AS June 2024 Paper 1 Q6
6 You are given that \(\mathbf{M} = \begin{pmatrix} 4 & -9 \\ 1 & -2 \end{pmatrix}\).
(a) Prove that \(\mathbf{M}^n = \begin{pmatrix} 1 + 3n & -9n \\ n & 1 - 3n \end{pmatrix}\) for all positive integers \(n\). [6]
(b) A student thinks that this formula, when \(n = 0\) and \(n = -1\), gives the identity matrix and the inverse matrix \(\mathbf{M}^{-1}\) respectively.
Determine whether the student is correct. [3]
Determine whether the student is correct. [3]
| Scheme | Marks | AO |
|---|---|---|
| When \(n = 1\), \(\mathbf{M}^1 = \begin{pmatrix} 1 + 3 & -9 \\ 1 & 1 - 3 \end{pmatrix} = \begin{pmatrix} 4 & -9 \\ 1 & -2 \end{pmatrix}\) as required | B1 | 2.1 |
| Assume true for \(n = k\), so \(\mathbf{M}^k = \begin{pmatrix} 1 + 3k & -9k \\ k & 1 - 3k \end{pmatrix}\) | M1 | 2.1 |
| \(\mathbf{M}^{k+1} = \begin{pmatrix} 1 + 3k & -9k \\ k & 1 - 3k \end{pmatrix}\begin{pmatrix} 4 & -9 \\ 1 & -2 \end{pmatrix}\) | M1 | 2.1 |
| \(= \begin{pmatrix} 4 + 3k & -9 - 9k \\ k + 1 & -2 - 3k \end{pmatrix}\) | A1 | 1.1 |
| \(= \begin{pmatrix} 1 + 3(k + 1) & -9(k + 1) \\ k + 1 & 1 - 3(k + 1) \end{pmatrix}\) so true for \(n = k + 1\) | A1 | 2.2a |
| As true for \(n = 1\), and if true for \(n = k\) then true for \(n = k + 1\), true for all positive integers \(n\) | A1 | 2.4 |
| [6] |
Notes
B1: check true for \(n = 1\)
A1: (final) cao dep previous A1
| Scheme | Marks | AO |
|---|---|---|
| \(n = 0\) gives \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathbf{I}\) so true for \(n = 0\) | B1 | 2.3 |
| Formula with \(n = -1\) gives \(\begin{pmatrix} -2 & 9 \\ -1 & 4 \end{pmatrix}\) | B1 | 1.1 |
| \(\det\mathbf{M} = 4 \times -2 - (-9) \times 1 = 1\), so \(\mathbf{M}^{-1} = \begin{pmatrix} -2 & 9 \\ -1 & 4 \end{pmatrix}\) so true for \(n = -1\) | B1 | 2.3 |
| [3] |
Notes
B1: [ \(= \mathbf{I}\) may be inferred from ‘student is correct’]
B1: (3rd) Must show evidence for inverse
e.g. calculate det or show \(\mathbf{MM}^{-1} = \mathbf{I}\)