Question 11 OR - A-Level Maths

CAIE FP1 2013 November — Question 11 OR

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2013
Session	November
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Invariant lines and eigenvalues and vectors
Type	Find eigenvalues/vectors of matrix combination
Difficulty	Standard +0.3 This is a structured multi-part question on eigenvalues/eigenvectors that guides students through standard procedures: verifying the eigenvalue property for matrix sums, computing eigenvalues by matrix-vector multiplication, applying linearity of eigenvalues, and diagonalization. While it involves multiple steps and 3×3 matrices, each part follows routine algorithms without requiring novel insight or complex problem-solving.
Spec	4.03a Matrix language: terminology and notation 4.03b Matrix operations: addition, multiplication, scalar

The vector $\mathbf { e }$ is an eigenvector of each of the $3 \times 3$ matrices $\mathbf { A }$ and $\mathbf { B }$, with corresponding eigenvalues $\lambda$ and $\mu$ respectively. Justifying your answer, state an eigenvalue of $\mathbf { A } + \mathbf { B }$. The matrix $\mathbf { A }$, where $$\mathbf { A } = \left( \begin{array} { r r r } 6 & - 1 & - 6 \\ 1 & 0 & - 2 \\ 3 & - 1 & - 3 \end{array} \right)$$ has eigenvectors $\left( \begin{array} { l } 1 \\ 1 \\ 1 \end{array} \right) , \left( \begin{array} { r } 1 \\ - 1 \\ 1 \end{array} \right) , \left( \begin{array} { l } 2 \\ 0 \\ 1 \end{array} \right)$. Find the corresponding eigenvalues. The matrix $\mathbf { B }$, where $$\mathbf { B } = \left( \begin{array} { r r r } 8 & - 2 & - 8 \\ 2 & 0 & - 4 \\ 4 & - 2 & - 4 \end{array} \right) ,$$ also has eigenvectors $\left( \begin{array} { l } 1 \\ 1 \\ 1 \end{array} \right) , \left( \begin{array} { r } 1 \\ - 1 \\ 1 \end{array} \right) , \left( \begin{array} { l } 2 \\ 0 \\ 1 \end{array} \right)$, for which $- 2,2,4$, respectively, are corresponding eigenvalues. The matrix $\mathbf { M }$ is given by $\mathbf { M } = \mathbf { A } + \mathbf { B } - 5 \mathbf { I }$, where $\mathbf { I }$ is the $3 \times 3$ identity matrix. State the eigenvalues of $\mathbf { M }$. Find matrices $\mathbf { R }$ and $\mathbf { S }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { M } ^ { 5 } = \mathbf { R D S }$.
[0pt] [You should show clearly all the elements of the matrices $\mathbf { R } , \mathbf { S }$ and $\mathbf { D }$.]

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Question 11 (EITHER – reduction formula):

Part (proving reduction formula):

Answer	Marks	Guidance
$I_n = \left[x(1+x^2)^n\right]_0^1 - 2\int_0^1 nx^2(1+x^2)^{n-1}dx$	M1A1
$= 2^n - 2n\int_0^1(1+x^2-1)(1+x^2)^{n-1}dx$	M1A1
$\Rightarrow I_n = 2^n - 2nI_n + 2nI_{n-1}$
$\Rightarrow (2n+1)I_n = 2nI_{n-1} + 2^n$ (AG)	A1	(5)

Part (finding $I_3$):

Answer	Marks	Guidance
$I_0 = \int_0^1 1\,dx = 1$	B1
$3I_1 = 2\times1\times1 + 2 \Rightarrow I_1 = \frac{4}{3}$	M1
$5I_2 = \frac{16}{3} + 4 \Rightarrow I_2 = \frac{28}{15}$	A1
$7I_3 = \frac{56}{5} + 8 \Rightarrow I_3 = \frac{96}{35}$ (2.74)	A1	(4)

Part (negative $n$):

Answer	Marks	Guidance
$2nI_{n-1} = (2n+1)I_n - 2^n$	M1
$\Rightarrow 2I_{-2} = I_{-1} + \frac{1}{2}$ and $4I_{-3} = 3I_{-2} + \frac{1}{4}$	A1A1
$\Rightarrow I_{-3} = \frac{3}{8}I_{-1} + \frac{1}{4} = \frac{3}{32}\pi + \frac{1}{4}$	M1A1	(5)

Question 11 (OR – eigenvalues/matrices):

Part (eigenvalue proof):

Answer	Marks	Guidance
$\mathbf{Ae} = \lambda\mathbf{e}$ and $\mathbf{Be} = \mu\mathbf{e}$	B1
Adding: $\mathbf{Ae} + \mathbf{Be} = \lambda\mathbf{e} + \mu\mathbf{e} \Rightarrow (\mathbf{A}+\mathbf{B})\mathbf{e} = (\lambda+\mu)\mathbf{e}$	M1
$\Rightarrow \lambda + \mu$ is an eigenvalue (since $\mathbf{e} \neq \mathbf{0}$)	A1	(3)

Part (eigenvalues of $\mathbf{A}$):

Answer	Marks	Guidance
$\mathbf{A}\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}-1\\-1\\-1\end{pmatrix} \Rightarrow \lambda_1 = -1$	M1A1
$\mathbf{A}\begin{pmatrix}1\\-1\\1\end{pmatrix} = \begin{pmatrix}1\\-1\\1\end{pmatrix} \Rightarrow \lambda^2 = 1$	A1
$\mathbf{A}\begin{pmatrix}2\\0\\1\end{pmatrix} = \begin{pmatrix}6\\0\\3\end{pmatrix} \Rightarrow \lambda^3 = 3$	A1	(4)

Part (eigenvalues of $\mathbf{M}$):

Answer	Marks	Guidance
Eigenvalues of $\mathbf{M}$ are $-8, -2, 2$	B1	(1)

Part (diagonalisation):

Answer	Marks	Guidance
$\mathbf{R} = \begin{pmatrix}1 & 1 & 2\\1 & -1 & 0\\1 & 1 & 1\end{pmatrix}$	B1
$\mathbf{S} = \mathbf{R}^{-1}$	M1
Determinant $= 2$, $\mathbf{S} = \frac{1}{2}\begin{pmatrix}-1 & 1 & 2\\-1 & -1 & 2\\2 & 0 & -2\end{pmatrix}$	M1A1
$\mathbf{D} = \begin{pmatrix}-32768 & 0 & 0\\0 & -32 & 0\\0 & 0 & 32\end{pmatrix}$	M1A1	(6)

# Question 11 (EITHER – reduction formula):

## Part (proving reduction formula):
| $I_n = \left[x(1+x^2)^n\right]_0^1 - 2\int_0^1 nx^2(1+x^2)^{n-1}dx$ | M1A1 | |
|---|---|---|
| $= 2^n - 2n\int_0^1(1+x^2-1)(1+x^2)^{n-1}dx$ | M1A1 | |
| $\Rightarrow I_n = 2^n - 2nI_n + 2nI_{n-1}$ | | |
| $\Rightarrow (2n+1)I_n = 2nI_{n-1} + 2^n$ (AG) | A1 | **(5)** |

## Part (finding $I_3$):
| $I_0 = \int_0^1 1\,dx = 1$ | B1 | |
|---|---|---|
| $3I_1 = 2\times1\times1 + 2 \Rightarrow I_1 = \frac{4}{3}$ | M1 | |
| $5I_2 = \frac{16}{3} + 4 \Rightarrow I_2 = \frac{28}{15}$ | A1 | |
| $7I_3 = \frac{56}{5} + 8 \Rightarrow I_3 = \frac{96}{35}$ (2.74) | A1 | **(4)** |

## Part (negative $n$):
| $2nI_{n-1} = (2n+1)I_n - 2^n$ | M1 | |
|---|---|---|
| $\Rightarrow 2I_{-2} = I_{-1} + \frac{1}{2}$ and $4I_{-3} = 3I_{-2} + \frac{1}{4}$ | A1A1 | |
| $\Rightarrow I_{-3} = \frac{3}{8}I_{-1} + \frac{1}{4} = \frac{3}{32}\pi + \frac{1}{4}$ | M1A1 | **(5)** |

---

# Question 11 (OR – eigenvalues/matrices):

## Part (eigenvalue proof):
| $\mathbf{Ae} = \lambda\mathbf{e}$ and $\mathbf{Be} = \mu\mathbf{e}$ | B1 | |
|---|---|---|
| Adding: $\mathbf{Ae} + \mathbf{Be} = \lambda\mathbf{e} + \mu\mathbf{e} \Rightarrow (\mathbf{A}+\mathbf{B})\mathbf{e} = (\lambda+\mu)\mathbf{e}$ | M1 | |
| $\Rightarrow \lambda + \mu$ is an eigenvalue (since $\mathbf{e} \neq \mathbf{0}$) | A1 | **(3)** |

## Part (eigenvalues of $\mathbf{A}$):
| $\mathbf{A}\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}-1\\-1\\-1\end{pmatrix} \Rightarrow \lambda_1 = -1$ | M1A1 | |
|---|---|---|
| $\mathbf{A}\begin{pmatrix}1\\-1\\1\end{pmatrix} = \begin{pmatrix}1\\-1\\1\end{pmatrix} \Rightarrow \lambda^2 = 1$ | A1 | |
| $\mathbf{A}\begin{pmatrix}2\\0\\1\end{pmatrix} = \begin{pmatrix}6\\0\\3\end{pmatrix} \Rightarrow \lambda^3 = 3$ | A1 | **(4)** |

## Part (eigenvalues of $\mathbf{M}$):
| Eigenvalues of $\mathbf{M}$ are $-8, -2, 2$ | B1 | **(1)** |
|---|---|---|

## Part (diagonalisation):
| $\mathbf{R} = \begin{pmatrix}1 & 1 & 2\\1 & -1 & 0\\1 & 1 & 1\end{pmatrix}$ | B1 | |
|---|---|---|
| $\mathbf{S} = \mathbf{R}^{-1}$ | M1 | |
| Determinant $= 2$, $\mathbf{S} = \frac{1}{2}\begin{pmatrix}-1 & 1 & 2\\-1 & -1 & 2\\2 & 0 & -2\end{pmatrix}$ | M1A1 | |
| $\mathbf{D} = \begin{pmatrix}-32768 & 0 & 0\\0 & -32 & 0\\0 & 0 & 32\end{pmatrix}$ | M1A1 | **(6)** |

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The vector $\mathbf { e }$ is an eigenvector of each of the $3 \times 3$ matrices $\mathbf { A }$ and $\mathbf { B }$, with corresponding eigenvalues $\lambda$ and $\mu$ respectively. Justifying your answer, state an eigenvalue of $\mathbf { A } + \mathbf { B }$.

The matrix $\mathbf { A }$, where

$$\mathbf { A } = \left( \begin{array} { r r r } 
6 & - 1 & - 6 \\
1 & 0 & - 2 \\
3 & - 1 & - 3
\end{array} \right)$$

has eigenvectors $\left( \begin{array} { l } 1 \\ 1 \\ 1 \end{array} \right) , \left( \begin{array} { r } 1 \\ - 1 \\ 1 \end{array} \right) , \left( \begin{array} { l } 2 \\ 0 \\ 1 \end{array} \right)$. Find the corresponding eigenvalues.

The matrix $\mathbf { B }$, where

$$\mathbf { B } = \left( \begin{array} { r r r } 
8 & - 2 & - 8 \\
2 & 0 & - 4 \\
4 & - 2 & - 4
\end{array} \right) ,$$

also has eigenvectors $\left( \begin{array} { l } 1 \\ 1 \\ 1 \end{array} \right) , \left( \begin{array} { r } 1 \\ - 1 \\ 1 \end{array} \right) , \left( \begin{array} { l } 2 \\ 0 \\ 1 \end{array} \right)$, for which $- 2,2,4$, respectively, are corresponding eigenvalues. The matrix $\mathbf { M }$ is given by $\mathbf { M } = \mathbf { A } + \mathbf { B } - 5 \mathbf { I }$, where $\mathbf { I }$ is the $3 \times 3$ identity matrix. State the eigenvalues of $\mathbf { M }$.

Find matrices $\mathbf { R }$ and $\mathbf { S }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { M } ^ { 5 } = \mathbf { R D S }$.\\[0pt]
[You should show clearly all the elements of the matrices $\mathbf { R } , \mathbf { S }$ and $\mathbf { D }$.]

\hfill \mbox{\textit{CAIE FP1 2013 Q11 OR}}

This paper (12 questions)

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Q1 6 Q2 6 Q3 7 Q4 7 Q5 8 Q6 8 Q7 9 Q8 11 Q9 11 Q10 13 Q11 EITHER Q11 OR

CAIE FP1 2013 November — Question 11 OR

Question 11 (EITHER – reduction formula):

Part (proving reduction formula):

Part (finding \(I_3\)):

Part (negative \(n\)):

Question 11 (OR – eigenvalues/matrices):

Part (eigenvalue proof):

Part (eigenvalues of \(\mathbf{A}\)):

Part (eigenvalues of \(\mathbf{M}\)):

Part (diagonalisation):

This paper (12 questions)

Answer	Marks	Guidance
\(I_n = \left[x(1+x^2)^n\right]_0^1 - 2\int_0^1 nx^2(1+x^2)^{n-1}dx\)	M1A1
\(= 2^n - 2n\int_0^1(1+x^2-1)(1+x^2)^{n-1}dx\)	M1A1
\(\Rightarrow I_n = 2^n - 2nI_n + 2nI_{n-1}\)
\(\Rightarrow (2n+1)I_n = 2nI_{n-1} + 2^n\) (AG)	A1	(5)

Answer	Marks	Guidance
\(I_0 = \int_0^1 1\,dx = 1\)	B1
\(3I_1 = 2\times1\times1 + 2 \Rightarrow I_1 = \frac{4}{3}\)	M1
\(5I_2 = \frac{16}{3} + 4 \Rightarrow I_2 = \frac{28}{15}\)	A1
\(7I_3 = \frac{56}{5} + 8 \Rightarrow I_3 = \frac{96}{35}\) (2.74)	A1	(4)

Answer	Marks	Guidance
\(2nI_{n-1} = (2n+1)I_n - 2^n\)	M1
\(\Rightarrow 2I_{-2} = I_{-1} + \frac{1}{2}\) and \(4I_{-3} = 3I_{-2} + \frac{1}{4}\)	A1A1
\(\Rightarrow I_{-3} = \frac{3}{8}I_{-1} + \frac{1}{4} = \frac{3}{32}\pi + \frac{1}{4}\)	M1A1	(5)

Answer	Marks	Guidance
\(\mathbf{Ae} = \lambda\mathbf{e}\) and \(\mathbf{Be} = \mu\mathbf{e}\)	B1
Adding: \(\mathbf{Ae} + \mathbf{Be} = \lambda\mathbf{e} + \mu\mathbf{e} \Rightarrow (\mathbf{A}+\mathbf{B})\mathbf{e} = (\lambda+\mu)\mathbf{e}\)	M1
\(\Rightarrow \lambda + \mu\) is an eigenvalue (since \(\mathbf{e} \neq \mathbf{0}\))	A1	(3)

Answer	Marks	Guidance
\(\mathbf{A}\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}-1\\-1\\-1\end{pmatrix} \Rightarrow \lambda_1 = -1\)	M1A1
\(\mathbf{A}\begin{pmatrix}1\\-1\\1\end{pmatrix} = \begin{pmatrix}1\\-1\\1\end{pmatrix} \Rightarrow \lambda^2 = 1\)	A1
\(\mathbf{A}\begin{pmatrix}2\\0\\1\end{pmatrix} = \begin{pmatrix}6\\0\\3\end{pmatrix} \Rightarrow \lambda^3 = 3\)	A1	(4)

Answer	Marks	Guidance
\(\mathbf{R} = \begin{pmatrix}1 & 1 & 2\\1 & -1 & 0\\1 & 1 & 1\end{pmatrix}\)	B1
\(\mathbf{S} = \mathbf{R}^{-1}\)	M1
Determinant \(= 2\), \(\mathbf{S} = \frac{1}{2}\begin{pmatrix}-1 & 1 & 2\\-1 & -1 & 2\\2 & 0 & -2\end{pmatrix}\)	M1A1
\(\mathbf{D} = \begin{pmatrix}-32768 & 0 & 0\\0 & -32 & 0\\0 & 0 & 32\end{pmatrix}\)	M1A1	(6)