AS June 2022 Paper 1 Q7
7. Prove by mathematical induction that, for \(n \in \mathbb{N}\)
\[\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}^n = \begin{pmatrix}1 - 6n & 9n\\ -4n & 1 + 6n\end{pmatrix}\](6)
| Scheme | Marks | AO |
|---|---|---|
| For \(n = 1\): \(\begin{pmatrix}1 - 6 \times 1 & 9 \times 1\\ -4 \times 1 & 1 + 6 \times 1\end{pmatrix} = \begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix} = \begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}^1\) So the statement is true for \(n = 1\) | B1 | 2.2a |
| Assume true for \(\boldsymbol{n = k}\), or Assume \(\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}^k = \begin{pmatrix}1 - 6k & 9k\\ -4k & 1 + 6k\end{pmatrix}\) | M1 | 2.5 |
| \(\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}^{k+1} = \begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}^k\times\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}\) OR \(\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}\times\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}^k\) | M1 | 2.1 |
| \[= \begin{pmatrix}1 - 6k & 9k\\ -4k & 1 + 6k\end{pmatrix}\times\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix} = \begin{pmatrix}-5 + 30k - 36k & 9 - 54k + 63k\\ 20k - 4 - 24k & -36k + 7 + 42k\end{pmatrix}\]OR\[= \begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}\times\begin{pmatrix}1 - 6k & 9k\\ -4k & 1 + 6k\end{pmatrix} = \begin{pmatrix}-5 + 30k - 36k & -45k + 9 + 54k\\ -4 + 24k - 28k & -36k + 7 + 42k\end{pmatrix}\] | M1 | 1.1b |
| Achieves from fully correct working \(= \begin{pmatrix}-5 - 6k & 9 + 9k\\ -4 - 4k & 7 + 6k\end{pmatrix}\) | A1 | 1.1b |
| \(= \begin{pmatrix}1 - 6(k + 1) & 9(k + 1)\\ -4(k + 1) & 1 + 6(k + 1)\end{pmatrix}\) Hence the result is true for \(n = k + 1\). Since it is true for \(n = 1\), and if true for \(n = k\) then true for \(n = k + 1\), thus by mathematical induction the result holds for all \(n \in \mathbb{N}\) | A1cso | 2.4 |
| (6) | ||
| (6 marks) |
Notes
B1: Shows the statement is true for \(n = 1\). Accept as minimum \(\begin{pmatrix}1 - 6 & 9\\ -4 & 1 + 6\end{pmatrix} = \begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}\)
M1: Makes the inductive assumption, assume true \(\boldsymbol{n = k}\). This may appear in the conclusion.
M1: A correct statement for \(\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}^{k+1}\) in terms of \(\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}^k\), can be either way round.
Can be implied by \(\begin{pmatrix}1 - 6k & 9k\\ -4k & 1 + 6k\end{pmatrix}\times\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}\) or \(\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}\times\begin{pmatrix}1 - 6k & 9k\\ -4k & 1 + 6k\end{pmatrix}\)
M1: Carries out the multiplication correctly, condone sign slips
A1: Correct simplified matrix from fully correct working
A1: Completes the inductive argument by showing clearly the matrix has the correct form (must have \((k + 1)\) factors in terms) or uses the result with \(n = k + 1\) and shows that their result is the same.
Conclusion conveying all three underlined points or equivalent at some point in their argument. Depends on all three M’s and A marks but can be scored without the B mark as long as it is stated true for \(n = 1\)