Standard +0.3 This is a straightforward induction proof with a simple matrix multiplication pattern. The base case is trivial, and the inductive step requires only one 2×2 matrix multiplication with a clear algebraic simplification. While it combines two A-level topics (matrices and induction), the structure is highly standard and the algebra is minimal, making it slightly easier than average.
6 You are given that \(\mathbf { A } = \left( \begin{array} { l l } 1 & a \\ 0 & 1 \end{array} \right)\) where \(a\) is a constant.
Prove by induction that \(\mathbf { A } ^ { \mathrm { n } } = \left( \begin{array} { c c } 1 & \text { an } \\ 0 & 1 \end{array} \right)\) for all integers \(n \geqslant 1\).
\(\mathbf{A}^1\) and \(a\times 1\) must both be seen explicitly. Accept "\(=\mathbf{A}\)" instead of "therefore true when \(n=1\)"
Assume true for \(n=k\): \(\mathbf{A}^k = \begin{pmatrix} 1 & ak \\ 0 & 1 \end{pmatrix}\)
M1
Setting up the inductive hypothesis properly.
\(\therefore \mathbf{A}^{k+1} = \mathbf{A}^k\mathbf{A} = \begin{pmatrix} 1 & ak \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}\) by inductive hypothesis
M1
Considering \(\mathbf{A}^{k+1}\) and using inductive hypothesis. Could also consider \(\mathbf{A}\mathbf{A}^k\). Jumping to \(\mathbf{A}^{k+1} = \begin{pmatrix} 1 & a(k+1) \\ 0 & 1 \end{pmatrix}\) without seeing two matrices multiplied scores M0.
So true for \(n=k \Rightarrow\) true for \(n=k+1\). But true for \(n=1\). Therefore true for all [integer] \(n \geq 1\).
A1
Clear conclusion for induction. 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 \(a\times 1\) not seen, allow A1 here (but must have attempt at base case).
[5]
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Basis case: $n=1$: $\mathbf{A}^1 = \begin{pmatrix} 1 & a\times 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}$ $[=\mathbf{A}]$ | B1 | $\mathbf{A}^1$ and $a\times 1$ must both be seen explicitly. Accept "$=\mathbf{A}$" instead of "therefore true when $n=1$" |
| Assume true for $n=k$: $\mathbf{A}^k = \begin{pmatrix} 1 & ak \\ 0 & 1 \end{pmatrix}$ | M1 | Setting up the inductive hypothesis properly. |
| $\therefore \mathbf{A}^{k+1} = \mathbf{A}^k\mathbf{A} = \begin{pmatrix} 1 & ak \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}$ by inductive hypothesis | M1 | Considering $\mathbf{A}^{k+1}$ and using inductive hypothesis. Could also consider $\mathbf{A}\mathbf{A}^k$. Jumping to $\mathbf{A}^{k+1} = \begin{pmatrix} 1 & a(k+1) \\ 0 & 1 \end{pmatrix}$ without seeing two matrices multiplied scores M0. |
| $\therefore \mathbf{A}^{k+1} = \begin{pmatrix} 1 & a+ak \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & a(k+1) \\ 0 & 1 \end{pmatrix}$ | A1 | Must see $a+ak$ appear before being factorised. |
| So true for $n=k \Rightarrow$ true for $n=k+1$. But true for $n=1$. Therefore true for all [integer] $n \geq 1$. | A1 | Clear conclusion for induction. 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 $a\times 1$ not seen, allow A1 here (but must have attempt at base case). |
| **[5]** | | |
6 You are given that $\mathbf { A } = \left( \begin{array} { l l } 1 & a \\ 0 & 1 \end{array} \right)$ where $a$ is a constant.\\
Prove by induction that $\mathbf { A } ^ { \mathrm { n } } = \left( \begin{array} { c c } 1 & \text { an } \\ 0 & 1 \end{array} \right)$ for all integers $n \geqslant 1$.
\hfill \mbox{\textit{OCR Further Pure Core AS 2024 Q6 [5]}}