to \((0, -1)\) and \((0, 1)\) goes to \((-1, 0)\), so the matrix is \(\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\). ✓
\(\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ -3 \end{pmatrix}\), so the image is \((-1, -3)\). ✓

Calculation

[4 marks]

The matrix \(\mathbf{M} = \begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}\) represents a transformation \(T\).

(a) The triangle with vertices \((0, 0)\), \((2, 0)\) and \((0, 3)\) is transformed by \(T\). Find the area of its image.

(b) Find the coordinates of the point that \(T\) maps to \((7, 6)\).

(a) The triangle has area \(\tfrac{1}{2} \times 2 \times 3 = 3\). \(\det \mathbf{M} = 6 - 2 = 4\), so the image has area \(3 \times 4 = 12\). ✓
(b) \(\mathbf{M}^{-1} = \dfrac{1}{4}\begin{pmatrix} 2 & -1 \\ -2 & 3 \end{pmatrix}\). ✓
\(\mathbf{M}^{-1}\begin{pmatrix} 7 \\ 6 \end{pmatrix} = \dfrac{1}{4}\begin{pmatrix} 14 - 6 \\ -14 + 18 \end{pmatrix} = \dfrac{1}{4}\begin{pmatrix} 8 \\ 4 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\). ✓
The point is \((2, 1)\); check: \(\mathbf{M}\begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 7 \\ 6 \end{pmatrix}\). ✓

Calculation

[4 marks]

\(\mathbf{S}\) is a stretch parallel to the \(x\)-axis with scale factor 3, and \(\mathbf{H}\) is the shear with the \(x\)-axis invariant that maps \((0, 1)\) to \((2, 1)\).

(a) Find the matrix for \(\mathbf{S}\) followed by \(\mathbf{H}\), and the matrix for \(\mathbf{H}\) followed by \(\mathbf{S}\).

(b) Find the image of \((1, 1)\) under each combined transformation, and state the area scale factor of each.

(a) \(\mathbf{S} = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}\), \(\mathbf{H} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\). ✓
\(\mathbf{S}\) then \(\mathbf{H}\): \(\mathbf{HS} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix}\). ✓
\(\mathbf{H}\) then \(\mathbf{S}\): \(\mathbf{SH} = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 6 \\ 0 & 1 \end{pmatrix}\). ✓
(b) \(\mathbf{HS}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}\) and \(\mathbf{SH}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 9 \\ 1 \end{pmatrix}\): the order matters. ✓
Both determinants are \(3 \times 1 - 0 = 3\), so both combined transformations multiply areas by 3 (\(\det \mathbf{S} \times \det \mathbf{H} = 3 \times 1\) in either order). ✓

Analysis

[5 marks]

A rotation through \(60^{\circ}\) anticlockwise about \(O\) is followed by a reflection in the \(x\)-axis.

(a) Find the matrix \(\mathbf{N}\) of the combined transformation, in exact form.

(b) Show that \(\mathbf{N}\) represents a single reflection, and find the equation of its mirror line.

(a) \(\mathbf{R} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\), \(\mathbf{X} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\); rotation first, so \(\mathbf{N} = \mathbf{XR}\). ✓
\(\mathbf{N} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}\). ✓
(b) Compare with \(\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}\): \(\cos 2\theta = \tfrac{1}{2}\) and \(\sin 2\theta = -\tfrac{\sqrt{3}}{2}\) fit every entry, with \(2\theta = -60^{\circ}\), so \(\theta = -30^{\circ}\). ✓
The mirror line is \(y = x\tan(-30^{\circ})\), that is \(y = -\dfrac{x}{\sqrt{3}}\) (and \(\det \mathbf{N} = -1\), as for any reflection). ✓
Check with the point \((\sqrt{3}, -1)\) on that line: \(\mathbf{N}\begin{pmatrix} \sqrt{3} \\ -1 \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \\ -\frac{3}{2} + \frac{1}{2} \end{pmatrix} = \begin{pmatrix} \sqrt{3} \\ -1 \end{pmatrix}\), unchanged. ✓

Analysis

[6 marks]

The matrix \(\mathbf{M}\) is \(\begin{pmatrix} 3 & 4 \\ 4 & -3 \end{pmatrix}\).

(a) Find \(\det \mathbf{M}\) and state what it tells you about the transformation.

(b) Show that \(\mathbf{M}\) maps \((2, 1)\) to a multiple of itself.

(d) Find \(\mathbf{M}^{2}\) and hence write down \(\mathbf{M}^{-1}\).

(a) \(\det \mathbf{M} = -9 - 16 = -25\): areas are multiplied by 25 and the orientation is reversed. ✓
(b) \(\mathbf{M}\begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 10 \\ 5 \end{pmatrix} = 5\begin{pmatrix} 2 \\ 1 \end{pmatrix}\). ✓
(c) \(\mathbf{M} = 5\begin{pmatrix} \frac{3}{5} & \frac{4}{5} \\ \frac{4}{5} & -\frac{3}{5} \end{pmatrix}\), a reflection matrix with \(\cos 2\theta = \tfrac{3}{5}\), \(\sin 2\theta = \tfrac{4}{5}\). Part (b) shows the line through \(O\) and \((2, 1)\) is mapped to itself, and \(\tan\theta = \dfrac{\sin 2\theta}{1 + \cos 2\theta} = \dfrac{4/5}{8/5} = \dfrac{1}{2}\). ✓
So \(\mathbf{M}\) is a reflection in the line \(y = \tfrac{1}{2}x\) combined with an enlargement, scale factor 5, centre \(O\) (in either order); \(5^{2} = 25 = |\det \mathbf{M}|\) agrees. ✓
(d) \(\mathbf{M}^{2} = \begin{pmatrix} 9 + 16 & 12 - 12 \\ 12 - 12 & 16 + 9 \end{pmatrix} = 25\mathbf{I}\). ✓
So \(\mathbf{M}\left(\tfrac{1}{25}\mathbf{M}\right) = \mathbf{I}\), giving \(\mathbf{M}^{-1} = \dfrac{1}{25}\begin{pmatrix} 3 & 4 \\ 4 & -3 \end{pmatrix}\): undo the enlargement with factor \(\tfrac{1}{5}\) and reflect again. ✓

Proof

[5 marks]

\(\mathbf{R}_{\alpha}\) denotes the reflection in the line \(y = x\tan\alpha\), with matrix \(\begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix}\).

Prove that the reflection \(\mathbf{R}_{\beta}\) followed by the reflection \(\mathbf{R}_{\alpha}\) is a rotation about \(O\), and find its angle.

\(\mathbf{R}_{\beta}\) first, so the combined matrix is \(\mathbf{R}_{\alpha}\mathbf{R}_{\beta}\). ✓
Top left: \(\cos 2\alpha\cos 2\beta + \sin 2\alpha\sin 2\beta = \cos(2\alpha - 2\beta)\). Top right: \(\cos 2\alpha\sin 2\beta - \sin 2\alpha\cos 2\beta = -\sin(2\alpha - 2\beta)\). ✓
Bottom left: \(\sin 2\alpha\cos 2\beta - \cos 2\alpha\sin 2\beta = \sin(2\alpha - 2\beta)\). Bottom right: \(\sin 2\alpha\sin 2\beta + \cos 2\alpha\cos 2\beta = \cos(2\alpha - 2\beta)\). ✓
So \(\mathbf{R}_{\alpha}\mathbf{R}_{\beta} = \begin{pmatrix} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \end{pmatrix}\) with \(\phi = 2(\alpha - \beta)\), the rotation matrix. ✓
The combination is a rotation about \(O\) through \(2(\alpha - \beta)\) anticlockwise: twice the angle from the first mirror line to the second. (Check: \(\det = (-1)(-1) = 1\).) ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.