A2 June 2019 Paper 1 Q11

OCR MEICurrent spec12 marksMatrices

(a) Specify fully the transformations represented by the following matrices.

\(\mathbf{M}_1 = {\def\arraystretch{1.6}\begin{pmatrix} \frac{3}{5} & -\frac{4}{5} \\ \frac{4}{5} & \frac{3}{5} \end{pmatrix}}\)
\(\mathbf{M}_2 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\)

[4]

(b) Find the equation of the mirror line of the reflection R represented by the matrix \(\mathbf{M}_3 = \mathbf{M}_1\mathbf{M}_2\). [5]

(c) It is claimed that the reflection represented by the matrix \(\mathbf{M}_4 = \mathbf{M}_2\mathbf{M}_1\) has the same mirror line as R. Explain whether or not this claim is correct. [3]

Mark scheme (a)

Scheme	Marks	AO
\(\mathbf{M}_1\) rotation	M1	3.1a
through \(\cos^{-1}(3/5)\) or \(53.1^\circ\) or 0.927 rads	A1	1.1b
anti-clockwise about O	A1	1.2
\(\mathbf{M}_2\) reflection in \(x\)-axis	B1	1.2
	[4]

Notes

A1: oe e.g. \(\sin^{-1}(4/5)\), \(\tan^{-1}(4/3)\); \(53^\circ\) or 0.93 rads or better

A1: or positive rotation about O

B1: or O\(x\) or \(y = 0\)

Mark scheme (b)

Scheme	Marks	AO
\(\mathbf{M}_3 = {\def\arraystretch{1.6}\begin{pmatrix} \frac{3}{5} & \frac{4}{5} \\ \frac{4}{5} & -\frac{3}{5} \end{pmatrix}}\)	B1	1.1b
\({\def\arraystretch{1.6}\begin{pmatrix} \frac{3}{5} & \frac{4}{5} \\ \frac{4}{5} & -\frac{3}{5} \end{pmatrix}}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\)	M1	3.1a
\(\Rightarrow \dfrac{3}{5}x + \dfrac{4}{5}y = x,\ \dfrac{4}{5}x - \dfrac{3}{5}y = y\)	A2	1.1b
\(\Rightarrow y = \frac{1}{2}x\) so \(y = \frac{1}{2}x\) is mirror line	A1	2.2a
	[5]

Notes

M1: attempt to find invariant points; or inv line \(y = mx\) [\(+c\)]: \(2m^2 + 3m - 2 = 0\) A1 \(\Rightarrow m = \frac{1}{2}, -2\) A1

A2: either or both

A1: accept valid geometric args

Alternative solution

Scheme	Marks
\(\dfrac{1}{1 + m^2}\begin{pmatrix} 1 - m^2 & 2m \\ 2m & m^2 - 1 \end{pmatrix} = {\def\arraystretch{1.6}\begin{pmatrix} \frac{3}{5} & \frac{4}{5} \\ \frac{4}{5} & -\frac{3}{5} \end{pmatrix}}\)	M1
\(\Rightarrow 2m^2 - 5m + 2 = 0\)	A1
\(\Rightarrow m = \frac{1}{2}\)	A2
	[5]

A2: must discount \(m = 2\)

Mark scheme (c)

Scheme	Marks	AO
\(\mathbf{M}_4 = {\def\arraystretch{1.6}\begin{pmatrix} \frac{3}{5} & -\frac{4}{5} \\ -\frac{4}{5} & -\frac{3}{5} \end{pmatrix}}\)	B1	1.1b
\(\mathbf{M}_4 \neq \mathbf{M}_3\) [so can’t represent same reflection]	M1	3.1a
so mirror line cannot be the same, and statement is incorrect	A1	2.4
	[3]

Notes

M1: or attempt to find mirror line as in part (b) \(\Rightarrow y = -\frac{1}{2}x\), so statement is incorrect. \(\mathbf{M}_4\) must be different