CAIE FP1 2013 November — Question 7 10 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeFind eigenvalues/vectors of matrix combination
DifficultyStandard +0.3 This question tests standard eigenvalue concepts with a 3×3 matrix. The proof that e is an eigenvector of A² is straightforward algebraic manipulation. Finding eigenvalues requires computing a 3×3 determinant and solving a cubic, which is routine for Further Maths. The final part applies the proven result to a polynomial of B, requiring pattern recognition but no novel insight. Slightly above average difficulty due to the 3×3 computation and multi-part structure, but all techniques are standard FP1 material.
Spec4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation

7 The square matrix \(\mathbf { A }\) has \(\lambda\) as an eigenvalue with \(\mathbf { e }\) as a corresponding eigenvector. Show that \(\mathbf { e }\) is an eigenvector of \(\mathbf { A } ^ { 2 }\) and state the corresponding eigenvalue. Find the eigenvalues of the matrix \(\mathbf { B }\), where $$\mathbf { B } = \left( \begin{array} { l l l } 1 & 3 & 0 \\ 2 & 0 & 2 \\ 1 & 1 & 2 \end{array} \right)$$ Find the eigenvalues of \(\mathbf { B } ^ { 4 } + 2 \mathbf { B } ^ { 2 } + 3 \mathbf { I }\), where \(\mathbf { I }\) is the \(3 \times 3\) identity matrix.

Question 7:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\mathbf{Ae}=\lambda\mathbf{e} \Rightarrow \mathbf{A}^2\mathbf{e} = \mathbf{A}\mathbf{A}\mathbf{e} = \mathbf{A}\lambda\mathbf{e} = \lambda\mathbf{A}\mathbf{e} = \lambda^2\mathbf{e} \Rightarrow\) result \((\mathbf{e}\neq\mathbf{0} \Rightarrow \lambda^2\) is an eigenvalue of \(\mathbf{A}^2)\)B1, M1A1 Proves first result
\((1-\lambda)(\lambda-4)(\lambda+2)=0 \Rightarrow \lambda = -2, 1, 4\)M1A1, A1A1 Obtains eigenvalues of \(\mathbf{B}\)
\(1^4\mathbf{e}+2\times1^2\mathbf{e}+3\mathbf{e} = \mathbf{6e} \Rightarrow 6\) is an eigenvalueM1A1 Obtains eigenvalues of related matrix
\((-2)^4\mathbf{e}+2\times(-2)^2\mathbf{e}+3\mathbf{e} = 27\mathbf{e} \Rightarrow 27\) is an eigenvalue
\(4^4\mathbf{e}+2\times4^2\mathbf{e}+3\mathbf{e} = 291\mathbf{e} \Rightarrow 291\) is an eigenvalueA1
## Question 7:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{Ae}=\lambda\mathbf{e} \Rightarrow \mathbf{A}^2\mathbf{e} = \mathbf{A}\mathbf{A}\mathbf{e} = \mathbf{A}\lambda\mathbf{e} = \lambda\mathbf{A}\mathbf{e} = \lambda^2\mathbf{e} \Rightarrow$ result $(\mathbf{e}\neq\mathbf{0} \Rightarrow \lambda^2$ is an eigenvalue of $\mathbf{A}^2)$ | B1, M1A1 | Proves first result |
| $(1-\lambda)(\lambda-4)(\lambda+2)=0 \Rightarrow \lambda = -2, 1, 4$ | M1A1, A1A1 | Obtains eigenvalues of $\mathbf{B}$ |
| $1^4\mathbf{e}+2\times1^2\mathbf{e}+3\mathbf{e} = \mathbf{6e} \Rightarrow 6$ is an eigenvalue | M1A1 | Obtains eigenvalues of related matrix |
| $(-2)^4\mathbf{e}+2\times(-2)^2\mathbf{e}+3\mathbf{e} = 27\mathbf{e} \Rightarrow 27$ is an eigenvalue | | |
| $4^4\mathbf{e}+2\times4^2\mathbf{e}+3\mathbf{e} = 291\mathbf{e} \Rightarrow 291$ is an eigenvalue | A1 | |
7 The square matrix $\mathbf { A }$ has $\lambda$ as an eigenvalue with $\mathbf { e }$ as a corresponding eigenvector. Show that $\mathbf { e }$ is an eigenvector of $\mathbf { A } ^ { 2 }$ and state the corresponding eigenvalue.

Find the eigenvalues of the matrix $\mathbf { B }$, where

$$\mathbf { B } = \left( \begin{array} { l l l } 
1 & 3 & 0 \\
2 & 0 & 2 \\
1 & 1 & 2
\end{array} \right)$$

Find the eigenvalues of $\mathbf { B } ^ { 4 } + 2 \mathbf { B } ^ { 2 } + 3 \mathbf { I }$, where $\mathbf { I }$ is the $3 \times 3$ identity matrix.

\hfill \mbox{\textit{CAIE FP1 2013 Q7 [10]}}