| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Matrix powers by induction |
| Difficulty | Standard +0.3 This is a straightforward multi-part further maths question testing standard techniques: (a) uses determinant for area scaling (routine), (b) is a textbook induction proof with simple matrix multiplication, (c) applies invariant line condition. All parts are mechanical applications of known methods with no novel insight required. Slightly easier than average A-level due to computational simplicity. |
| Spec | 4.01a Mathematical induction: construct proofs4.03g Invariant points and lines4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\det \mathbf{A}^{-1} = (\det \mathbf{A})^{-1} = \frac{1}{2}\) | M1 A1 | |
| \(d = 15\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{A} = \begin{pmatrix}2&0\\1&1\end{pmatrix} = \begin{pmatrix}2^1&0\\2^1-1&1\end{pmatrix}\) so true when \(n=1\) | B1 | |
| Assume true for \(n=k\), so \(\mathbf{A}^k = \begin{pmatrix}2^k&0\\2^k-1&1\end{pmatrix}\) | B1 | |
| \(\mathbf{A}^{k+1} = \begin{pmatrix}2^k&0\\2^k-1&1\end{pmatrix}\begin{pmatrix}2&0\\1&1\end{pmatrix} = \begin{pmatrix}2^{k+1}&0\\2(2^k-1)+1&1\end{pmatrix} = \begin{pmatrix}2^{k+1}&0\\2^{k+1}-1&1\end{pmatrix}\) | M1 A1 | |
| So it is also true for \(n=k+1\). Hence, by induction, true for all positive integers. | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{A}^n\mathbf{B} = \begin{pmatrix}2^n&0\\2^n-1&1\end{pmatrix}\begin{pmatrix}1&0\\33&0\end{pmatrix} = \begin{pmatrix}2^n&0\\2^n+32&0\end{pmatrix}\) | M1 A1 | |
| \(\begin{pmatrix}2^n&0\\2^n+32&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}2^n x\\(2^n+32)x\end{pmatrix}\) | B1 | |
| \((2^n+32)x = 2^{n+1}x \Rightarrow 2^n = 32 \Rightarrow n=5\) | M1 A1 |
## Question 6:
### Part 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\det \mathbf{A}^{-1} = (\det \mathbf{A})^{-1} = \frac{1}{2}$ | M1 A1 | |
| $d = 15$ | A1 | |
### Part 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{A} = \begin{pmatrix}2&0\\1&1\end{pmatrix} = \begin{pmatrix}2^1&0\\2^1-1&1\end{pmatrix}$ so true when $n=1$ | B1 | |
| Assume true for $n=k$, so $\mathbf{A}^k = \begin{pmatrix}2^k&0\\2^k-1&1\end{pmatrix}$ | B1 | |
| $\mathbf{A}^{k+1} = \begin{pmatrix}2^k&0\\2^k-1&1\end{pmatrix}\begin{pmatrix}2&0\\1&1\end{pmatrix} = \begin{pmatrix}2^{k+1}&0\\2(2^k-1)+1&1\end{pmatrix} = \begin{pmatrix}2^{k+1}&0\\2^{k+1}-1&1\end{pmatrix}$ | M1 A1 | |
| So it is also true for $n=k+1$. Hence, by induction, true for all positive integers. | A1 | |
### Part 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{A}^n\mathbf{B} = \begin{pmatrix}2^n&0\\2^n-1&1\end{pmatrix}\begin{pmatrix}1&0\\33&0\end{pmatrix} = \begin{pmatrix}2^n&0\\2^n+32&0\end{pmatrix}$ | M1 A1 | |
| $\begin{pmatrix}2^n&0\\2^n+32&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}2^n x\\(2^n+32)x\end{pmatrix}$ | B1 | |
| $(2^n+32)x = 2^{n+1}x \Rightarrow 2^n = 32 \Rightarrow n=5$ | M1 A1 | |
---6 Let $\mathbf { A } = \left( \begin{array} { l l } 2 & 0 \\ 1 & 1 \end{array} \right)$.
\begin{enumerate}[label=(\alph*)]
\item The transformation in the $x - y$ plane represented by $\mathbf { A } ^ { - 1 }$ transforms a triangle of area $30 \mathrm {~cm} ^ { 2 }$ into a triangle of area $d \mathrm {~cm} ^ { 2 }$.
Find the value of $d$.
\item Prove by mathematical induction that, for all positive integers $n$,
$$\mathbf { A } ^ { n } = \left( \begin{array} { c c }
2 ^ { n } & 0 \\
2 ^ { n } - 1 & 1
\end{array} \right)$$
\item The line $y = 2 x$ is invariant under the transformation in the $x - y$ plane represented by $\mathbf { A } ^ { n } \mathbf { B }$, where $\mathbf { B } = \left( \begin{array} { r l } 1 & 0 \\ 33 & 0 \end{array} \right)$.
Find the value of $n$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q6 [13]}}