AS June 2022 Paper 1 Q2
2 Matrices \(\mathbf{A}\) and \(\mathbf{B}\) are given by \(\mathbf{A} = \begin{pmatrix} a & 1 \\ -1 & 3 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} -2 & 5 \\ -1 & 0 \end{pmatrix}\) where \(a\) is a constant.
(a) Find the following matrices.
- \(\mathbf{A} + \mathbf{B}\)
- \(\mathbf{AB}\)
- \(\mathbf{A}^2\) [3]
(b)
(i) Given that the determinant of \(\mathbf{A}\) is 25 find the value of \(a\). [2]
(ii) You are given instead that the following system of equations does not have a unique solution.\[\begin{aligned} ax + y &= -2 \\ -x + 3y &= -6 \end{aligned}\]Determine the value of \(a\). [2]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{A} + \mathbf{B} = \begin{pmatrix} a - 2 & 6 \\ -2 & 3 \end{pmatrix}\) | B1 | 1.1 |
| \(\mathbf{AB} = \begin{pmatrix} -2a - 1 & 5a \\ -1 & -5 \end{pmatrix}\) | B1 | 1.1 |
| \(\mathbf{A}^2 = \begin{pmatrix} a^2 - 1 & a + 3 \\ -a - 3 & 8 \end{pmatrix}\) | B1 | 1.1 |
| [3] |
Notes
Any double signs must be simplified correctly
| Scheme | Marks | AO |
|---|---|---|
| (i) \((\det\mathbf{A}) = a \times 3 - 1 \times -1\) | M1 | 1.1 |
| \(3a + 1 = 25 \Rightarrow a = 8\) | A1 | 1.1 |
| [2] | ||
| (ii) (System reduces to \(\mathbf{A}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -2 \\ -6 \end{pmatrix}\) so no unique solution \(\Rightarrow\)) \((\det\mathbf{A}) = 3a + 1 = 0\) | M1 | 3.1a |
| \(\therefore a = -\tfrac{1}{3}\) | A1 | 1.1 |
| [2] |
Notes
(b)(i)
M1: Correct expansion of determinant of \(\mathbf{A}\)
(b)(ii)
M1: Setting their determinant to 0 if it is a linear function of \(a\).
Answer only is ok here
(b)(ii) Alternate solution
| Scheme | Marks |
|---|---|
| Multiplying the first equation by 3 gives: \(3ax + 3y = -6\) \(-x + 3y = -6\) | M1 |
| These two equations are the same if \(3a = -1 \to a = \dfrac{-1}{3}\) | A1 |
A1: Or subtracting gives \((3a + 1)x = 0 \to a = \frac{-1}{3}\)