June 2019 Paper 1 Q6
6 The transformation matrix \(\begin{pmatrix} 2a & b \\ -b & -a \end{pmatrix}\) maps the point \((3,\ 4)\) onto the point \((8,\ -7)\)
Work out the values of \(a\) and \(b\). [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\begin{pmatrix} 2a & b \\ -b & -a \end{pmatrix}\begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 8 \\ -7 \end{pmatrix}\) | M1 | |
| \(6a + 4b = 8\) and \(-3b - 4a = -7\) | M1dep | oe allow these to be written as a matrix equation in all likelihood this will imply M2 as the matrices may not be seen |
| Solve eg \(12a + 8b = 16\) and \(-12a - 9b = -21\) or \(18a + 12b = 24\) and \(-16a - 12b = -28\) or substitution eg \(a = \dfrac{8 - 4b}{6}\) and \(\dfrac{-4(8 - 4b)}{6} - 3b = -7\) or \(b = \dfrac{4 - 3a}{2}\) and \(-4a - \dfrac{3(4 - 3a)}{2} = -7\) | M1dep | oe for making coefficients of \(a\) or \(b\) equal dependent on first M1 only oe |
| \(a = -2\) or \(b = 5\) | A1 | |
| \(a = -2\) and \(b = 5\) | A1 |
Additional guidance
| Matrices wrong way round can be recovered by correct equations in second M | |
| Point written as coordinates rather than a matrix can be recovered by correct equations in second M | |
| \(a\) or \(b\) correct with no incorrect working | M1, M1, M1, A1, A0 |