CAIE FP1 2014 June — Question 8 11 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2014
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeProve eigenvalue/eigenvector properties
DifficultyChallenging +1.2 This is a structured multi-part question on eigenvalues/eigenvectors that guides students through standard proofs and computations. The first proof (showing λ^(-1) is an eigenvalue of A^(-1)) is a classic textbook result requiring straightforward algebraic manipulation. Finding eigenvectors and eigenvalues involves routine calculations. The final part requires understanding diagonalization and applying the earlier result, but follows a clear path. While it requires multiple techniques and careful bookkeeping across several parts, each individual step is standard Further Maths material with no novel insights required.
Spec4.03a Matrix language: terminology and notation4.03h Determinant 2x2: calculation4.03i Determinant: area scale factor and orientation4.03n Inverse 2x2 matrix4.03o Inverse 3x3 matrix

8 It is given that \(\lambda\) is an eigenvalue of the non-singular square matrix \(\mathbf { A }\), with corresponding eigenvector
e. Show that \(\lambda ^ { - 1 }\) is an eigenvalue of \(\mathbf { A } ^ { - 1 }\) for which \(\mathbf { e }\) is a corresponding eigenvector. Deduce that \(\lambda + \lambda ^ { - 1 }\) is an eigenvalue of \(\mathbf { A } + \mathbf { A } ^ { - 1 }\). It is given that 1 is an eigenvalue of the matrix \(\mathbf { A }\), where $$\mathbf { A } = \left( \begin{array} { r r r } 2 & 0 & 1 \\ - 1 & 2 & 3 \\ 1 & 0 & 2 \end{array} \right)$$ Find a corresponding eigenvector. It is also given that \(\left( \begin{array} { l } 0 \\ 1 \\ 0 \end{array} \right)\) and \(\left( \begin{array} { l } 1 \\ 2 \\ 1 \end{array} \right)\) are eigenvectors of the matrix \(\mathbf { A }\). Find the corresponding eigenvalues.
Hence find a matrix \(\mathbf { P }\) and a diagonal matrix \(\mathbf { D }\) such that $$\left( \mathbf { A } + \mathbf { A } ^ { - 1 } \right) ^ { 3 } = \mathbf { P D P } \mathbf { P } ^ { - 1 }$$

Question 8:
AnswerMarks Guidance
Working/AnswerMark Guidance
\(\mathbf{A}\mathbf{e} = \lambda\mathbf{e} \Rightarrow \mathbf{A}^{-1}\mathbf{A}\mathbf{e} = \mathbf{A}^{-1}\lambda\mathbf{e}\)M1
\(\therefore \mathbf{e} = \mathbf{A}^{-1}\lambda\mathbf{e} \Rightarrow \frac{1}{\lambda}\mathbf{e} = \mathbf{A}^{-1}\mathbf{e}\)A1
\(\mathbf{A}\mathbf{e} + \mathbf{A}^{-1}\mathbf{e} = \lambda\mathbf{e} + \frac{1}{\lambda}\mathbf{e} \Rightarrow (\mathbf{A} + \mathbf{A}^{-1})\mathbf{e} = \left(\lambda + \frac{1}{\lambda}\right)\mathbf{e}\)B1
\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ -1 & 1 & 3 \end{vmatrix} = \begin{pmatrix}-1\\-4\\1\end{pmatrix}\)M1A1 OE
\(\begin{pmatrix}2&0&1\\-1&2&3\\1&0&2\end{pmatrix}\begin{pmatrix}0\\1\\0\end{pmatrix} = \begin{pmatrix}0\\2\\0\end{pmatrix} \Rightarrow \lambda = 2\)B1
\(\begin{pmatrix}2&0&1\\-1&2&3\\1&0&2\end{pmatrix}\begin{pmatrix}1\\2\\1\end{pmatrix} = \begin{pmatrix}3\\6\\3\end{pmatrix} \Rightarrow \lambda = 3\)B1 S.C. Award B1 for eigenvalues from characteristic equation not matched to eigenvectors
\(\mathbf{P} = \begin{pmatrix}-1&0&1\\-4&1&2\\1&0&1\end{pmatrix}\)B1✓ OE, F.T. on their calculated eigenvector
Eigenvalues of \((\mathbf{A}+\mathbf{A}^{-1})^3\): \(\left(1+\frac{1}{1}\right)^3 = 8\), \(\left(2+\frac{1}{2}\right)^3 = \frac{125}{8}\), \(\left(3+\frac{1}{3}\right)^3 = \frac{1000}{27}\)M1A1
\(\mathbf{D} = \begin{pmatrix}8&0&0\\0&\frac{125}{8}&0\\0&0&\frac{1000}{27}\end{pmatrix}\)B1✓ F.T. requires decent attempt at \(\left(\lambda+\frac{1}{\lambda}\right)^3\)
Total[4]
# Question 8:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\mathbf{A}\mathbf{e} = \lambda\mathbf{e} \Rightarrow \mathbf{A}^{-1}\mathbf{A}\mathbf{e} = \mathbf{A}^{-1}\lambda\mathbf{e}$ | M1 | |
| $\therefore \mathbf{e} = \mathbf{A}^{-1}\lambda\mathbf{e} \Rightarrow \frac{1}{\lambda}\mathbf{e} = \mathbf{A}^{-1}\mathbf{e}$ | A1 | |
| $\mathbf{A}\mathbf{e} + \mathbf{A}^{-1}\mathbf{e} = \lambda\mathbf{e} + \frac{1}{\lambda}\mathbf{e} \Rightarrow (\mathbf{A} + \mathbf{A}^{-1})\mathbf{e} = \left(\lambda + \frac{1}{\lambda}\right)\mathbf{e}$ | B1 | |
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ -1 & 1 & 3 \end{vmatrix} = \begin{pmatrix}-1\\-4\\1\end{pmatrix}$ | M1A1 | OE |
| $\begin{pmatrix}2&0&1\\-1&2&3\\1&0&2\end{pmatrix}\begin{pmatrix}0\\1\\0\end{pmatrix} = \begin{pmatrix}0\\2\\0\end{pmatrix} \Rightarrow \lambda = 2$ | B1 | |
| $\begin{pmatrix}2&0&1\\-1&2&3\\1&0&2\end{pmatrix}\begin{pmatrix}1\\2\\1\end{pmatrix} = \begin{pmatrix}3\\6\\3\end{pmatrix} \Rightarrow \lambda = 3$ | B1 | S.C. Award B1 for eigenvalues from characteristic equation not matched to eigenvectors |
| $\mathbf{P} = \begin{pmatrix}-1&0&1\\-4&1&2\\1&0&1\end{pmatrix}$ | B1✓ | OE, F.T. on their calculated eigenvector |
| Eigenvalues of $(\mathbf{A}+\mathbf{A}^{-1})^3$: $\left(1+\frac{1}{1}\right)^3 = 8$, $\left(2+\frac{1}{2}\right)^3 = \frac{125}{8}$, $\left(3+\frac{1}{3}\right)^3 = \frac{1000}{27}$ | M1A1 | |
| $\mathbf{D} = \begin{pmatrix}8&0&0\\0&\frac{125}{8}&0\\0&0&\frac{1000}{27}\end{pmatrix}$ | B1✓ | F.T. requires decent attempt at $\left(\lambda+\frac{1}{\lambda}\right)^3$ |
| Total | [4] | |

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8 It is given that $\lambda$ is an eigenvalue of the non-singular square matrix $\mathbf { A }$, with corresponding eigenvector\\
e. Show that $\lambda ^ { - 1 }$ is an eigenvalue of $\mathbf { A } ^ { - 1 }$ for which $\mathbf { e }$ is a corresponding eigenvector.

Deduce that $\lambda + \lambda ^ { - 1 }$ is an eigenvalue of $\mathbf { A } + \mathbf { A } ^ { - 1 }$.

It is given that 1 is an eigenvalue of the matrix $\mathbf { A }$, where

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

Find a corresponding eigenvector.

It is also given that $\left( \begin{array} { l } 0 \\ 1 \\ 0 \end{array} \right)$ and $\left( \begin{array} { l } 1 \\ 2 \\ 1 \end{array} \right)$ are eigenvectors of the matrix $\mathbf { A }$. Find the corresponding eigenvalues.\\
Hence find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that

$$\left( \mathbf { A } + \mathbf { A } ^ { - 1 } \right) ^ { 3 } = \mathbf { P D P } \mathbf { P } ^ { - 1 }$$

\hfill \mbox{\textit{CAIE FP1 2014 Q8 [11]}}