CAIE FP1 2010 June — Question 1 4 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2010
SessionJune
Marks4
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TopicInvariant lines and eigenvalues and vectors
TypeFind eigenvalues/vectors of matrix combination
DifficultyStandard +0.3 This is a straightforward eigenvalue/eigenvector question with a given eigenvalue (eliminating the characteristic equation step). Finding the eigenvector requires solving a system of linear equations, which is routine. The second part uses the standard result that if λ is an eigenvalue of A with eigenvector v, then λ + λ² is an eigenvalue of A + A² with the same eigenvector—this is a direct application of a known theorem requiring minimal additional work.
Spec4.03a Matrix language: terminology and notation
1 Given that 5 is an eigenvalue of the matrix $$\mathbf { A } = \left( \begin{array} { r r r } 5 & - 3 & 0 \\ 1 & 2 & 1 \\ - 1 & 3 & 4 \end{array} \right)$$ find a corresponding eigenvector. Hence find an eigenvalue and a corresponding eigenvector of the matrix \(\mathbf { A } + \mathbf { A } ^ { 2 }\).
AnswerMarks Guidance
Relevant working from \(A - 5I = \begin{pmatrix} 0 & -3 & 0 \\ 1 & -3 & 1 \\ -1 & 3 & -1 \end{pmatrix}\) or equivalent in equationsM1
to obtain an eigenvector of the form \(\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}\)A1 [2]
An eigenvalue of \(A + A^2\) is \(5 + 25 = 30\)B1
Corresponding eigenvector, as aboveB1
[2]
No penalty for not hence methods.
AnswerMarks
Accept \(\lambda = 6\) with \(\begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}\) and \(\lambda = 20\) with \(\begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}\)Accept linear scaling of eigenvectors 
Relevant working from $A - 5I = \begin{pmatrix} 0 & -3 & 0 \\ 1 & -3 & 1 \\ -1 & 3 & -1 \end{pmatrix}$ or equivalent in equations | M1 |

to obtain an eigenvector of the form $\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$ | A1 | [2]

An eigenvalue of $A + A^2$ is $5 + 25 = 30$ | B1 |

Corresponding eigenvector, as above | B1 |

| | [2] |

No penalty for not hence methods.

Accept $\lambda = 6$ with $\begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}$ and $\lambda = 20$ with $\begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}$ | Accept linear scaling of eigenvectors |

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1 Given that 5 is an eigenvalue of the matrix

$$\mathbf { A } = \left( \begin{array} { r r r } 
5 & - 3 & 0 \\
1 & 2 & 1 \\
- 1 & 3 & 4
\end{array} \right)$$

find a corresponding eigenvector.

Hence find an eigenvalue and a corresponding eigenvector of the matrix $\mathbf { A } + \mathbf { A } ^ { 2 }$.

\hfill \mbox{\textit{CAIE FP1 2010 Q1 [4]}}
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