FP1 January 2013 Q6
6. \[\mathbf{X} = \begin{pmatrix} 1 & a \\ 3 & 2 \end{pmatrix}, \text{ where } a \text{ is a constant.}\]
\[\mathbf{Y} = \begin{pmatrix} 1 & -1 \\ 3 & 2 \end{pmatrix}\]
The transformation represented by \(\mathbf{Y}\) maps the point \(A\) onto the point \(B\).
Given that \(B\) has coordinates \((1 - \lambda,\ 7\lambda - 2)\), where \(\lambda\) is a constant,
| Scheme | Marks |
|---|---|
| Determinant: \(2 - 3a = 0\) and solve for \(a = \) | M1 |
| So \(a = \tfrac{2}{3}\) or equivalent | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| Determinant: \((1 \times 2) - (3 \times -1) = 5\quad (\Delta)\) | |
| \(\mathrm{Y}^{-1} = \dfrac{1}{5}\begin{pmatrix} 2 & 1 \\ -3 & 1 \end{pmatrix}\quad \left[= \begin{pmatrix} 0.4 & 0.2 \\ -0.6 & 0.2 \end{pmatrix}\right]\) | M1A1 |
| (2) |
Notes
(b) M for \(\dfrac{1}{\text{their det}}\begin{pmatrix} 2 & 1 \\ -3 & 1 \end{pmatrix}\)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{5}\begin{pmatrix} 2 & 1 \\ -3 & 1 \end{pmatrix}\begin{pmatrix} 1 - \lambda \\ 7\lambda - 2 \end{pmatrix} = \dfrac{1}{5}\begin{pmatrix} 2 - 2\lambda + 7\lambda - 2 \\ -3 + 3\lambda + 7\lambda - 2 \end{pmatrix} = \begin{pmatrix} \lambda \\ 2\lambda - 1 \end{pmatrix}\) | M1depM1A1 A1 |
| (4) | |
| [8] |
Notes
(c) First M for their \(\mathbf{Y}^{-1}\mathbf{B}\) in correct order with \(\mathbf{B}\) written as a 2x1 matrix, second M dependent on first for attempt at multiplying their matrices resulting in a 2x1 matrix, first A for \(\lambda\), second A for \(2\lambda - 1\)
Alternative method for (c)
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} 1 & -1 \\ 3 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 - \lambda \\ 7\lambda - 2 \end{pmatrix}\) so \(x - y = 1 - \lambda\) and \(3x + 2y = 7\lambda - 2\) | M1M1 |
| Solve to give \(x = \lambda\) and \(y = 2\lambda - 1\) | A1A1 |
Alternative for (c)
First M to obtain two linear equations in \(x, y, \lambda\)
Second M for attempting to solve for \(x\) or \(y\) in terms of \(\lambda\)