FP1 June 2009 Q7
7. \(\mathbf{A} = \begin{pmatrix} a & -2 \\ -1 & 4 \end{pmatrix}\), where \(a\) is a constant.
(a) Find the value of \(a\) for which the matrix \(\mathbf{A}\) is singular. (2)
\[\mathbf{B} = \begin{pmatrix} 3 & -2 \\ -1 & 4 \end{pmatrix}\](b) Find \(\mathbf{B}^{-1}\). (3)
The transformation represented by \(\mathbf{B}\) maps the point \(P\) onto the point \(Q\).
Given that \(Q\) has coordinates \((k - 6,\ 3k + 12)\), where \(k\) is a constant,
(c) show that \(P\) lies on the line with equation \(y = x + 3\). (3)
| Scheme | Marks |
|---|---|
| Use \(4a - (-2 \times -1) = 0 \quad \Rightarrow \quad a = \dfrac{1}{2}\) | M1, A1 |
| (2) |
Notes
(a) Allow sign slips for first M1
| Scheme | Marks |
|---|---|
| Determinant: \((3 \times 4) - (-2 \times -1) = 10 \quad (\Delta)\) | M1 |
| \(\mathbf{B}^{-1} = \dfrac{1}{10}\begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}\) | M1 A1cso |
| (3) |
Notes
(b) Allow sign slip for determinant for first M1 (This mark may be awarded for 1/10 appearing in inverse matrix.)
Second M1 is for correctly treating the 2 by 2 matrix, ie for \(\begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}\)
Watch out for determinant (3 + 4) – (−1 + −2) = 10 – M0 then final answer is A0
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{10}\begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} k - 6 \\ 3k + 12 \end{pmatrix}, \quad = \dfrac{1}{10}\begin{pmatrix} 4(k - 6) + 2(3k + 12) \\ (k - 6) + 3(3k + 12) \end{pmatrix}\) | M1, A1ft |
| \(\begin{pmatrix} k \\ k + 3 \end{pmatrix}\) Lies on \(y = x + 3\) | A1 |
| (3) | |
| [8] |
Alternatives
| Scheme | Marks |
|---|---|
| (c) \(\begin{pmatrix} 3 & -2 \\ -1 & 4 \end{pmatrix}\begin{pmatrix} x \\ x + 3 \end{pmatrix}, \quad = \begin{pmatrix} 3x - 2(x + 3) \\ -x + 4(x + 3) \end{pmatrix}\), | M1, A1, |
| \(= \begin{pmatrix} x - 6 \\ 3x + 12 \end{pmatrix}\), which was of the form \((k - 6,\ 3k + 12)\) | A1 |
| Scheme | Marks |
|---|---|
| Or \(\begin{pmatrix} 3 & -2 \\ -1 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}, \quad = \begin{pmatrix} 3x - 2y \\ -x + 4y \end{pmatrix} = \begin{pmatrix} k - 6 \\ 3k + 12 \end{pmatrix}\), and solves simultaneous equations | M1 |
| Both equations correct and eliminate one letter to get \(x = k\) or \(y = k + 3\) or \(10x - 10y = -30\) or equivalent. | A1 |
| Completely correct work (to \(x = k\) and \(y = k + 3\)), and conclusion lies on \(y = x + 3\) | A1 |
Notes
(c) M1 for multiplying matrix by appropriate column vector
A1 correct work (ft wrong determinant)
A1 for conclusion