AS June 2024 Paper 1 Q6
6 You are given that \(\mathbf{A} = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}\) where \(a\) is a constant.
Prove by induction that \(\mathbf{A}^n = \begin{pmatrix} 1 & an \\ 0 & 1 \end{pmatrix}\) for all integers \(n \geqslant 1\). [5]
| Scheme | Marks | AO |
|---|---|---|
| Basis case: \(n = 1\) \(\mathbf{A}^1 = \begin{pmatrix} 1 & a \times 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}\quad [= \mathbf{A}]\) [So it’s true when \(n = 1\)] | B1 | 2.1 |
| Assume true for \(n = k\) so assume \(\mathbf{A}^k = \begin{pmatrix} 1 & ak \\ 0 & 1 \end{pmatrix}\) | M1 | 2.1 |
| \(\therefore \mathbf{A}^{k+1} = \mathbf{A}^k\mathbf{A} = \begin{pmatrix} 1 & ak \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}\) by our inductive hypothesis. | M1 | 1.1 |
| \(\therefore \mathbf{A}^{k+1} = \begin{pmatrix} 1 & a + ak \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & a(k + 1) \\ 0 & 1 \end{pmatrix}\) | A1 | 2.2a |
| So true for \(n = k \Rightarrow\) true for \(n = k + 1\). But true for \(n = 1\). Therefore true for all [integer] \(n \geqslant 1\). | A1 | 2.4 |
| [5] |
Notes
B1: \(\mathbf{A}^1\) and \(a \times 1\) must both be seen explicitly.
Accept “\(= \mathbf{A}\)” instead of “therefore true when \(n = 1\)”
M1: Setting up the inductive hypothesis properly.
M1: Considering \(\mathbf{A}^{k+1}\) and using their inductive hypothesis. Could also consider \(\mathbf{AA}^k\)
Jumping to \(\mathbf{A}^{k+1} = \begin{pmatrix} 1 & a(k + 1) \\ 0 & 1 \end{pmatrix}\) without seeing the two matrices multiplied together first scores M0.
A1: Must see \(a + ak\) appear before being factorised.
A1: Clear conclusion for induction process. Must mention both basis case and that statement true for \(k\) implies true for \(k + 1\). BOD missing the word integer
If B0 not awarded as \(\mathbf{A}^1\) or \(a \times 1\) (or both) not seen allow A1 here (but must have attempt at base case)