A2 June 2021 Paper 1 Q5
5 The matrix \(\mathbf{M}\) is defined by \(\mathbf{M} = \begin{bmatrix} 3 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
Prove by induction that \(\mathbf{M}^n = \begin{bmatrix} 3^{n} & 3^{n} - 1 & -3^{n} + 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) for all integers \(n \geqslant 1\) [5 marks]
| Scheme | Marks | AO |
|---|---|---|
| Demonstrates the result for \(n = 1\) and states that it is true for \(n = 1\) | B1 | 1.1b |
| States the assumption that the result true for \(n = k\) Condone use of \(n\) instead of \(k\) | B1 | 2.4 |
| Writes \(\mathbf{M}^{k+1}\) as \(\mathbf{M}\mathbf{M}^k\) or \(\mathbf{M}^k\mathbf{M}\) Condone use of \(n\) instead of \(k\) | M1 | 3.1a |
| Calculates \(\mathbf{M}^{k+1}\) correctly (fully simplified) Condone use of \(n\) instead of \(k\) | A1 | 1.1b |
| Completes a rigorous argument by stating that It is true for \(n = 1\); that if it is true for \(n = k\) then it is true for \(n = k + 1\) And hence (by induction) true for all integers \(n \geqslant 1\) Do not condone use of \(n\) instead of \(k\) in the inductive step | R1 | 2.1 |
| (5 marks) |
Typical solution
Let \(n = 1\) then \(\mathbf{M}^1 = \begin{bmatrix} 3 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \mathbf{M}\) so the result is true for \(n = 1\)
Assume the result if true for \(n = k\):
\[\mathbf{M}^{k+1} = \begin{bmatrix} 3 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 3^{k} & 3^{k} - 1 & -3^{k} + 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]\[= \begin{bmatrix} 3^{k+1} & 3^{k+1} - 1 & -3^{k+1} + 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]Hence true for \(n = k + 1\)
It is true for \(n = 1\). If it is true for \(n = k\) then it is true for \(n = k + 1\). Hence true by induction for all integers \(n \geqslant 1\)