June 2024 Paper 1 Q7
7 A point \(P\ (x, y)\) is transformed using the transformation represented by \(\quad \begin{pmatrix} 4 & 0 \\ -2 & 3 \end{pmatrix}\)
The image of \(P\) is \((-8, 7)\)
Work out the values of \(x\) and \(y\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\begin{pmatrix} 4 & 0 \\ -2 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -8 \\ 7 \end{pmatrix}\) | M1 | may be inferred from correct second M |
| \(4x = -8\) and \(-2x + 3y = 7\) | M1dep | oe may be seen in vector notation |
| \((x =)\ -2\) and \((y =)\ 1\) | A1 |
Additional guidance
| \(x = -2\) if no other marks scored | SC1 |
| \(\begin{pmatrix} 4 & 0 \\ -2 & 3 \end{pmatrix}\begin{pmatrix} -8 \\ 7 \end{pmatrix} = \begin{pmatrix} -32 \\ 37 \end{pmatrix}\) or \((x =)\ -32\) and \((y =)\ 37\) | SC1 |
| Condone \(\begin{pmatrix} x \\ y \end{pmatrix}\begin{pmatrix} 4 & 0 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} -8 \\ 7 \end{pmatrix}\) if recovered to get correct equations | |
| Condone answer left as column vector \(\begin{pmatrix} -2 \\ 1 \end{pmatrix}\) or as coordinates \((-2, 1)\) |