Standard +0.8 This is a proof by induction on matrix powers requiring students to verify a formula, perform matrix multiplication with algebraic entries, and manipulate expressions involving powers of 2. While the structure is standard for Further Maths, it requires careful algebraic manipulation and understanding of induction—moderately above average difficulty.
8 In this question you must show detailed reasoning.
\(\mathbf { M }\) is the matrix \(\left( \begin{array} { l l } 1 & 6 \\ 0 & 2 \end{array} \right)\).
Prove that \(\mathbf { M } ^ { n } = \left( \begin{array} { c c } 1 & 3 \left( 2 ^ { n + 1 } - 2 \right) \\ 0 & 2 ^ { n } \end{array} \right)\), for any positive integer \(n\).
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AO 2.2a — Simplification with sufficient working to establish truth for \(k+1\). Must see at least one stage of working between \(\mathbf{M}^k\mathbf{M}\) and required expression for \(\mathbf{M}^{k+1}\)
So true for \(n=k \Rightarrow\) true for \(n=k+1\). But true for \(n=1\). So true for all positive integers \(n\)
E1
AO 2.4 — Clear conclusion for induction process. Needs a fully correct proof. Cannot be awarded if there are mistakes in the proof. SC: If B1M1M1M1 gained but A0, allow SC B1 for fully correct ending statement
[6]
## Question 8:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{M}^1 = \begin{pmatrix}1 & 3(2^{1+1}-2)\\0 & 2^1\end{pmatrix} = \begin{pmatrix}1 & 3\times2\\0 & 2\end{pmatrix} = \begin{pmatrix}1 & 6\\0 & 2\end{pmatrix} = \mathbf{M}$ | **B1** | AO 3.1a — Full details must be shown. DR: Detailed reasoning required |
| Assume true for $n=k$, i.e. $\mathbf{M}^k = \begin{pmatrix}1 & 3(2^{k+1}-2)\\0 & 2^k\end{pmatrix}$ | **M1** | AO 2.1 — Must have statement in terms of some variable other than $n$. Watch out for $2^{k+1}$ appearing prematurely as bottom right element |
| $\mathbf{M}^{k+1} = \mathbf{M}\mathbf{M}^k = \begin{pmatrix}1 & 6\\0 & 2\end{pmatrix}\begin{pmatrix}1 & 3(2^{k+1}-2)\\0 & 2^k\end{pmatrix}$ | **M1** | AO 1.1 — Uses inductive hypothesis properly |
| $\begin{pmatrix}1 & 3(2^{k+1}-2)+6\times2^k\\0 & 2\times2^k\end{pmatrix}$ | **M1** | AO 1.1 — Genuine attempt at matrix multiplication (columns multiplied into rows) |
| $\begin{pmatrix}1 & 3(2^{k+1}-2)+3\times2^{k+1}\\0 & 2\times2^k\end{pmatrix} = \begin{pmatrix}1 & 3(2^{(k+1)+1}-2)\\0 & 2^{(k+1)}\end{pmatrix}$ | **A1 (AG)** | AO 2.2a — Simplification with sufficient working to establish truth for $k+1$. Must see at least one stage of working between $\mathbf{M}^k\mathbf{M}$ and required expression for $\mathbf{M}^{k+1}$ |
| So true for $n=k \Rightarrow$ true for $n=k+1$. But true for $n=1$. So true for all positive integers $n$ | **E1** | AO 2.4 — Clear conclusion for induction process. Needs a **fully correct proof**. Cannot be awarded if there are mistakes in the proof. SC: If **B1M1M1M1** gained but **A0**, allow **SC B1** for fully correct ending statement |
| **[6]** | | |
8 In this question you must show detailed reasoning.
$\mathbf { M }$ is the matrix $\left( \begin{array} { l l } 1 & 6 \\ 0 & 2 \end{array} \right)$.\\
Prove that $\mathbf { M } ^ { n } = \left( \begin{array} { c c } 1 & 3 \left( 2 ^ { n + 1 } - 2 \right) \\ 0 & 2 ^ { n } \end{array} \right)$, for any positive integer $n$.
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\hfill \mbox{\textit{OCR Further Pure Core AS 2019 Q8 [6]}}