June 2018 Paper 2 Q10
10 The transformation matrix \(\ \begin{pmatrix} a & b \\ 2a & 3b \end{pmatrix}\ \) maps the point \((1,\ -3)\) onto the point \((1,\ 4)\)
Work out the values of \(a\) and \(b\).
You must show your working. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\begin{pmatrix} a & b \\ 2a & 3b \end{pmatrix}\begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}\) | M1 | oe implied by a correct equation |
| \(a - 3b = 1\) | A1 | oe may be implied by correct answers |
| \(2a - 9b = 4\) | A1 | |
| Correct elimination of a variable from their 2 linear equations with both equations having the same two variables | M1 | eg \(3a - 2a = 3 - 4\) or \(-6b - -9b = 2 - 4\) |
| \(a = -1 \qquad b = -\dfrac{2}{3}\) | A1 | must be exact values |
Additional guidance
| \(\begin{pmatrix} a - 3b \\ 2a - 9b \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}\) implies M1 but does not imply any correct equations | |
| If the same method is used for both \(a\) and \(b\) (eg equates coefficients and eliminates a variable), mark the attempt that favours the student | |
| Ignore commas and lines within matrices | |
| Allow missing brackets if arrays are correct dimensions | |
| Answers only \(\ a = -1\) and \(b = -\dfrac{2}{3}\) | 5 marks |
| Allow use of \(\begin{pmatrix} 1 \\ 3 \end{pmatrix}\) as a misread If solution \(a = -1\) and \(b = \dfrac{2}{3}\) | M1A0A0 M1A1ft (A1ft after misread) |
| Transposing \(\begin{pmatrix} 1 \\ -3 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 4 \end{pmatrix}\) is not a misread (could still score 2nd M1) | |
| \(\begin{pmatrix} 1 \\ -3 \end{pmatrix}\begin{pmatrix} a & b \\ 2a & 3b \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}\) followed by \(a - 3b = 1\) and \(2a - 9b = 4\) (could still score 2nd M1) | M1A1A1 |
| \(\begin{pmatrix} 1 \\ -3 \end{pmatrix}\begin{pmatrix} a & b \\ 2a & 3b \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}\) with only one of \(a - 3b = 1\) and \(2a - 9b = 4\) (could still score 2nd M1) | M1A1A0 |
| \(\begin{pmatrix} 1 \\ -3 \end{pmatrix}\begin{pmatrix} a & b \\ 2a & 3b \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}\) with neither equation correct # (could still score 2nd M1) | M0A0A0 |