AS June 2023 Paper 1 Q6
6 The matrix \(\mathbf{M}\) is \(\begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}\).
(a) Calculate \(\mathbf{M}^2\), \(\mathbf{M}^3\) and \(\mathbf{M}^4\). [2]
(b) Hence make a conjecture about the matrix \(\mathbf{M}^n\). [1]
(c) Prove your conjecture. [5]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{M}^2 = \begin{pmatrix} 3 & 2 \\ -2 & -1 \end{pmatrix},\ \mathbf{M}^3 = \begin{pmatrix} 4 & 3 \\ -3 & -2 \end{pmatrix},\ \mathbf{M}^4 = \begin{pmatrix} 5 & 4 \\ -4 & -3 \end{pmatrix}\) | B2 | 1.1 |
| [2] |
Notes
B2: Allow B1 if \(\mathbf{M}^2\) correct
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{M}^n = \begin{pmatrix} n + 1 & n \\ -n & -n + 1 \end{pmatrix}\) | B1 | 1.1 |
| [1] |
Notes
B1: oe – allow correct unsimplified expressions
allow if correct expression is seen in part (c)
| Scheme | Marks | AO |
|---|---|---|
| \(n = 1\): \(\mathbf{M}^1 = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}\ \left[= \begin{pmatrix} 1 + 1 & 1 \\ -1 & -1 + 1 \end{pmatrix}\right]\) so true | B1 | 2.1 |
| Assume true for \(n = k\), so \(\mathbf{M}^k = \begin{pmatrix} k + 1 & k \\ -k & -k + 1 \end{pmatrix}\) | 2.1 | |
| \(\mathbf{M}^{k+1} = \begin{pmatrix} k + 1 & k \\ -k & -k + 1 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}\) | M1 | 1.1 |
| \(= \begin{pmatrix} k + 2 & k + 1 \\ -k - 1 & -k \end{pmatrix}\) | A1 | |
| \(= \begin{pmatrix} (k + 1) + 1 & k + 1 \\ -(k + 1) & -(k + 1) + 1 \end{pmatrix}\) So true for \(n = k + 1\) | A1 | |
| As true for \(n = 1\), and if true for \(n = k\) then true for \(n = k + 1\), true for all \(n\). | B1 | 2.2a |
| [5] |
Notes
M1: or \(\mathbf{M} \times \mathbf{M}^k\)
A1: (2nd) or using target expression
B1: (2nd) dep first three marks gained. Must have if ... then... (oe)