Question 9 - A-Level Maths

CAIE FP1 2013 June — Question 9 11 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2013
Session	June
Marks	11
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 similarity transformation properties, finding eigenvectors of an upper triangular matrix, then applying the theory. While the proof is conceptually straightforward (substitute definitions and manipulate), students must handle abstract matrix notation carefully. The eigenvalue/eigenvector calculations are routine for upper triangular matrices, but applying the transformation Me to find B's eigenvectors requires connecting theory to practice. This is moderately challenging for FP1 level but follows standard patterns.
Spec	4.03a Matrix language: terminology and notation 4.03b Matrix operations: addition, multiplication, scalar

9 The square matrix $\mathbf { A }$ has an eigenvalue $\lambda$ with corresponding eigenvector $\mathbf { e }$. The non-singular matrix $\mathbf { M }$ is of the same order as $\mathbf { A }$. Show that $\mathbf { M e }$ is an eigenvector of the matrix $\mathbf { B }$, where $\mathbf { B } = \mathbf { M } \mathbf { A } \mathbf { M } ^ { - 1 }$, and that $\lambda$ is the corresponding eigenvalue. Let $$\mathbf { A } = \left( \begin{array} { r r r } - 1 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 0 & 2 \end{array} \right)$$ Write down the eigenvalues of $\mathbf { A }$ and obtain corresponding eigenvectors. Given that $$\mathbf { M } = \left( \begin{array} { l l l } 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$$ find the eigenvalues and corresponding eigenvectors of $\mathbf { B }$.

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Question 9:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\mathbf{A}\mathbf{e} = \lambda\mathbf{e}$	B1
$\mathbf{B}\mathbf{M}\mathbf{e} = \mathbf{M}\mathbf{A}\mathbf{M}^{-1}\mathbf{M}\mathbf{e} = (\mathbf{M}\mathbf{A}\mathbf{I})\mathbf{e} = \mathbf{M}\mathbf{A}\mathbf{e} = \mathbf{M}\lambda\mathbf{e} = \lambda\mathbf{M}\mathbf{e}$ (CWO)	M1A1	3 marks; ($\mathbf{M}\mathbf{e} \neq \mathbf{0}$ since $\mathbf{M}$ non-singular $\Rightarrow \lambda$ is an eigenvalue)
Eigenvalues are: $-1, 1, 2$	B1
$\lambda = -1$: $\begin{vmatrix} i & j & k \\ 0 & 2 & 1 \\ 0 & 2 & 4 \end{vmatrix} = \begin{pmatrix}6\\0\\0\end{pmatrix} \sim \begin{pmatrix}1\\0\\0\end{pmatrix}$	M1A1
$\lambda = 1$: $\begin{vmatrix} i & j & k \\ -2 & 2 & 1 \\ 0 & 2 & 4 \end{vmatrix} = \begin{pmatrix}8\\8\\0\end{pmatrix} \sim \begin{pmatrix}1\\1\\0\end{pmatrix}$
$\lambda = 2$: $\begin{vmatrix} i & j & k \\ -3 & 2 & 1 \\ 0 & -1 & 4 \end{vmatrix} = \begin{pmatrix}9\\12\\3\end{pmatrix} \sim \begin{pmatrix}3\\4\\1\end{pmatrix}$	A1	4 marks
Eigenvalues of $\mathbf{B}$ are $-1, 1, 2$	B1
Corresponding eigenvectors: $\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}1\\1\\0\end{pmatrix}, \begin{pmatrix}4\\4\\1\end{pmatrix}$	M1A1 A1	4 marks, [11]; N.B. Same as $\mathbf{A}$'s M0

# Question 9:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{A}\mathbf{e} = \lambda\mathbf{e}$ | B1 | |
| $\mathbf{B}\mathbf{M}\mathbf{e} = \mathbf{M}\mathbf{A}\mathbf{M}^{-1}\mathbf{M}\mathbf{e} = (\mathbf{M}\mathbf{A}\mathbf{I})\mathbf{e} = \mathbf{M}\mathbf{A}\mathbf{e} = \mathbf{M}\lambda\mathbf{e} = \lambda\mathbf{M}\mathbf{e}$ (CWO) | M1A1 | 3 marks; ($\mathbf{M}\mathbf{e} \neq \mathbf{0}$ since $\mathbf{M}$ non-singular $\Rightarrow \lambda$ is an eigenvalue) |
| Eigenvalues are: $-1, 1, 2$ | B1 | |
| $\lambda = -1$: $\begin{vmatrix} i & j & k \\ 0 & 2 & 1 \\ 0 & 2 & 4 \end{vmatrix} = \begin{pmatrix}6\\0\\0\end{pmatrix} \sim \begin{pmatrix}1\\0\\0\end{pmatrix}$ | M1A1 | |
| $\lambda = 1$: $\begin{vmatrix} i & j & k \\ -2 & 2 & 1 \\ 0 & 2 & 4 \end{vmatrix} = \begin{pmatrix}8\\8\\0\end{pmatrix} \sim \begin{pmatrix}1\\1\\0\end{pmatrix}$ | | |
| $\lambda = 2$: $\begin{vmatrix} i & j & k \\ -3 & 2 & 1 \\ 0 & -1 & 4 \end{vmatrix} = \begin{pmatrix}9\\12\\3\end{pmatrix} \sim \begin{pmatrix}3\\4\\1\end{pmatrix}$ | A1 | 4 marks |
| Eigenvalues of $\mathbf{B}$ are $-1, 1, 2$ | B1 | |
| Corresponding eigenvectors: $\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}1\\1\\0\end{pmatrix}, \begin{pmatrix}4\\4\\1\end{pmatrix}$ | M1A1 A1 | 4 marks, **[11]**; N.B. Same as $\mathbf{A}$'s M0 |

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9 The square matrix $\mathbf { A }$ has an eigenvalue $\lambda$ with corresponding eigenvector $\mathbf { e }$. The non-singular matrix $\mathbf { M }$ is of the same order as $\mathbf { A }$. Show that $\mathbf { M e }$ is an eigenvector of the matrix $\mathbf { B }$, where $\mathbf { B } = \mathbf { M } \mathbf { A } \mathbf { M } ^ { - 1 }$, and that $\lambda$ is the corresponding eigenvalue.

Let

$$\mathbf { A } = \left( \begin{array} { r r r } 
- 1 & 2 & 1 \\
0 & 1 & 4 \\
0 & 0 & 2
\end{array} \right)$$

Write down the eigenvalues of $\mathbf { A }$ and obtain corresponding eigenvectors.

Given that

$$\mathbf { M } = \left( \begin{array} { l l l } 
1 & 0 & 1 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array} \right)$$

find the eigenvalues and corresponding eigenvectors of $\mathbf { B }$.

\hfill \mbox{\textit{CAIE FP1 2013 Q9 [11]}}

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