A2 October 2021 Paper 2 Q9
9 The matrix \(\mathbf{A}\) is given by \(\mathbf{A} = \begin{pmatrix} 2 & 3 \\ 0 & 2 \end{pmatrix}\).
(a) By considering \(\mathbf{A}\), \(\mathbf{A}^2\), \(\mathbf{A}^3\) and \(\mathbf{A}^4\) make a conjecture about the form of the matrix \(\mathbf{A}^n\) in terms of \(n\) for \(n \geqslant 1\). [2]
(b) Use induction to prove the conjecture made in part (a). [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{A}^2 = \begin{pmatrix} 4 & 12 \\ 0 & 4 \end{pmatrix},\ \mathbf{A}^3 = \begin{pmatrix} 8 & 36 \\ 0 & 8 \end{pmatrix},\ \mathbf{A}^4 = \begin{pmatrix} 16 & 96 \\ 0 & 16 \end{pmatrix}\) | B1 | 2.2a |
| Conjecture: \(\mathbf{A}^n = \begin{pmatrix} 2^n & 3n \times 2^{n-1} \\ 0 & 2^n \end{pmatrix}\) | B1 | 2.2b |
| [2] |
Notes
B1: (first) BC
B1: (second) Allow this mark for any conjecture which works for \(n = 1\), 2, 3 and 4.
| Scheme | Marks | AO |
|---|---|---|
| Basis case: \(n = 1\): \(\mathbf{A}^1 = \begin{pmatrix} 2^1 & 3 \times 1 \times 2^0 \\ 0 & 2^1 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 0 & 2 \end{pmatrix} = \mathbf{A}\) so true for \(n = 1\) | B1 | 2.1 |
| Assume true for \(n = k\) ie \(\mathbf{A}^k = \begin{pmatrix} 2^k & 3k \times 2^{k-1} \\ 0 & 2^k \end{pmatrix}\) | M1 | 2.1 |
| \(\mathbf{A}^{k+1} = \mathbf{A}^k\mathbf{A} = \begin{pmatrix} 2^k & 3k \times 2^{k-1} \\ 0 & 2^k \end{pmatrix}\begin{pmatrix} 2 & 3 \\ 0 & 2 \end{pmatrix}\) \(= \begin{pmatrix} 2^{k+1} & 3 \times 2^k + 3k \times 2^k \\ 0 & 2^{k+1} \end{pmatrix}\) | M1 | 2.2a |
| \(= \begin{pmatrix} 2^{k+1} & 3(k + 1) \times 2^k \\ 0 & 2^{k+1} \end{pmatrix}\) So true for \(n = k \Rightarrow\) true for \(n = k + 1\). But true for \(n = 1\). So true for all positive integer \(n\) | A1 | 2.4 |
| [4] |
Notes
B1: Allow this mark even if the conjecture is wrong, provided that it works for \(n = 1\)
M1: (first) Must have statement in terms of some other variable than \(n\). Conjecture need not be correct.
M1: (second) Uses inductive hypothesis properly & expands
A1: AG. Manipulating terms correctly and convincingly to obtain required form. Some intermediate working must be seen and a clear conclusion must be given for the induction process.
A formal proof by induction is required for full marks.