June 2016 Paper 2 Q20

AQAOld spec4 marksMatrices

20 Under the transformation represented by \(\ \begin{pmatrix} -1 & -3 \\ 2 & 4 \end{pmatrix}\ \) the image of point \(P(a, 2)\) is point \(Q\).

Can point \(Q\) be the same as point \(P\)?

You must show your working. [4 marks]

Mark scheme

Answer	Mark	Comments
\(\begin{pmatrix} -1 & -3 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} a \\ 2 \end{pmatrix}\) or \(\begin{pmatrix} -a - 6 \\ 2a + 8 \end{pmatrix}\)	M1	Allow \((-a - 6 \quad 2a + 8)\)
\(-a - 6 = a\) or \(2a + 8 = 2\)	M1	oe linear equation(s) (not \(a = -3\)) Implies M1 M1
\(-a - 6 = a\) and \(2a + 8 = 2\)	A1	oe equations (not \(a = -3\))
Shows both equations have a common solution (\(a = -3\)) and Yes	A1ft	ft M1 M1 A0 Must show that their two linear equations do not have a common solution and No SC4 \(\begin{pmatrix} -1 & -3 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} -3 \\ 2 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}\) and Yes SC3 \(\begin{pmatrix} -1 & -3 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} -3 \\ 2 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}\)

\(\begin{pmatrix} a \\ 2 \end{pmatrix}\begin{pmatrix} -1 & -3 \\ 2 & 4 \end{pmatrix}\) is first M0 unless recovered
In matrices, allow missing brackets or inclusion of ‘fraction’ lines
Only one equation can score a maximum of M1 M1 A0 A0
\(a = -3\) with no correct working	Zero
\(\begin{pmatrix} -a - 6 \\ 2a + 8 \end{pmatrix} = \begin{pmatrix} a \\ 2 \end{pmatrix}\) with no further valid work	M1 M0 A0 A0
The final A mark may be seen in various ways eg1 Solves both equations obtaining \(a = -3\) each time and Yes (or shows that both equations simplify to \(2a = -6\) and Yes) eg2 Solves one equation obtaining \(a = -3\) and shows by substitution that \(a = -3\) satisfies the other equation and Yes eg3 Adds the two equations to obtain a correct statement and Yes \(\quad \begin{aligned} -2a - 6 &= 0 \\ \underline{2a + 8} &\underline{{}= 2} \\ 2 &= 2 \end{aligned}\)