Question 6 - A-Level Maths

CAIE Further Paper 2 2021 June — Question 6 11 marks

Exam Board	CAIE
Module	Further Paper 2 (Further Paper 2)
Year	2021
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Invariant lines and eigenvalues and vectors
Type	Find P and D for A² = PDP⁻¹ or A⁻¹ = PDP⁻¹
Difficulty	Challenging +1.2 This is a structured Further Maths question requiring eigenvalue/eigenvector calculation and diagonalization, but the triangular form makes finding eigenvalues trivial (just read off the diagonal). Part (b) using Cayley-Hamilton is a standard technique. More routine than typical diagonalization problems due to the simplified matrix structure, but still requires multiple technical steps beyond basic A-level.
Spec	4.03o Inverse 3x3 matrix 4.03t Plane intersection: geometric interpretation

6 The matrix $\mathbf { A }$ is given by $$A = \left( \begin{array} { r r r } 5 & - \frac { 22 } { 3 } & 8 \\ 0 & - 6 & 0 \\ 0 & 0 & 1 \end{array} \right)$$

Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { A } ^ { 2 } = \mathbf { P D P } ^ { - 1 }$.
Use the characteristic equation of $\mathbf { A }$ to find $\mathbf { A } ^ { 3 }$.

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Question 6(a):

Answer	Marks	Guidance
Eigenvalues of A are $5$, $-6$ and $1$.	B1	Upper diagonal matrix or characteristic equation.
$\lambda = 5$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -\frac{22}{3} & 8 \\ 0 & -11 & 0 \end{vmatrix} = \begin{pmatrix}88\\0\\0\end{pmatrix} \sim \begin{pmatrix}1\\0\\0\end{pmatrix}$	M1	Uses vector product (or equations) to find corresponding eigenvectors.
$\lambda = -6$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 11 & -\frac{22}{3} & 8 \\ 0 & 0 & 7 \end{vmatrix} = \begin{pmatrix}-\frac{154}{3}\\-77\\0\end{pmatrix} \sim \begin{pmatrix}2\\3\\0\end{pmatrix}$	A1 A1 A1	A1 for each correct eigenvector.

$\lambda = 1$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -\frac{22}{3} & 8 \\ 0 & -7 & 0 \end{vmatrix} = \begin{pmatrix}56\\0\\-28\end{pmatrix} \sim \begin{pmatrix}2\\0\\-1\end{pmatrix}$

Answer	Marks	Guidance
$\mathbf{P} = \begin{pmatrix}1&2&2\\0&3&0\\0&0&-1\end{pmatrix}$ and $\mathbf{D} = \begin{pmatrix}25&0&0\\0&36&0\\0&0&1\end{pmatrix}$	M1 A1	Or correctly matched permutations of columns. (Accept scalar multiples of the eigenvectors shown.) P must have at least two non-zero columns.

Question 6(b):

Answer	Marks	Guidance
$(\lambda-5)(\lambda+6)(\lambda-1) = \lambda^3 - 31\lambda + 30 = 0$	B1	Finds characteristic equation.
$\mathbf{A}^3 - 31\mathbf{A} + 30\mathbf{I} = \mathbf{0} \Rightarrow \mathbf{A}^3 = 31\mathbf{A} - 30\mathbf{I}$	M1	Finds $\mathbf{A}^3$ in terms of A. Allow missing I.
$\mathbf{A}^3 = \begin{pmatrix}125 & -\frac{682}{3} & 248\\ 0 & -216 & 0\\ 0 & 0 & 1\end{pmatrix}$	M1A1	Substitutes for A.

## Question 6(a):

Eigenvalues of **A** are $5$, $-6$ and $1$. | **B1** | Upper diagonal matrix or characteristic equation.

$\lambda = 5$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -\frac{22}{3} & 8 \\ 0 & -11 & 0 \end{vmatrix} = \begin{pmatrix}88\\0\\0\end{pmatrix} \sim \begin{pmatrix}1\\0\\0\end{pmatrix}$ | **M1** | Uses vector product (or equations) to find corresponding eigenvectors.

$\lambda = -6$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 11 & -\frac{22}{3} & 8 \\ 0 & 0 & 7 \end{vmatrix} = \begin{pmatrix}-\frac{154}{3}\\-77\\0\end{pmatrix} \sim \begin{pmatrix}2\\3\\0\end{pmatrix}$ | **A1 A1 A1** | A1 for each correct eigenvector.

$\lambda = 1$: $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -\frac{22}{3} & 8 \\ 0 & -7 & 0 \end{vmatrix} = \begin{pmatrix}56\\0\\-28\end{pmatrix} \sim \begin{pmatrix}2\\0\\-1\end{pmatrix}$

$\mathbf{P} = \begin{pmatrix}1&2&2\\0&3&0\\0&0&-1\end{pmatrix}$ and $\mathbf{D} = \begin{pmatrix}25&0&0\\0&36&0\\0&0&1\end{pmatrix}$ | **M1 A1** | Or correctly matched permutations of columns. (Accept scalar multiples of the eigenvectors shown.) **P** must have at least two non-zero columns.

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## Question 6(b):

$(\lambda-5)(\lambda+6)(\lambda-1) = \lambda^3 - 31\lambda + 30 = 0$ | **B1** | Finds characteristic equation.

$\mathbf{A}^3 - 31\mathbf{A} + 30\mathbf{I} = \mathbf{0} \Rightarrow \mathbf{A}^3 = 31\mathbf{A} - 30\mathbf{I}$ | **M1** | Finds $\mathbf{A}^3$ in terms of **A**. Allow missing **I**.

$\mathbf{A}^3 = \begin{pmatrix}125 & -\frac{682}{3} & 248\\ 0 & -216 & 0\\ 0 & 0 & 1\end{pmatrix}$ | **M1A1** | Substitutes for **A**.

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6 The matrix $\mathbf { A }$ is given by

$$A = \left( \begin{array} { r r r } 
5 & - \frac { 22 } { 3 } & 8 \\
0 & - 6 & 0 \\
0 & 0 & 1
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Find a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { A } ^ { 2 } = \mathbf { P D P } ^ { - 1 }$.
\item Use the characteristic equation of $\mathbf { A }$ to find $\mathbf { A } ^ { 3 }$.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q6 [11]}}

This paper (8 questions)

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Q1 5 Q2 7 Q3 10 Q4 9 Q5 10 Q6 11 Q7 10 Q8 13

Answer	Marks	Guidance
Eigenvalues of A are \(5\), \(-6\) and \(1\).	B1	Upper diagonal matrix or characteristic equation.
\(\lambda = 5\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -\frac{22}{3} & 8 \\ 0 & -11 & 0 \end{vmatrix} = \begin{pmatrix}88\\0\\0\end{pmatrix} \sim \begin{pmatrix}1\\0\\0\end{pmatrix}\)	M1	Uses vector product (or equations) to find corresponding eigenvectors.
\(\lambda = -6\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 11 & -\frac{22}{3} & 8 \\ 0 & 0 & 7 \end{vmatrix} = \begin{pmatrix}-\frac{154}{3}\\-77\\0\end{pmatrix} \sim \begin{pmatrix}2\\3\\0\end{pmatrix}\)	A1 A1 A1	A1 for each correct eigenvector.

Answer	Marks	Guidance
\((\lambda-5)(\lambda+6)(\lambda-1) = \lambda^3 - 31\lambda + 30 = 0\)	B1	Finds characteristic equation.
\(\mathbf{A}^3 - 31\mathbf{A} + 30\mathbf{I} = \mathbf{0} \Rightarrow \mathbf{A}^3 = 31\mathbf{A} - 30\mathbf{I}\)	M1	Finds \(\mathbf{A}^3\) in terms of A. Allow missing I.
\(\mathbf{A}^3 = \begin{pmatrix}125 & -\frac{682}{3} & 248\\ 0 & -216 & 0\\ 0 & 0 & 1\end{pmatrix}\)	M1A1	Substitutes for A.

CAIE Further Paper 2 2021 June — Question 6 11 marks

Question 6(a):

\(\lambda = 1\): \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -\frac{22}{3} & 8 \\ 0 & -7 & 0 \end{vmatrix} = \begin{pmatrix}56\\0\\-28\end{pmatrix} \sim \begin{pmatrix}2\\0\\-1\end{pmatrix}\)

Question 6(b):

This paper (8 questions)