| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove matrix power formula |
| Difficulty | Standard +0.3 This is a standard proof by induction for matrix powers with a straightforward structure. Students verify the base case n=1 by direct calculation, assume the result for n=k, then multiply both sides by the original matrix to prove n=k+1. The matrix multiplication is routine (2×2 matrices with simple linear expressions), and the algebraic simplification is mechanical. While it requires careful bookkeeping, it follows the standard induction template without requiring insight or novel problem-solving, making it slightly easier than average. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For \(n=1\): \(\begin{pmatrix}1-6\times1 & 9\times1\\-4\times1 & 1+6\times1\end{pmatrix} = \begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix} = \begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}^1\) so true for \(n=1\) | B1 | Shows statement true for \(n=1\); accept as minimum \(\begin{pmatrix}1-6 & 9\\-4 & 1+6\end{pmatrix} = \begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}\) |
| Assume true for \(n=k\): \(\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}^k = \begin{pmatrix}1-6k & 9k\\-4k & 1+6k\end{pmatrix}\) | M1 | Makes the inductive assumption, assume true for \(n=k\) |
| \(\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}^{k+1} = \begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}^k \times \begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}\) OR \(\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}\times\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}^k\) | M1 | 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 |
| \(= \begin{pmatrix}1-6k & 9k\\-4k & 1+6k\end{pmatrix}\times\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix} = \begin{pmatrix}-5+30k-36k & 9-54k+63k\\20k-4-24k & -36k+7+42k\end{pmatrix}\) | M1 | Carries out the multiplication correctly; condone sign slips |
| \(= \begin{pmatrix}-5-6k & 9+9k\\-4-4k & 7+6k\end{pmatrix}\) | A1 | Correct simplified matrix from fully correct working |
| \(= \begin{pmatrix}1-6(k+1) & 9(k+1)\\-4(k+1) & 1+6(k+1)\end{pmatrix}\) Hence true for \(n=k+1\). Since 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 | Completes inductive argument showing matrix has correct form with \((k+1)\) factors; conclusion conveying all three underlined points; depends on all three M marks and A mark |
# Question 7:
| Answer/Working | Mark | Guidance |
|---|---|---|
| For $n=1$: $\begin{pmatrix}1-6\times1 & 9\times1\\-4\times1 & 1+6\times1\end{pmatrix} = \begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix} = \begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}^1$ so true for $n=1$ | B1 | Shows statement true for $n=1$; accept as minimum $\begin{pmatrix}1-6 & 9\\-4 & 1+6\end{pmatrix} = \begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}$ |
| **Assume** true for $n=k$: $\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}^k = \begin{pmatrix}1-6k & 9k\\-4k & 1+6k\end{pmatrix}$ | M1 | Makes the inductive assumption, **assume** true for $n=k$ |
| $\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}^{k+1} = \begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}^k \times \begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}$ OR $\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}\times\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}^k$ | M1 | 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 |
| $= \begin{pmatrix}1-6k & 9k\\-4k & 1+6k\end{pmatrix}\times\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix} = \begin{pmatrix}-5+30k-36k & 9-54k+63k\\20k-4-24k & -36k+7+42k\end{pmatrix}$ | M1 | Carries out the multiplication correctly; condone sign slips |
| $= \begin{pmatrix}-5-6k & 9+9k\\-4-4k & 7+6k\end{pmatrix}$ | A1 | Correct simplified matrix **from fully correct working** |
| $= \begin{pmatrix}1-6(k+1) & 9(k+1)\\-4(k+1) & 1+6(k+1)\end{pmatrix}$ Hence true for $n=k+1$. Since 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 | Completes inductive argument showing matrix has correct form with $(k+1)$ factors; conclusion conveying all three underlined points; depends on all three M marks and A mark |
---\begin{enumerate}
\item Prove by mathematical induction that, for $n \in \mathbb { N }$
\end{enumerate}
$$\left( \begin{array} { l l }
- 5 & 9 \\
- 4 & 7
\end{array} \right) ^ { n } = \left( \begin{array} { c c }
1 - 6 n & 9 n \\
- 4 n & 1 + 6 n
\end{array} \right)$$
\hfill \mbox{\textit{Edexcel CP AS 2022 Q7 [6]}}