Question 12 EITHER - A-Level Maths

CAIE FP1 2012 November — Question 12 EITHER

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2012
Session	November
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Invariant lines and eigenvalues and vectors
Type	Prove eigenvalue/eigenvector properties
Difficulty	Standard +0.8 This is a multi-part Further Maths question requiring proof of eigenvalue properties, computation of eigenvalues/eigenvectors, and matrix diagonalization. While the proof is straightforward application of definitions, finding the third eigenvector and constructing the diagonalization requires solid technique and several computational steps. The conceptual demand is moderate for FP1 level, placing it somewhat above average difficulty.
Spec	4.03h Determinant 2x2: calculation 4.03i Determinant: area scale factor and orientation 4.03o Inverse 3x3 matrix 4.03p Inverse properties: (AB)^(-1) = B^(-1)*A^(-1)

The vector $\mathbf { e }$ is an eigenvector of each of the $n \times n$ matrices $\mathbf { A }$ and $\mathbf { B }$, with corresponding eigenvalues $\lambda$ and $\mu$ respectively. Prove that $\mathbf { e }$ is an eigenvector of the matrix $\mathbf { A B }$ with eigenvalue $\lambda \mu$. It is given that the matrix $\mathbf { A }$, where $$\mathbf { A } = \left( \begin{array} { r r r } 3 & 2 & 2 \\ - 2 & - 2 & - 2 \\ 1 & 2 & 2 \end{array} \right) ,$$ has eigenvectors $\left( \begin{array} { r } 0 \\ 1 \\ - 1 \end{array} \right)$ and $\left( \begin{array} { r } 1 \\ 0 \\ - 1 \end{array} \right)$. Find the corresponding eigenvalues. Given that 2 is also an eigenvalue of $\mathbf { A }$, find a corresponding eigenvector. The matrix $\mathbf { B }$, where $$\mathbf { B } = \left( \begin{array} { r r r } - 1 & 2 & 2 \\ 2 & 2 & 2 \\ - 3 & - 6 & - 6 \end{array} \right) ,$$ has the same eigenvectors as $\mathbf { A }$. Given that $\mathbf { A B } = \mathbf { C }$, find a non-singular matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $$\mathbf { P } ^ { - 1 } \mathbf { C } ^ { 2 } \mathbf { P } = \mathbf { D }$$

Show mark scheme Show mark scheme source

Question 12 (EITHER):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\mathbf{A}\mathbf{e}=\lambda\mathbf{e}$ and $\mathbf{B}\mathbf{e}=\mu\mathbf{e}$; $\mathbf{AB}\mathbf{e} = \mathbf{A}\mu\mathbf{e} = \mu\mathbf{A}\mathbf{e} = \mu\lambda\mathbf{e} = \lambda\mu\mathbf{e}$	M1A1	Use of eigenvalue results
$\begin{pmatrix}3&2&2\\-2&-2&-2\\1&2&2\end{pmatrix}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix} \Rightarrow \lambda=0$	B1	Missing eigenvalue of $\mathbf{A}$
$\begin{pmatrix}3&2&2\\-2&-2&-2\\1&2&2\end{pmatrix}\begin{pmatrix}1\\0\\-1\end{pmatrix}=\begin{pmatrix}1\\0\\-1\end{pmatrix} \Rightarrow \lambda=1$	B1	Part marks: 2
$\lambda=2 \Rightarrow \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&2&2\\-2&-4&-2\end{vmatrix}=\begin{pmatrix}4\\-2\\0\end{pmatrix}\sim\begin{pmatrix}2\\-1\\0\end{pmatrix}$	M1A1	Finds missing eigenvector of $\mathbf{A}$ — Part marks: 2
$\begin{pmatrix}-1&2&2\\2&2&2\\-3&-6&-6\end{pmatrix}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix} \Rightarrow \mu=0$	B1
$\begin{pmatrix}-1&2&2\\2&2&2\\-3&-6&-6\end{pmatrix}\begin{pmatrix}1\\0\\-1\end{pmatrix}=\begin{pmatrix}-3\\0\\3\end{pmatrix} \Rightarrow \mu=-3$	B1
$\begin{pmatrix}-1&2&2\\2&2&2\\-3&-6&-6\end{pmatrix}\begin{pmatrix}2\\-1\\0\end{pmatrix}=\begin{pmatrix}-4\\2\\0\end{pmatrix} \Rightarrow \mu=-2$	B1
$\mathbf{C}$ has eigenvalues: $0\times0=0$, $1\times(-3)=-3$, $2\times(-2)=-4$	B2,1,0	1 mark for one correct, 2 for all three
$\mathbf{P}=\begin{pmatrix}0&1&2\\1&0&-1\\-1&-1&0\end{pmatrix}$, $\mathbf{D}=\begin{pmatrix}0&0&0\\0&9&0\\0&0&16\end{pmatrix}$	B1 M1A1	Part marks: 8

Question 12 (OR):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$m^2+6m+13=0 \Rightarrow m=-3\pm2i$; $x=e^{-3t}(A\cos 2t+B\sin 2t)$	M1A1	Finds complementary function
$x=p\cos 2t+q\sin 2t$; $\dot{x}=-2p\sin 2t+2q\cos 2t$; $\ddot{x}=-4p\cos 2t-4q\sin 2t$	M1	Finds particular integral
$(9p+12q)\cos 2t+(9q-12p)\sin 2t=75\cos 2t \Rightarrow p=3,\; q=4$	M1A1A1
$x=e^{-3t}(A\cos 2t+B\sin 2t)+3\cos 2t+4\sin 2t$	A1B1	General solution — Part marks: 7
$x=5$ when $t=0 \Rightarrow 5=A+3 \Rightarrow A=2$
$\dot{x}=-3e^{-3t}(A\cos 2t+B\sin 2t)+e^{-3t}(-2A\sin 2t+2B\cos 2t)-6\sin 2t+8\cos 2t$	M1A1	Uses initial conditions
$\dot{x}=0$ when $t=0 \Rightarrow 0=-6+8+2B \Rightarrow B=-1$	A1	Part marks: 4
$x=e^{-3t}(2\cos 2t-\sin 2t)+3\cos 2t+4\sin 2t$
As $t\to\infty$, $e^{-3t}\to 0$; $\therefore x\approx 3\cos 2t+4\sin 2t$	M1	Obtains limit
$\therefore x\approx 5\!\left(\frac{3}{5}\cos 2t+\frac{4}{5}\sin 2t\right)=5\cos\!\left(2t-\tan^{-1}\!\tfrac{4}{3}\right)$	M1A1	AG — Part marks: 3

# Question 12 (EITHER):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{A}\mathbf{e}=\lambda\mathbf{e}$ and $\mathbf{B}\mathbf{e}=\mu\mathbf{e}$; $\mathbf{AB}\mathbf{e} = \mathbf{A}\mu\mathbf{e} = \mu\mathbf{A}\mathbf{e} = \mu\lambda\mathbf{e} = \lambda\mu\mathbf{e}$ | M1A1 | Use of eigenvalue results |
| $\begin{pmatrix}3&2&2\\-2&-2&-2\\1&2&2\end{pmatrix}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix} \Rightarrow \lambda=0$ | B1 | Missing eigenvalue of $\mathbf{A}$ |
| $\begin{pmatrix}3&2&2\\-2&-2&-2\\1&2&2\end{pmatrix}\begin{pmatrix}1\\0\\-1\end{pmatrix}=\begin{pmatrix}1\\0\\-1\end{pmatrix} \Rightarrow \lambda=1$ | B1 | **Part marks: 2** |
| $\lambda=2 \Rightarrow \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&2&2\\-2&-4&-2\end{vmatrix}=\begin{pmatrix}4\\-2\\0\end{pmatrix}\sim\begin{pmatrix}2\\-1\\0\end{pmatrix}$ | M1A1 | Finds missing eigenvector of $\mathbf{A}$ — **Part marks: 2** |
| $\begin{pmatrix}-1&2&2\\2&2&2\\-3&-6&-6\end{pmatrix}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix} \Rightarrow \mu=0$ | B1 | |
| $\begin{pmatrix}-1&2&2\\2&2&2\\-3&-6&-6\end{pmatrix}\begin{pmatrix}1\\0\\-1\end{pmatrix}=\begin{pmatrix}-3\\0\\3\end{pmatrix} \Rightarrow \mu=-3$ | B1 | |
| $\begin{pmatrix}-1&2&2\\2&2&2\\-3&-6&-6\end{pmatrix}\begin{pmatrix}2\\-1\\0\end{pmatrix}=\begin{pmatrix}-4\\2\\0\end{pmatrix} \Rightarrow \mu=-2$ | B1 | |
| $\mathbf{C}$ has eigenvalues: $0\times0=0$, $1\times(-3)=-3$, $2\times(-2)=-4$ | B2,1,0 | 1 mark for one correct, 2 for all three |
| $\mathbf{P}=\begin{pmatrix}0&1&2\\1&0&-1\\-1&-1&0\end{pmatrix}$, $\mathbf{D}=\begin{pmatrix}0&0&0\\0&9&0\\0&0&16\end{pmatrix}$ | B1 M1A1 | **Part marks: 8** |

---

# Question 12 (OR):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $m^2+6m+13=0 \Rightarrow m=-3\pm2i$; $x=e^{-3t}(A\cos 2t+B\sin 2t)$ | M1A1 | Finds complementary function |
| $x=p\cos 2t+q\sin 2t$; $\dot{x}=-2p\sin 2t+2q\cos 2t$; $\ddot{x}=-4p\cos 2t-4q\sin 2t$ | M1 | Finds particular integral |
| $(9p+12q)\cos 2t+(9q-12p)\sin 2t=75\cos 2t \Rightarrow p=3,\; q=4$ | M1A1A1 | |
| $x=e^{-3t}(A\cos 2t+B\sin 2t)+3\cos 2t+4\sin 2t$ | A1B1 | General solution — **Part marks: 7** |
| $x=5$ when $t=0 \Rightarrow 5=A+3 \Rightarrow A=2$ | | |
| $\dot{x}=-3e^{-3t}(A\cos 2t+B\sin 2t)+e^{-3t}(-2A\sin 2t+2B\cos 2t)-6\sin 2t+8\cos 2t$ | M1A1 | Uses initial conditions |
| $\dot{x}=0$ when $t=0 \Rightarrow 0=-6+8+2B \Rightarrow B=-1$ | A1 | **Part marks: 4** |
| $x=e^{-3t}(2\cos 2t-\sin 2t)+3\cos 2t+4\sin 2t$ | | |
| As $t\to\infty$, $e^{-3t}\to 0$; $\therefore x\approx 3\cos 2t+4\sin 2t$ | M1 | Obtains limit |
| $\therefore x\approx 5\!\left(\frac{3}{5}\cos 2t+\frac{4}{5}\sin 2t\right)=5\cos\!\left(2t-\tan^{-1}\!\tfrac{4}{3}\right)$ | M1A1 | AG — **Part marks: 3** |

Show LaTeX source

The vector $\mathbf { e }$ is an eigenvector of each of the $n \times n$ matrices $\mathbf { A }$ and $\mathbf { B }$, with corresponding eigenvalues $\lambda$ and $\mu$ respectively. Prove that $\mathbf { e }$ is an eigenvector of the matrix $\mathbf { A B }$ with eigenvalue $\lambda \mu$.

It is given that the matrix $\mathbf { A }$, where

$$\mathbf { A } = \left( \begin{array} { r r r } 
3 & 2 & 2 \\
- 2 & - 2 & - 2 \\
1 & 2 & 2
\end{array} \right) ,$$

has eigenvectors $\left( \begin{array} { r } 0 \\ 1 \\ - 1 \end{array} \right)$ and $\left( \begin{array} { r } 1 \\ 0 \\ - 1 \end{array} \right)$. Find the corresponding eigenvalues.

Given that 2 is also an eigenvalue of $\mathbf { A }$, find a corresponding eigenvector.

The matrix $\mathbf { B }$, where

$$\mathbf { B } = \left( \begin{array} { r r r } 
- 1 & 2 & 2 \\
2 & 2 & 2 \\
- 3 & - 6 & - 6
\end{array} \right) ,$$

has the same eigenvectors as $\mathbf { A }$. Given that $\mathbf { A B } = \mathbf { C }$, find a non-singular matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that

$$\mathbf { P } ^ { - 1 } \mathbf { C } ^ { 2 } \mathbf { P } = \mathbf { D }$$

\hfill \mbox{\textit{CAIE FP1 2012 Q12 EITHER}}

This paper (13 questions)

View full paper

Q1 4 Q2 4 Q3 5 Q4 6 Q5 6 Q6 7 Q7 8 Q8 9 Q9 12 Q10 12 Q11 13 Q12 EITHER Q12 OR