AS June 2019 Paper 1 Q8
8 In this question you must show detailed reasoning.
\(\mathbf{M}\) is the matrix \(\begin{pmatrix} 1 & 6 \\ 0 & 2 \end{pmatrix}\).
Prove that \(\mathbf{M}^n = \begin{pmatrix} 1 & 3(2^{n+1} - 2) \\ 0 & 2^n \end{pmatrix}\), for any positive integer \(n\). [6]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{M}^1 = \begin{pmatrix} 1 & 3(2^{1+1} - 2) \\ 0 & 2^1 \end{pmatrix} = \begin{pmatrix} 1 & 3 \times 2 \\ 0 & 2 \end{pmatrix}\) \(= \begin{pmatrix} 1 & 6 \\ 0 & 2 \end{pmatrix} = \mathbf{M}\) | B1 | 3.1a |
| Assume true for \(n = k\) ie \(\mathbf{M}^k = \begin{pmatrix} 1 & 3(2^{k+1} - 2) \\ 0 & 2^k \end{pmatrix}\). | M1 | 2.1 |
| \(\mathbf{M}^{k+1} = \mathbf{MM}^k = \begin{pmatrix} 1 & 6 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 1 & 3(2^{k+1} - 2) \\ 0 & 2^k \end{pmatrix}\) | M1 | 1.1 |
| \(\begin{pmatrix} 1 & 3(2^{k+1} - 2) + 6 \times 2^k \\ 0 & 2 \times 2^k \end{pmatrix}\) | M1 | 1.1 |
| \(\begin{pmatrix} 1 & 3(2^{k+1} - 2) + 3 \times 2^{k+1} \\ 0 & 2 \times 2^k \end{pmatrix} = \begin{pmatrix} 1 & 3(2^{(k+1)+1} - 2) \\ 0 & 2^{(k+1)} \end{pmatrix}\) | A1 (AG) | 2.2a |
| So true for \(n = k \Rightarrow\) true for \(n = k + 1\). But true for \(n = 1\). So true for all positive integer \(n\) | E1 | 2.4 |
| [6] |
Notes
DR: Detailed reasoning required for this question
B1: Full details must be shown.
M1: (1st) Must have statement in terms of some other variable than \(n\).
Watch out for \(2^{k+1}\) appearing prematurely as bottom right element
M1: (2nd) Uses inductive hypothesis properly
or
\(\mathbf{M}^k\mathbf{M} = \begin{pmatrix} 1 & 3(2^{k+1} - 2) \\ 0 & 2^k \end{pmatrix}\begin{pmatrix} 1 & 6 \\ 0 & 2 \end{pmatrix}\)
M1: (3rd) Genuine attempt at matrix multiplication (ie columns multiplied into rows)
\(\begin{pmatrix} 1 & 6 + 2 \times 3(2^{k+1} - 2) \\ 0 & 2^k \times 2 \end{pmatrix}\)
A1: Simplification with sufficient working to establish truth for \(k + 1\)
Must see at least one stage of working between expression for \(\mathbf{M}^k\mathbf{M}\) (or \(\mathbf{MM}^k\)) and required expression for \(\mathbf{M}^{k+1}\) (e.g.:
\(\begin{pmatrix} 1 & 3(2 + 2 \times 2^{k+1} - 4) \\ 0 & 2^k \times 2 \end{pmatrix}\)
\(= \begin{pmatrix} 1 & 3(2^{(k+2)} - 2) \\ 0 & 2^{(k+1)} \end{pmatrix}\)
E1: Clear conclusion for induction process. Needs a fully correct proof which usually means all other marks awarded. Cannot be awarded if there are mistakes in the proof.
A formal proof by induction is required for full marks.
SC If B1M1M1M1 gained, but A0 because of lack of intermediate step of working then allow SC B1 for fully correct ending statement