| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove matrix power formula |
| Difficulty | Standard +0.3 This is a straightforward matrix induction proof with standard structure: verify base case n=1, assume true for n=k, prove for n=k+1 by multiplying A^k by A. Part (b) requires finding the inverse of a 2×2 matrix using the standard formula, which is routine. While it's a Further Maths topic, the execution is mechanical with no novel insight required, making it slightly easier than average. |
| Spec | 4.01a Mathematical induction: construct proofs4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2\mathbf{A} = \begin{pmatrix} 6 & 0 \\ 2 & 2 \end{pmatrix} = \begin{pmatrix} 2\times3 & 0 \\ 3-1 & 2 \end{pmatrix}\) so true when \(n=1\) | B1 | States base case |
| Assume true for \(n=k\), so \(2\mathbf{A}^k = \begin{pmatrix} 2\times3^k & 0 \\ 3^k-1 & 2 \end{pmatrix}\) | B1 | States inductive hypothesis |
| \(2\mathbf{A}^{k+1} = \begin{pmatrix} 2\times3^k & 0 \\ 3^k-1 & 2 \end{pmatrix}\begin{pmatrix} 3 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2\times3^{k+1} & 0 \\ 3^{k+1}-3+2 & 2 \end{pmatrix}\) | M1A1 | Multiplies \(2\mathbf{A}^k\) with \(\mathbf{A}\) |
| So true for \(n=k+1\). Hence, by induction, true for all positive integers. | A1 | States conclusion |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\det \mathbf{A}^n = \det\begin{pmatrix} 3^n & 0 \\ \frac{1}{2}(3^n-1) & 1 \end{pmatrix} = 3^n\) or a multiple of \(\mathbf{A}^{-n} = 2^{-1}3^{-n}\begin{pmatrix} 2 & 0 \\ 1-3^n & 2\times3^n \end{pmatrix}\) seen | B1 | |
| \(\mathbf{A}^{-n} = 3^{-n}\begin{pmatrix} 1 & 0 \\ \frac{1}{2}(1-3^n) & 3^n \end{pmatrix}\) | B1 | OE \(\mathbf{A}^{-n} = 2^{-1}3^{-n}\begin{pmatrix} 2 & 0 \\ 1-3^n & 2\times3^n \end{pmatrix}\) |
| Total: 2 |
## Question 1:
### Part 1(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\mathbf{A} = \begin{pmatrix} 6 & 0 \\ 2 & 2 \end{pmatrix} = \begin{pmatrix} 2\times3 & 0 \\ 3-1 & 2 \end{pmatrix}$ so true when $n=1$ | B1 | States base case |
| Assume true for $n=k$, so $2\mathbf{A}^k = \begin{pmatrix} 2\times3^k & 0 \\ 3^k-1 & 2 \end{pmatrix}$ | B1 | States inductive hypothesis |
| $2\mathbf{A}^{k+1} = \begin{pmatrix} 2\times3^k & 0 \\ 3^k-1 & 2 \end{pmatrix}\begin{pmatrix} 3 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2\times3^{k+1} & 0 \\ 3^{k+1}-3+2 & 2 \end{pmatrix}$ | M1A1 | Multiplies $2\mathbf{A}^k$ with $\mathbf{A}$ |
| So true for $n=k+1$. Hence, by induction, true for all positive integers. | A1 | States conclusion |
| **Total: 5** | | |
---
### Part 1(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\det \mathbf{A}^n = \det\begin{pmatrix} 3^n & 0 \\ \frac{1}{2}(3^n-1) & 1 \end{pmatrix} = 3^n$ or a multiple of $\mathbf{A}^{-n} = 2^{-1}3^{-n}\begin{pmatrix} 2 & 0 \\ 1-3^n & 2\times3^n \end{pmatrix}$ seen | B1 | |
| $\mathbf{A}^{-n} = 3^{-n}\begin{pmatrix} 1 & 0 \\ \frac{1}{2}(1-3^n) & 3^n \end{pmatrix}$ | B1 | OE $\mathbf{A}^{-n} = 2^{-1}3^{-n}\begin{pmatrix} 2 & 0 \\ 1-3^n & 2\times3^n \end{pmatrix}$ |
| **Total: 2** | | |1 Let $\mathbf { A } = \left( \begin{array} { l l } 3 & 0 \\ 1 & 1 \end{array} \right)$.
\begin{enumerate}[label=(\alph*)]
\item Prove by mathematical induction that, for all positive integers $n$,
$$2 \mathbf { A } ^ { n } = \left( \begin{array} { l l }
2 \times 3 ^ { n } & 0 \\
3 ^ { n } - 1 & 2
\end{array} \right)$$
\item Find, in terms of $n$, the inverse of $\mathbf { A } ^ { n }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q1}}