A2 June 2022 Paper 2 Q13

AQACurrent spec16 marksMatrices

(a) The matrix \(\mathbf{A}\) represents a reflection in the line \(y = mx\), where \(m\) is a constant.

Show that \(\mathbf{A} = \left(\dfrac{1}{m^2 + 1}\right)\begin{bmatrix} 1 - m^2 & 2m \\ 2m & m^2 - 1 \end{bmatrix}\)

You may use the result in the formulae booklet. [5 marks]

(b) The matrix \(\mathbf{B}\) is defined as \(\mathbf{B} = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\)

Show that \((\mathbf{BA})^2 = k\mathbf{I}\)

where \(\mathbf{I}\) is the \(2 \times 2\) identity matrix and \(k\) is an integer. [3 marks]

(c)

(i) The diagram below shows a point \(P\) and the line \(y = mx\)

Draw four lines on the diagram to demonstrate the result proved in part (b).

Label as \(P^{\prime}\) the image of \(P\) under the transformation represented by \((\mathbf{BA})^2\) [2 marks]

Axes x and y through O, with the line y = mx drawn from O and a point P just below the line, close to O

(ii) Explain how your completed diagram shows the result proved in part (b). [2 marks]

(d) The matrix \(\mathbf{C}\) is defined as \(\mathbf{C} = \begin{bmatrix} \dfrac{12}{5} & \dfrac{9}{5} \\[6pt] \dfrac{9}{5} & -\dfrac{12}{5} \end{bmatrix}\)

Find the value of \(m\) such that \(\mathbf{C} = \mathbf{BA}\)

Fully justify your answer. [4 marks]

Mark scheme (a)

Scheme	Marks	AO
States that \(m = \tan\theta\) and \(\mathbf{A} = \begin{bmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{bmatrix}\) Condone omission of A=	B1	1.2
Replaces \(\sin 2\theta\) and \(\cos 2\theta\) in the matrix with expressions in terms of trigonometric ratios of \(\theta\) or in terms of \(m\)	M1	3.1a
Deduces that one element of the matrix can be expressed as \(\cos^2\theta(1 - \tan^2\theta)\) or \(\cos^2\theta(1 - m^2)\) or \(2\tan\theta(\cos^2\theta)\) or \(2m\cos^2\theta\) or uses \(\cos 2\theta = \dfrac{1 - m^2}{1 + m^2}\) or \(\sin 2\theta = \dfrac{2m}{1 + m^2}\)	M1	2.2a
Uses \(\cos^2\theta = \dfrac{1}{m^2 + 1}\)	M1	1.1a
Completes a reasoned argument to obtain the required result Must include A=…	R1	2.1
	(5)

Typical solution

\[\mathbf{A} = \begin{bmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{bmatrix} \quad \text{where } \tan\theta = m\]\[\begin{aligned}\mathbf{A} &= \begin{bmatrix} \cos^2\theta - \sin^2\theta & 2\sin\theta\cos\theta \\ 2\sin\theta\cos\theta & \sin^2\theta - \cos^2\theta \end{bmatrix} \\ &= \cos^2\theta\begin{bmatrix} 1 - \tan^2\theta & 2\tan\theta \\ 2\tan\theta & \tan^2\theta - 1 \end{bmatrix} \\ &= \frac{1}{\sec^2\theta}\begin{bmatrix} 1 - m^2 & 2m \\ 2m & m^2 - 1 \end{bmatrix} \\ &= \frac{1}{\tan^2\theta + 1}\begin{bmatrix} 1 - m^2 & 2m \\ 2m & m^2 - 1 \end{bmatrix} \\ &= \left(\dfrac{1}{m^2 + 1}\right)\begin{bmatrix} 1 - m^2 & 2m \\ 2m & m^2 - 1 \end{bmatrix}\end{aligned}\]

Mark scheme (b)

Scheme	Marks	AO
Obtains BA (condone AB) or Uses the fact that B = 3I	M1	1.1a
Squares their BA or AB or 3IA and simplifies	M1	1.1a
Completes a reasoned argument to show the required result, using BA not AB or Uses and states the fact that \(\mathbf{A}^2 = \mathbf{I}\) because it is a reflection Condone missing \((\mathbf{BA})^2 =\)	R1	2.1
	(3)

Typical solution

\[\begin{aligned}\mathbf{BA} &= \left(\dfrac{1}{m^2 + 1}\right)\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\begin{bmatrix} 1 - m^2 & 2m \\ 2m & m^2 - 1 \end{bmatrix} \\ &= \left(\dfrac{1}{m^2 + 1}\right)\begin{bmatrix} 3(1 - m^2) & 6m \\ 6m & 3(m^2 - 1) \end{bmatrix}\end{aligned}\]\[\begin{aligned}(\mathbf{BA})^2 &= \left(\dfrac{1}{m^2 + 1}\right)^2\begin{bmatrix} 3(1 - m^2) & 6m \\ 6m & 3(m^2 - 1) \end{bmatrix}\begin{bmatrix} 3(1 - m^2) & 6m \\ 6m & 3(m^2 - 1) \end{bmatrix} \\ &= \left(\dfrac{1}{m^2 + 1}\right)^2\begin{bmatrix} 9(1 - m^2)^2 + 36m^2 & 0 \\ 0 & 36m^2 + 9(m^2 - 1)^2 \end{bmatrix} \\ &= \left(\dfrac{1}{m^2 + 1}\right)^2\begin{bmatrix} 9 + 18m^2 + 9m^4 & 0 \\ 0 & 9 + 18m^2 + 9m^4 \end{bmatrix} \\ &= \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} \\ &= 9\mathbf{I}\end{aligned}\]

The diagram with P reflected in y = mx, the image enlarged by scale factor 3 from O, that point reflected in y = mx, then enlarged again by scale factor 3 to give P prime, which lies on the line OP extended — Mark scheme (c)

Scheme	Marks	AO
(i) Uses the fact that B represents an enlargement, scale factor 3 or that A represents a reflection in \(y = mx\)	M1	1.2
Draws four correct lines on the diagram (the arrows and the dashed line are not needed), and labels \(P^{\prime}\) Condone A and B reversed	A1	1.1b
	(2)
(ii) Explains how at least one of their lines represents a relevant transformation.	E1	2.4
Explains how their lines represent the result of the transformations.	E1	2.4
	(2)

Mark scheme (d)

Scheme	Marks	AO
Expresses BA correctly in terms of \(m\) or calculates \(\mathbf{B}^{-1}\mathbf{C}\) correctly	B1	1.1b
Sets up an equation in \(m\)	M1	3.1a
Finds a correct solution of their correct equation in \(m\) (condone other solution(s) not rejected)	A1	1.1b
Uses a rigorous argument to obtain the required result, including clear reason for choosing \(m = \dfrac{1}{3}\)	R1	2.1
	(4)
	(16 marks)

Typical solution

\(\mathbf{C} = \mathbf{BA}\)

\[\begin{bmatrix} \dfrac{12}{5} & \dfrac{9}{5} \\[6pt] \dfrac{9}{5} & -\dfrac{12}{5} \end{bmatrix} = \left(\dfrac{1}{m^2 + 1}\right)\begin{bmatrix} 3(1 - m^2) & 6m \\ 6m & 3(m^2 - 1) \end{bmatrix}\]\[\frac{9}{5} = \frac{6m}{m^2 + 1}\]\[3m^2 - 10m + 3 = 0\]\[m = 3 \text{ or } m = \frac{1}{3}\]

\(m = 3\) gives \(\dfrac{3(1 - m^2)}{m^2 + 1} = -\dfrac{12}{5} \neq \dfrac{12}{5}\)

so discard \(m = 3\)

Correct value is \(m = \dfrac{1}{3}\)