| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove matrix power formula |
| Difficulty | Standard +0.3 This is a straightforward proof by induction with a simple 2×2 matrix. The matrix multiplication is elementary (upper triangular with 1s on diagonal), the inductive step requires only one matrix multiplication, and the pattern is immediately apparent. Slightly easier than average as it's a standard template application with minimal computational complexity. |
| Spec | 4.01a Mathematical induction: construct proofs4.03b Matrix operations: addition, multiplication, scalar |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}^1 = \begin{pmatrix} 1 & -2\times1 \\ 0 & 1 \end{pmatrix}\) (so true when \(n=1\)) | B1 | Shows true for \(n=1\); minimum acceptable is seeing \(n=1\) substituted into rhs giving \(\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}\) |
| Assume true for \(n=k\), then \(\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & -2k \\ 0 & 1 \end{pmatrix}\) or \(\begin{pmatrix} 1 & -2k \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}\) | M1 | Assumes for \(n=k\) and multiplies original matrix by \(k\)th power matrix either way round; assumption statement not needed for this mark |
| \(= \begin{pmatrix} 1 & -2k-2 \\ 0 & 1 \end{pmatrix}\) or \(= \begin{pmatrix} 1 & -2-2k \\ 0 & 1 \end{pmatrix}\) | A1 | Correct unsimplified matrix |
| \(= \begin{pmatrix} 1 & -2(k+1) \\ 0 & 1 \end{pmatrix}\) | A1 | Correct form with \(k+1\) factored out, no errors; result may be proved by equivalence |
| Hence result is true for \(n=k+1\). As true for \(n=1\) and have shown if true for \(n=k\) then it is true for \(n=k+1\), so it is true for all \(n\). | A1 | Correct conclusion with assumption made; must convey "if true for \(n=k\) then true for \(n=k+1\)"; allow "correct" for "true"; must be dependent on all previous marks (except B1) |
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}^1 = \begin{pmatrix} 1 & -2\times1 \\ 0 & 1 \end{pmatrix}$ (so true when $n=1$) | B1 | Shows true for $n=1$; minimum acceptable is seeing $n=1$ substituted into rhs giving $\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}$ |
| Assume true for $n=k$, then $\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & -2k \\ 0 & 1 \end{pmatrix}$ or $\begin{pmatrix} 1 & -2k \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}$ | M1 | Assumes for $n=k$ and multiplies original matrix by $k$th power matrix either way round; assumption statement not needed for this mark |
| $= \begin{pmatrix} 1 & -2k-2 \\ 0 & 1 \end{pmatrix}$ or $= \begin{pmatrix} 1 & -2-2k \\ 0 & 1 \end{pmatrix}$ | A1 | Correct unsimplified matrix |
| $= \begin{pmatrix} 1 & -2(k+1) \\ 0 & 1 \end{pmatrix}$ | A1 | Correct form with $k+1$ factored out, no errors; result may be proved by equivalence |
| Hence result is true for $n=k+1$. As true for $n=1$ and have shown if true for $n=k$ then it is true for $n=k+1$, so it is true for all $n$. | A1 | Correct conclusion **with assumption made**; must convey "if true for $n=k$ then true for $n=k+1$"; allow "correct" for "true"; must be dependent on all previous marks (except B1) |
---\begin{enumerate}
\item Prove by induction that for $n \in \mathbb { N }$
\end{enumerate}
$$\left( \begin{array} { c c }
1 & - 2 \\
0 & 1
\end{array} \right) ^ { n } = \left( \begin{array} { c c }
1 & - 2 n \\
0 & 1
\end{array} \right)$$
\hfill \mbox{\textit{Edexcel CP1 2023 Q4 [5]}}