OCR MEI FP2 2010 June — Question 3 19 marks

Exam BoardOCR MEI
ModuleFP2 (Further Pure Mathematics 2)
Year2010
SessionJune
Marks19
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInvariant lines and eigenvalues and vectors
TypeUse Cayley-Hamilton for matrix power
DifficultyStandard +0.8 This is a substantial Further Maths question requiring multiple techniques: verifying eigenvalues by substitution, factorizing a cubic, applying eigenvector properties, solving a matrix equation, and crucially using Cayley-Hamilton theorem to express M^4 in terms of lower powers. Part (b) requires reconstructing a matrix from eigenvalues/eigenvectors. While each individual step is methodical, the combination of techniques, the 18-mark length, and the Cayley-Hamilton application (which students often find non-trivial) place this above average difficulty even for FP2 students.
Spec4.03g Invariant points and lines4.03i Determinant: area scale factor and orientation4.03l Singular/non-singular matrices4.03n Inverse 2x2 matrix4.03q Inverse transformations
3
    1. A \(3 \times 3\) matrix \(\mathbf { M }\) has characteristic equation $$2 \lambda ^ { 3 } + \lambda ^ { 2 } - 13 \lambda + 6 = 0$$ Show that \(\lambda = 2\) is an eigenvalue of \(\mathbf { M }\). Find the other eigenvalues.
    2. An eigenvector corresponding to \(\lambda = 2\) is \(\left( \begin{array} { r } 3 \\ - 3 \\ 1 \end{array} \right)\). Evaluate \(\mathbf { M } \left( \begin{array} { r } 3 \\ - 3 \\ 1 \end{array} \right)\) and \(\mathbf { M } ^ { 2 } \left( \begin{array} { r } 1 \\ - 1 \\ \frac { 1 } { 3 } \end{array} \right)\).
      Solve the equation \(\mathbf { M } \left( \begin{array} { l } x \\ y \\ z \end{array} \right) = \left( \begin{array} { r } 3 \\ - 3 \\ 1 \end{array} \right)\).
    3. Find constants \(A , B , C\) such that $$\mathbf { M } ^ { 4 } = A \mathbf { M } ^ { 2 } + B \mathbf { M } + C \mathbf { I }$$
  2. A \(2 \times 2\) matrix \(\mathbf { N }\) has eigenvalues -1 and 2, with eigenvectors \(\binom { 1 } { 2 }\) and \(\binom { - 1 } { 1 }\) respectively. Find \(\mathbf { N }\). Section B (18 marks)
Question 3:
Part (a)(i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(2\lambda^3 + \lambda^2 - 13\lambda + 6 = 0 \Rightarrow (\lambda-2)(2\lambda^2+5\lambda-3)=0\)B1 Substituting \(\lambda=2\) or factorising
\(\lambda=2\) or \(2\lambda^2+5\lambda-3=0\)M1 Obtaining and solving a quadratic
\((2\lambda-1)(\lambda+3)=0\)
\(\lambda=\frac{1}{2}\), \(\lambda=-3\)A1A1
Part (a)(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\mathbf{M}\begin{pmatrix}3\\-3\\1\end{pmatrix} = 2\begin{pmatrix}3\\-3\\1\end{pmatrix} = \begin{pmatrix}6\\-6\\2\end{pmatrix}\)B1
\(\mathbf{M}^2\mathbf{v} = 2^2\mathbf{v} = 4\begin{pmatrix}1\\-1\\\frac{1}{3}\end{pmatrix} = \begin{pmatrix}4\\-4\\\frac{4}{3}\end{pmatrix}\)B2 Give B1 for one component with wrong sign
\(\mathbf{M}\begin{pmatrix}\frac{3}{2}\\-\frac{3}{2}\\\frac{1}{2}\end{pmatrix} = 2\begin{pmatrix}\frac{3}{2}\\-\frac{3}{2}\\\frac{1}{2}\end{pmatrix} = \begin{pmatrix}3\\-3\\1\end{pmatrix}\)M1 Recognising solution is a multiple of given RHS
\(x=\frac{3}{2}\), \(y=-\frac{3}{2}\), \(z=\frac{1}{2}\)A1 Correct multiple
Part (a)(iii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(2\lambda^3+\lambda^2-13\lambda+6=0 \Rightarrow 2\mathbf{M}^3+\mathbf{M}^2-13\mathbf{M}+6\mathbf{I}=\mathbf{0}\)M1 Using Cayley-Hamilton Theorem
\(\Rightarrow \mathbf{M}^3 = -\frac{1}{2}\mathbf{M}^2+\frac{13}{2}\mathbf{M}-3\mathbf{I}\)M1 Multiplying by \(\mathbf{M}\)
\(\Rightarrow \mathbf{M}^4 = -\frac{1}{2}\mathbf{M}^3+\frac{13}{2}\mathbf{M}^2-3\mathbf{M}\)M1 Substituting for \(\mathbf{M}^3\)
\(\Rightarrow \mathbf{M}^4 = -\frac{1}{2}(-\frac{1}{2}\mathbf{M}^2+\frac{13}{2}\mathbf{M}-3\mathbf{I})+\frac{13}{2}\mathbf{M}^2-3\mathbf{M}\)
\(\Rightarrow \mathbf{M}^4 = \frac{27}{4}\mathbf{M}^2-\frac{25}{4}\mathbf{M}+\frac{3}{2}\mathbf{I}\)A1
\(A=\frac{27}{4}\), \(B=-\frac{25}{4}\), \(C=\frac{3}{2}\)
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\mathbf{N}=\mathbf{PDP}^{-1}\)B1 Order must be correct
\(\mathbf{D}=\begin{pmatrix}-1&0\\0&2\end{pmatrix}\)B1
\(\mathbf{P}=\begin{pmatrix}1&-1\\2&1\end{pmatrix}\)B1 For B1B1, order must be consistent
\(\mathbf{P}^{-1}=\frac{1}{3}\begin{pmatrix}1&1\\-2&1\end{pmatrix}\)B1ft Ft their \(\mathbf{P}\)
\(\mathbf{N}=\frac{1}{3}\begin{pmatrix}1&-1\\2&1\end{pmatrix}\begin{pmatrix}-1&0\\0&2\end{pmatrix}\begin{pmatrix}1&1\\-2&1\end{pmatrix}\)
\(=\frac{1}{3}\begin{pmatrix}-1&-2\\-2&2\end{pmatrix}\begin{pmatrix}1&1\\-2&1\end{pmatrix}\)M1 Attempting matrix product
\(=\frac{1}{3}\begin{pmatrix}3&-3\\-6&0\end{pmatrix}=\begin{pmatrix}1&-1\\-2&0\end{pmatrix}\)A1
OR Let \(\mathbf{N}=\begin{pmatrix}a&c\\b&d\end{pmatrix}\):
\(\begin{pmatrix}a&c\\b&d\end{pmatrix}\begin{pmatrix}1\\2\end{pmatrix}=-1\begin{pmatrix}1\\2\end{pmatrix}\)B1 Or \(\begin{pmatrix}a+1&c\\b&d+1\end{pmatrix}\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\)
\(\begin{pmatrix}a&c\\b&d\end{pmatrix}\begin{pmatrix}-1\\1\end{pmatrix}=2\begin{pmatrix}-1\\1\end{pmatrix}\)B1 Or \(\begin{pmatrix}a-2&c\\b&d-2\end{pmatrix}\begin{pmatrix}-1\\1\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\)
\(a+2c=-1\), \(-a+c=-2\); \(b+2d=-2\), \(-b+d=2\)B1, B1
\(a=1, c=-1; b=-2, d=0\)M1A1 Solving both pairs of equations 
# Question 3:

## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\lambda^3 + \lambda^2 - 13\lambda + 6 = 0 \Rightarrow (\lambda-2)(2\lambda^2+5\lambda-3)=0$ | B1 | Substituting $\lambda=2$ or factorising |
| $\lambda=2$ or $2\lambda^2+5\lambda-3=0$ | M1 | Obtaining and solving a quadratic |
| $(2\lambda-1)(\lambda+3)=0$ | | |
| $\lambda=\frac{1}{2}$, $\lambda=-3$ | A1A1 | |

## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{M}\begin{pmatrix}3\\-3\\1\end{pmatrix} = 2\begin{pmatrix}3\\-3\\1\end{pmatrix} = \begin{pmatrix}6\\-6\\2\end{pmatrix}$ | B1 | |
| $\mathbf{M}^2\mathbf{v} = 2^2\mathbf{v} = 4\begin{pmatrix}1\\-1\\\frac{1}{3}\end{pmatrix} = \begin{pmatrix}4\\-4\\\frac{4}{3}\end{pmatrix}$ | B2 | Give B1 for one component with wrong sign |
| $\mathbf{M}\begin{pmatrix}\frac{3}{2}\\-\frac{3}{2}\\\frac{1}{2}\end{pmatrix} = 2\begin{pmatrix}\frac{3}{2}\\-\frac{3}{2}\\\frac{1}{2}\end{pmatrix} = \begin{pmatrix}3\\-3\\1\end{pmatrix}$ | M1 | Recognising solution is a multiple of given RHS |
| $x=\frac{3}{2}$, $y=-\frac{3}{2}$, $z=\frac{1}{2}$ | A1 | Correct multiple |

## Part (a)(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\lambda^3+\lambda^2-13\lambda+6=0 \Rightarrow 2\mathbf{M}^3+\mathbf{M}^2-13\mathbf{M}+6\mathbf{I}=\mathbf{0}$ | M1 | Using Cayley-Hamilton Theorem |
| $\Rightarrow \mathbf{M}^3 = -\frac{1}{2}\mathbf{M}^2+\frac{13}{2}\mathbf{M}-3\mathbf{I}$ | M1 | Multiplying by $\mathbf{M}$ |
| $\Rightarrow \mathbf{M}^4 = -\frac{1}{2}\mathbf{M}^3+\frac{13}{2}\mathbf{M}^2-3\mathbf{M}$ | M1 | Substituting for $\mathbf{M}^3$ |
| $\Rightarrow \mathbf{M}^4 = -\frac{1}{2}(-\frac{1}{2}\mathbf{M}^2+\frac{13}{2}\mathbf{M}-3\mathbf{I})+\frac{13}{2}\mathbf{M}^2-3\mathbf{M}$ | | |
| $\Rightarrow \mathbf{M}^4 = \frac{27}{4}\mathbf{M}^2-\frac{25}{4}\mathbf{M}+\frac{3}{2}\mathbf{I}$ | A1 | |
| $A=\frac{27}{4}$, $B=-\frac{25}{4}$, $C=\frac{3}{2}$ | | |

## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{N}=\mathbf{PDP}^{-1}$ | B1 | Order must be correct |
| $\mathbf{D}=\begin{pmatrix}-1&0\\0&2\end{pmatrix}$ | B1 | |
| $\mathbf{P}=\begin{pmatrix}1&-1\\2&1\end{pmatrix}$ | B1 | For B1B1, order must be consistent |
| $\mathbf{P}^{-1}=\frac{1}{3}\begin{pmatrix}1&1\\-2&1\end{pmatrix}$ | B1ft | Ft their $\mathbf{P}$ |
| $\mathbf{N}=\frac{1}{3}\begin{pmatrix}1&-1\\2&1\end{pmatrix}\begin{pmatrix}-1&0\\0&2\end{pmatrix}\begin{pmatrix}1&1\\-2&1\end{pmatrix}$ | | |
| $=\frac{1}{3}\begin{pmatrix}-1&-2\\-2&2\end{pmatrix}\begin{pmatrix}1&1\\-2&1\end{pmatrix}$ | M1 | Attempting matrix product |
| $=\frac{1}{3}\begin{pmatrix}3&-3\\-6&0\end{pmatrix}=\begin{pmatrix}1&-1\\-2&0\end{pmatrix}$ | A1 | |
| **OR** Let $\mathbf{N}=\begin{pmatrix}a&c\\b&d\end{pmatrix}$: | | |
| $\begin{pmatrix}a&c\\b&d\end{pmatrix}\begin{pmatrix}1\\2\end{pmatrix}=-1\begin{pmatrix}1\\2\end{pmatrix}$ | B1 | Or $\begin{pmatrix}a+1&c\\b&d+1\end{pmatrix}\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$ |
| $\begin{pmatrix}a&c\\b&d\end{pmatrix}\begin{pmatrix}-1\\1\end{pmatrix}=2\begin{pmatrix}-1\\1\end{pmatrix}$ | B1 | Or $\begin{pmatrix}a-2&c\\b&d-2\end{pmatrix}\begin{pmatrix}-1\\1\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$ |
| $a+2c=-1$, $-a+c=-2$; $b+2d=-2$, $-b+d=2$ | B1, B1 | |
| $a=1, c=-1; b=-2, d=0$ | M1A1 | Solving both pairs of equations |

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3
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item A $3 \times 3$ matrix $\mathbf { M }$ has characteristic equation

$$2 \lambda ^ { 3 } + \lambda ^ { 2 } - 13 \lambda + 6 = 0$$

Show that $\lambda = 2$ is an eigenvalue of $\mathbf { M }$. Find the other eigenvalues.
\item An eigenvector corresponding to $\lambda = 2$ is $\left( \begin{array} { r } 3 \\ - 3 \\ 1 \end{array} \right)$.

Evaluate $\mathbf { M } \left( \begin{array} { r } 3 \\ - 3 \\ 1 \end{array} \right)$ and $\mathbf { M } ^ { 2 } \left( \begin{array} { r } 1 \\ - 1 \\ \frac { 1 } { 3 } \end{array} \right)$.\\
Solve the equation $\mathbf { M } \left( \begin{array} { l } x \\ y \\ z \end{array} \right) = \left( \begin{array} { r } 3 \\ - 3 \\ 1 \end{array} \right)$.
\item Find constants $A , B , C$ such that

$$\mathbf { M } ^ { 4 } = A \mathbf { M } ^ { 2 } + B \mathbf { M } + C \mathbf { I }$$
\end{enumerate}\item A $2 \times 2$ matrix $\mathbf { N }$ has eigenvalues -1 and 2, with eigenvectors $\binom { 1 } { 2 }$ and $\binom { - 1 } { 1 }$ respectively. Find $\mathbf { N }$.

Section B (18 marks)
\end{enumerate}

\hfill \mbox{\textit{OCR MEI FP2 2010 Q3 [19]}}
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Q1 19 Q2 16 Q3 19 Q4 18 Q5 18