CAIE FP1 2011 November — Question 8 11 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeProve eigenvalue/eigenvector properties
DifficultyStandard +0.3 This is a structured multi-part question that guides students through standard eigenvalue/eigenvector techniques. The proof is straightforward algebraic manipulation (Ae=λe, Be=μe → ABe=λμe), the triangular matrix C has eigenvalues immediately visible on the diagonal, and the final part applies the proven result. While it requires understanding of eigenvector concepts and matrix multiplication, each step is routine for Further Maths students with no novel problem-solving required.
Spec4.03h Determinant 2x2: calculation4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices
8 The vector \(\mathbf { e }\) is an eigenvector of the matrix \(\mathbf { A }\), with corresponding eigenvalue \(\lambda\), and is also an eigenvector of the matrix \(\mathbf { B }\), with corresponding eigenvalue \(\mu\). Show that \(\mathbf { e }\) is an eigenvector of the matrix \(\mathbf { A B }\) with corresponding eigenvalue \(\lambda \mu\). State the eigenvalues of the matrix \(\mathbf { C }\), where $$\mathbf { C } = \left( \begin{array} { r r r } - 1 & - 1 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{array} \right) ,$$ and find corresponding eigenvectors. Show that \(\left( \begin{array} { l } 1 \\ 6 \\ 3 \end{array} \right)\) is an eigenvector of the matrix \(\mathbf { D }\), where $$\mathbf { D } = \left( \begin{array} { r r r } 1 & - 1 & 1 \\ - 6 & - 3 & 4 \\ - 9 & - 3 & 7 \end{array} \right) ,$$ and state the corresponding eigenvalue. Hence state an eigenvector of the matrix CD and give the corresponding eigenvalue.
Question 8:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\mathbf{AB}\mathbf{e} = \mathbf{A}\,\mu\mathbf{e} = \mu\mathbf{A}\mathbf{e} = \mu\lambda\mathbf{e} = \lambda\mu\mathbf{e}\)M1A1 Shows required result using \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\)
Eigenvalues of \(\mathbf{C}\) are \(-1\), \(1\) and \(2\)B1 States eigenvalues from leading diagonal
\(\lambda = -1\): \(\mathbf{e}_1 = \begin{vmatrix} i & j & k \\ 0 & -1 & 3 \\ 0 & 2 & 2\end{vmatrix} = \begin{pmatrix}-8\\0\\0\end{pmatrix} \sim \begin{pmatrix}1\\0\\0\end{pmatrix}\)M1A1 Finds eigenvectors using cross-product or equations; M1A1 for first correct, A1 for other two
\(\lambda = 1\): \(\mathbf{e}_2 = \begin{vmatrix} i & j & k \\ -2 & -1 & 3 \\ 0 & 0 & 2\end{vmatrix} = \begin{pmatrix}-2\\4\\0\end{pmatrix} \sim \begin{pmatrix}1\\-2\\0\end{pmatrix}\)
\(\lambda = 2\): \(\mathbf{e}_3 = \begin{vmatrix} i & j & k \\ -3 & 1 & 3 \\ 0 & -1 & 2\end{vmatrix} = \begin{pmatrix}1\\6\\3\end{pmatrix}\)A1 Part total: 4
\(\mathbf{D}\begin{pmatrix}1\\6\\3\end{pmatrix} = \begin{pmatrix}-2\\-12\\-6\end{pmatrix} = -2\begin{pmatrix}1\\6\\3\end{pmatrix}\)M1A1 Uses \(\mathbf{D}\mathbf{e} = \mu\mathbf{e}\)
Eigenvalue of \(\mathbf{D}\) is \(-2\)A1 States eigenvalue
\(\mathbf{CD}\) has an eigenvector \(\begin{pmatrix}1\\6\\3\end{pmatrix}\)B1 Recognises eigenvector common to \(\mathbf{C}\) and \(\mathbf{D}\)
Corresponding eigenvalue is \(-2 \times 2 = -4\)B1√ States corresponding eigenvalue 
# Question 8:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{AB}\mathbf{e} = \mathbf{A}\,\mu\mathbf{e} = \mu\mathbf{A}\mathbf{e} = \mu\lambda\mathbf{e} = \lambda\mu\mathbf{e}$ | M1A1 | Shows required result using $\mathbf{A}\mathbf{e} = \lambda\mathbf{e}$ | **Part total: 2** |
| Eigenvalues of $\mathbf{C}$ are $-1$, $1$ and $2$ | B1 | States eigenvalues from leading diagonal |
| $\lambda = -1$: $\mathbf{e}_1 = \begin{vmatrix} i & j & k \\ 0 & -1 & 3 \\ 0 & 2 & 2\end{vmatrix} = \begin{pmatrix}-8\\0\\0\end{pmatrix} \sim \begin{pmatrix}1\\0\\0\end{pmatrix}$ | M1A1 | Finds eigenvectors using cross-product or equations; M1A1 for first correct, A1 for other two |
| $\lambda = 1$: $\mathbf{e}_2 = \begin{vmatrix} i & j & k \\ -2 & -1 & 3 \\ 0 & 0 & 2\end{vmatrix} = \begin{pmatrix}-2\\4\\0\end{pmatrix} \sim \begin{pmatrix}1\\-2\\0\end{pmatrix}$ | |
| $\lambda = 2$: $\mathbf{e}_3 = \begin{vmatrix} i & j & k \\ -3 & 1 & 3 \\ 0 & -1 & 2\end{vmatrix} = \begin{pmatrix}1\\6\\3\end{pmatrix}$ | A1 | **Part total: 4** |
| $\mathbf{D}\begin{pmatrix}1\\6\\3\end{pmatrix} = \begin{pmatrix}-2\\-12\\-6\end{pmatrix} = -2\begin{pmatrix}1\\6\\3\end{pmatrix}$ | M1A1 | Uses $\mathbf{D}\mathbf{e} = \mu\mathbf{e}$ |
| Eigenvalue of $\mathbf{D}$ is $-2$ | A1 | States eigenvalue | **Part total: 3** |
| $\mathbf{CD}$ has an eigenvector $\begin{pmatrix}1\\6\\3\end{pmatrix}$ | B1 | Recognises eigenvector common to $\mathbf{C}$ and $\mathbf{D}$ |
| Corresponding eigenvalue is $-2 \times 2 = -4$ | B1√ | States corresponding eigenvalue | **Part total: 2** |

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8 The vector $\mathbf { e }$ is an eigenvector of the matrix $\mathbf { A }$, with corresponding eigenvalue $\lambda$, and is also an eigenvector of the matrix $\mathbf { B }$, with corresponding eigenvalue $\mu$. Show that $\mathbf { e }$ is an eigenvector of the matrix $\mathbf { A B }$ with corresponding eigenvalue $\lambda \mu$.

State the eigenvalues of the matrix $\mathbf { C }$, where

$$\mathbf { C } = \left( \begin{array} { r r r } 
- 1 & - 1 & 3 \\
0 & 1 & 2 \\
0 & 0 & 2
\end{array} \right) ,$$

and find corresponding eigenvectors.

Show that $\left( \begin{array} { l } 1 \\ 6 \\ 3 \end{array} \right)$ is an eigenvector of the matrix $\mathbf { D }$, where

$$\mathbf { D } = \left( \begin{array} { r r r } 
1 & - 1 & 1 \\
- 6 & - 3 & 4 \\
- 9 & - 3 & 7
\end{array} \right) ,$$

and state the corresponding eigenvalue.

Hence state an eigenvector of the matrix CD and give the corresponding eigenvalue.

\hfill \mbox{\textit{CAIE FP1 2011 Q8 [11]}}
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