Question 9 - A-Level Maths

CAIE FP1 2010 November — Question 9 10 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2010
Session	November
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	3x3 Matrices
Type	Find P and D for diagonalization / matrix powers
Difficulty	Challenging +1.2 This is a standard diagonalization problem requiring eigenvalue/eigenvector calculation for a symmetric 3×3 matrix, followed by applying the diagonalization to (A-2I)³. While it involves multiple steps and matrix manipulation, it follows a completely routine algorithmic procedure taught in Further Maths with no novel insight required. The symmetric structure simplifies the eigenvalue calculation, and the transformation to (A-2I)³ is a direct application of diagonalization properties.
Spec	4.03a Matrix language: terminology and notation 4.03b Matrix operations: addition, multiplication, scalar 4.03c Matrix multiplication: properties (associative, not commutative)

9 Find the eigenvalues and corresponding eigenvectors of the matrix $$\mathbf { A } = \left( \begin{array} { r r r } 3 & - 1 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 1 & 3 \end{array} \right)$$ Find a non-singular matrix $\mathbf { M }$ and a diagonal matrix $\mathbf { D }$ such that $( \mathbf { A } - 2 \mathbf { I } ) ^ { 3 } = \mathbf { M D M } ^ { - 1 }$, where $\mathbf { I }$ is the $3 \times 3$ identity matrix.

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Answer	Marks	Guidance
$(3 - \lambda)[(2 - \lambda)(3 - \lambda) - 1] + [(3 - \lambda)] = 0$	M1	characteristic equation
$(3 - \lambda)(\lambda - 1)(\lambda - 4) = 0$	M1	factorise
$\lambda = 1, 3, 4$	A1
$\begin{pmatrix} 3-\lambda & -1 & 0 \\ -1 & 2-\lambda & -1 \\ 0 & -1 & 3-\lambda \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$	M1A1
$\text{Solve for } \lambda = 1: (1, 2, 1)$	A1
$\text{Solve for } \lambda = 3: (1, 0, -1)$	A1
$\text{Solve for } \lambda = 4: (1, -1, 1)$	A1
$\mathbf{M} = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 0 & -1 \\ 1 & -1 & 1 \end{pmatrix}$	B1∨	eigenvectors as columns
(except $\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$)
$\mathbf{D} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 8 \end{pmatrix}$	M1A1∨	ft on eigenvalues

$(3 - \lambda)[(2 - \lambda)(3 - \lambda) - 1] + [(3 - \lambda)] = 0$ | M1 | characteristic equation
$(3 - \lambda)(\lambda - 1)(\lambda - 4) = 0$ | M1 | factorise
$\lambda = 1, 3, 4$ | A1

$\begin{pmatrix} 3-\lambda & -1 & 0 \\ -1 & 2-\lambda & -1 \\ 0 & -1 & 3-\lambda \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ | M1A1

$\text{Solve for } \lambda = 1: (1, 2, 1)$ | A1
$\text{Solve for } \lambda = 3: (1, 0, -1)$ | A1
$\text{Solve for } \lambda = 4: (1, -1, 1)$ | A1 | | [7]

$\mathbf{M} = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 0 & -1 \\ 1 & -1 & 1 \end{pmatrix}$ | B1∨ | eigenvectors as columns
| | | (except $\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$)

$\mathbf{D} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 8 \end{pmatrix}$ | M1A1∨ | ft on eigenvalues | [3]

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9 Find the eigenvalues and corresponding eigenvectors of the matrix

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

Find a non-singular matrix $\mathbf { M }$ and a diagonal matrix $\mathbf { D }$ such that $( \mathbf { A } - 2 \mathbf { I } ) ^ { 3 } = \mathbf { M D M } ^ { - 1 }$, where $\mathbf { I }$ is the $3 \times 3$ identity matrix.

\hfill \mbox{\textit{CAIE FP1 2010 Q9 [10]}}

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Q1 4 Q2 5 Q3 5 Q4 5 Q5 8 Q6 8 Q7 9 Q8 10 Q9 10 Q10 10 Q11 12 Q12 EITHER Q12 OR

Answer	Marks	Guidance
\((3 - \lambda)[(2 - \lambda)(3 - \lambda) - 1] + [(3 - \lambda)] = 0\)	M1	characteristic equation
\((3 - \lambda)(\lambda - 1)(\lambda - 4) = 0\)	M1	factorise
\(\lambda = 1, 3, 4\)	A1
\(\begin{pmatrix} 3-\lambda & -1 & 0 \\ -1 & 2-\lambda & -1 \\ 0 & -1 & 3-\lambda \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\)	M1A1
\(\text{Solve for } \lambda = 1: (1, 2, 1)\)	A1
\(\text{Solve for } \lambda = 3: (1, 0, -1)\)	A1
\(\text{Solve for } \lambda = 4: (1, -1, 1)\)	A1
\(\mathbf{M} = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 0 & -1 \\ 1 & -1 & 1 \end{pmatrix}\)	B1∨	eigenvectors as columns
(except \(\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\))
\(\mathbf{D} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 8 \end{pmatrix}\)	M1A1∨	ft on eigenvalues