CAIE FP1 2012 June — Question 5 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionJune
Marks9
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 on eigenvalues/eigenvectors that guides students through standard techniques. The proof is a straightforward algebraic manipulation, finding eigenvectors involves routine row reduction, and the final part applies the proven result to relate matrices B and C (where C = B - 3I). While it requires multiple steps and understanding of the theory, each component is a standard textbook exercise with clear signposting, making it slightly easier than average for Further Maths.
Spec1.01a Proof: structure of mathematical proof and logical steps4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix

5 The matrix \(\mathbf { A }\) has an eigenvalue \(\lambda\) with corresponding eigenvector \(\mathbf { e }\). Prove that the matrix \(( \mathbf { A } + k \mathbf { I } )\), where \(k\) is a real constant and \(\mathbf { I }\) is the identity matrix, has an eigenvalue ( \(\lambda + k\) ) with corresponding eigenvector \(\mathbf { e }\). The matrix \(\mathbf { B }\) is given by $$\mathbf { B } = \left( \begin{array} { r r r } 2 & 2 & - 3 \\ 2 & 2 & 3 \\ - 3 & 3 & 3 \end{array} \right) .$$ Two of the eigenvalues of \(\mathbf { B }\) are - 3 and 4 . Find corresponding eigenvectors. Given that \(\left( \begin{array} { r } 1 \\ - 1 \\ - 2 \end{array} \right)\) is an eigenvector of \(\mathbf { B }\), find the corresponding eigenvalue. Hence find the eigenvalues of \(\mathbf { C }\), where $$\mathbf { C } = \left( \begin{array} { r r r } - 1 & 2 & - 3 \\ 2 & - 1 & 3 \\ - 3 & 3 & 0 \end{array} \right) ,$$ and state corresponding eigenvectors.

Question 5:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\((\mathbf{A}+k\mathbf{I})\mathbf{e} = \mathbf{A}\mathbf{e}+k\mathbf{I}\mathbf{e} = \lambda\mathbf{e}+k\mathbf{e} = (\lambda+k)\mathbf{e}\) \(\therefore (\mathbf{A}+k\mathbf{I})\) has eigenvalue \((\lambda+k)\) with eigenvector \(\mathbf{e}\)M1A1 Part Mark: 2
Eigenvalues of \(\mathbf{B}\) are \(-3\) and \(4\) (given). Eigenvectors are \(\begin{pmatrix}1\\-1\\1\end{pmatrix}, \begin{pmatrix}1\\1\\0\end{pmatrix}\)M1A1, A1 Part Mark: 3
Third eigenvalue is \(6\)B1 Part Mark: 1
\(\mathbf{C} = \mathbf{B} - 3\mathbf{I}\) (stated or implied)M1 Recognises result proved initially
Eigenvalues of \(\mathbf{C}\): \(-6, 1, 3\)A1
Eigenvectors: \(\begin{pmatrix}1\\-1\\1\end{pmatrix}, \begin{pmatrix}1\\1\\0\end{pmatrix}, \begin{pmatrix}1\\-1\\-2\end{pmatrix}\) (OE)A1\(\sqrt{}\) Part Mark: 3; *For 'non hence' method using characteristic equation, award B1 rather than M1A1 for eigenvalues, followed by B1 for eigenvectors*
## Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(\mathbf{A}+k\mathbf{I})\mathbf{e} = \mathbf{A}\mathbf{e}+k\mathbf{I}\mathbf{e} = \lambda\mathbf{e}+k\mathbf{e} = (\lambda+k)\mathbf{e}$ $\therefore (\mathbf{A}+k\mathbf{I})$ has eigenvalue $(\lambda+k)$ with eigenvector $\mathbf{e}$ | M1A1 | Part Mark: 2 |
| Eigenvalues of $\mathbf{B}$ are $-3$ and $4$ (given). Eigenvectors are $\begin{pmatrix}1\\-1\\1\end{pmatrix}, \begin{pmatrix}1\\1\\0\end{pmatrix}$ | M1A1, A1 | Part Mark: 3 |
| Third eigenvalue is $6$ | B1 | Part Mark: 1 |
| $\mathbf{C} = \mathbf{B} - 3\mathbf{I}$ (stated or implied) | M1 | Recognises result proved initially |
| Eigenvalues of $\mathbf{C}$: $-6, 1, 3$ | A1 | |
| Eigenvectors: $\begin{pmatrix}1\\-1\\1\end{pmatrix}, \begin{pmatrix}1\\1\\0\end{pmatrix}, \begin{pmatrix}1\\-1\\-2\end{pmatrix}$ (OE) | A1$\sqrt{}$ | Part Mark: 3; *For 'non hence' method using characteristic equation, award B1 rather than M1A1 for eigenvalues, followed by B1 for eigenvectors* |

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5 The matrix $\mathbf { A }$ has an eigenvalue $\lambda$ with corresponding eigenvector $\mathbf { e }$. Prove that the matrix $( \mathbf { A } + k \mathbf { I } )$, where $k$ is a real constant and $\mathbf { I }$ is the identity matrix, has an eigenvalue ( $\lambda + k$ ) with corresponding eigenvector $\mathbf { e }$.

The matrix $\mathbf { B }$ is given by

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

Two of the eigenvalues of $\mathbf { B }$ are - 3 and 4 . Find corresponding eigenvectors.

Given that $\left( \begin{array} { r } 1 \\ - 1 \\ - 2 \end{array} \right)$ is an eigenvector of $\mathbf { B }$, find the corresponding eigenvalue.

Hence find the eigenvalues of $\mathbf { C }$, where

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

and state corresponding eigenvectors.

\hfill \mbox{\textit{CAIE FP1 2012 Q5 [9]}}