A2 June 2022 Paper 1 Q6
6 Prove by mathematical induction that \(\begin{pmatrix} 2 & 0 \\ -1 & 1 \end{pmatrix}^n = \begin{pmatrix} 2^n & 0 \\ 1 - 2^n & 1 \end{pmatrix}\) for all positive integers \(n\). [5]
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} 2 & 0 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 1 - 2 & 1 \end{pmatrix}\) so true when \(n = 1\) | B1 | 2.1 |
| [Assume true for \(n = k\)] \(\begin{pmatrix} 2 & 0 \\ -1 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 2^k & 0 \\ 1 - 2^k & 1 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ -1 & 1 \end{pmatrix}\) | M1 | 2.1 |
| \(= \begin{pmatrix} 2^{k+1} & 0 \\ 2 - 2^{k+1} - 1 & 1 \end{pmatrix}\) | M1 | 2.1 |
| \(= \begin{pmatrix} 2^{k+1} & 0 \\ 1 - 2^{k+1} & 1 \end{pmatrix}\) [so true for \(n = k + 1\)] | A1 | 2.3 |
| As true for \(n = 1\), and if true for \(n = k\) then true for \(n = k + 1\), true for all \(n\) | B1cao | 2.4 |
| [5] |
Notes
M1: Intermediate step seen
B1cao: Must receive all 4 previous marks for this to be awarded