CAIE FP1 2010 June — Question 8 10 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2010
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeFind P and D for A = PDP⁻¹
DifficultyStandard +0.3 This is a structured, multi-part eigenvalue/eigenvector question where most information is given. Students must perform routine matrix multiplication to find one eigenvalue, solve (A-4I)v=0 for an eigenvector, then construct the diagonalization A=PDP⁻¹ and compute A⁵=PD⁵P⁻¹. While it involves multiple steps, each is standard FP1 technique with no novel insight required, making it slightly easier than average.
Spec4.03a Matrix language: terminology and notation

8 The matrix \(\mathbf { A }\) is given by $$\mathbf { A } = \left( \begin{array} { r r r } 4 & 1 & - 1 \\ - 4 & - 1 & 4 \\ 0 & - 1 & 5 \end{array} \right)$$ Given that one eigenvector of \(\mathbf { A }\) is \(\left( \begin{array} { r } 1 \\ - 2 \\ - 1 \end{array} \right)\), find the corresponding eigenvalue. Given also that another eigenvalue of \(\mathbf { A }\) is 4, find a corresponding eigenvector. Given further that \(\left( \begin{array} { r } 1 \\ - 4 \\ - 1 \end{array} \right)\) is an eigenvector of \(\mathbf { A }\), with corresponding eigenvalue 1 , find matrices \(\mathbf { P }\) and \(\mathbf { Q }\), together with a diagonal matrix \(\mathbf { D }\), such that \(\mathbf { A } ^ { 5 } = \mathbf { P D Q }\).

AnswerMarks
\(\begin{pmatrix}4 & 1 & -1 \\ -4 & -1 & 4 \\ 0 & -1 & 5\end{pmatrix}\begin{pmatrix}1 \\ -2 \\ -1\end{pmatrix}\begin{pmatrix}3 \\ -6 \\ -3\end{pmatrix}\)M1
\(\Rightarrow\) eigenvalue \(= 3\)A1
[2]
Eigenvector corresponding to 4 is \(\begin{pmatrix}1 \\ -4 \\ -4\end{pmatrix}\)M1A1
[2]
\(\mathbf{D} = \text{diag}(1 \, 243 \, 1024)\)B1
\(\mathbf{P} = \begin{pmatrix}1 & 1 & 1 \\ -4 & -2 & -4 \\ -1 & -1 & -4\end{pmatrix}\) (f.t.)A1
\(\mathbf{Q} = \mathbf{P}^{-1}\)B1
\(= \begin{pmatrix}-2/3 & -1/2 & 1/3 \\ 2 & 1/2 & 0 \\ -1/3 & 0 & -1/3\end{pmatrix}\) ft on PM1A2∇
M1 for any valid method; A2 if completely correct; A1 if exactly 1 error; A0 if > 1 errors[6]
$\begin{pmatrix}4 & 1 & -1 \\ -4 & -1 & 4 \\ 0 & -1 & 5\end{pmatrix}\begin{pmatrix}1 \\ -2 \\ -1\end{pmatrix}\begin{pmatrix}3 \\ -6 \\ -3\end{pmatrix}$ | M1 |

$\Rightarrow$ eigenvalue $= 3$ | A1 |

| [2] |

Eigenvector corresponding to 4 is $\begin{pmatrix}1 \\ -4 \\ -4\end{pmatrix}$ | M1A1 |

| [2] |

$\mathbf{D} = \text{diag}(1 \, 243 \, 1024)$ | B1 |

$\mathbf{P} = \begin{pmatrix}1 & 1 & 1 \\ -4 & -2 & -4 \\ -1 & -1 & -4\end{pmatrix}$ (f.t.) | A1 |

$\mathbf{Q} = \mathbf{P}^{-1}$ | B1 |

$= \begin{pmatrix}-2/3 & -1/2 & 1/3 \\ 2 & 1/2 & 0 \\ -1/3 & 0 & -1/3\end{pmatrix}$ ft on P | M1A2∇ |

M1 for any valid method; A2 if completely correct; A1 if exactly 1 error; A0 if > 1 errors | [6] |
8 The matrix $\mathbf { A }$ is given by

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

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

Given also that another eigenvalue of $\mathbf { A }$ is 4, find a corresponding eigenvector.

Given further that $\left( \begin{array} { r } 1 \\ - 4 \\ - 1 \end{array} \right)$ is an eigenvector of $\mathbf { A }$, with corresponding eigenvalue 1 , find matrices $\mathbf { P }$ and $\mathbf { Q }$, together with a diagonal matrix $\mathbf { D }$, such that $\mathbf { A } ^ { 5 } = \mathbf { P D Q }$.

\hfill \mbox{\textit{CAIE FP1 2010 Q8 [10]}}