| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Invariant lines and eigenvalues and vectors |
| Type | Find P and D for A = PDP⁻¹ |
| Difficulty | Standard +0.3 This is a standard Further Maths diagonalization question with routine steps: finding the characteristic polynomial (given form to verify), solving a cubic that factors nicely, finding eigenvectors by row reduction, and writing down P and D. While it involves 3×3 matrices (more computation than 2×2), all steps are algorithmic with no novel insight required, making it slightly easier than average for FP2. |
| Spec | 4.03j Determinant 3x3: calculation4.03l Singular/non-singular matrices |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{M} - \lambda\mathbf{I} = \begin{pmatrix} 1-\lambda & 3 & 0 \\ 3 & -2-\lambda & -1 \\ 0 & -1 & 1-\lambda \end{pmatrix}\) | ||
| \(\det(\mathbf{M}-\lambda\mathbf{I})\) | M1 | Forming \(\det(\mathbf{M}-\lambda\mathbf{I})\) |
| \(= (1-\lambda)[(-2-\lambda)(1-\lambda)-1] - 3[3(1-\lambda)]\) | A1 | Any correct form; Sarrus: \((1-\lambda)^2(-2-\lambda)-10(1-\lambda)\) or e.g. \(\lambda - 1 + (1-\lambda)(\lambda^2+\lambda-11)\) |
| \(= (1-\lambda)(\lambda^2+\lambda-3) - 9(1-\lambda)\) | ||
| \(\Rightarrow \lambda^3 - 13\lambda + 12 = 0\) | A1(ag) | Condone omission of \(= 0\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\lambda-1)(\lambda^2+\lambda-12) = 0\) | M1 | Factorising as far as quadratic; allow one error |
| \(\Rightarrow (\lambda-1)(\lambda-3)(\lambda+4) = 0\) | A1 | |
| \(\Rightarrow\) eigenvalues are \(1, 3, -4\) | A1 | |
| \(\lambda=1\): \(\begin{pmatrix}0&3&0\\3&-3&-1\\0&-1&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\) | M2 | For any one of \(\lambda = 1, 3, -4\): obtaining two independent equations; from which an eigenvector could be found |
| M1 | Obtaining a non-zero eigenvector | |
| \(\Rightarrow y=0, 3x-z=0\) | A1 | o.e.; allow e.g. \(3y=0, 3x-3y-z=0\) |
| \(\Rightarrow\) eigenvector is \(\begin{pmatrix}1\\0\\3\end{pmatrix}\) | A1 | |
| \(\lambda=3\): \(\begin{pmatrix}-2&3&0\\3&-5&-1\\0&-1&-2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\) | ||
| \(\Rightarrow -2x+3y=0, -y-2z=0\) | A1 | o.e. |
| \(\Rightarrow y=-2z, x=-3z\) | ||
| \(\Rightarrow\) eigenvector is \(\begin{pmatrix}-3\\-2\\1\end{pmatrix}\) | A1 | |
| \(\lambda=-4\): \(\begin{pmatrix}5&3&0\\3&2&-1\\0&-1&5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\) | ||
| \(\Rightarrow 5x+3y=0, -y+5z=0\) | A1 | o.e. |
| \(\Rightarrow y=5z, x=-3z\) | ||
| \(\Rightarrow\) eigenvector is \(\begin{pmatrix}-3\\5\\1\end{pmatrix}\) | A1 | |
| [12] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| e.g. \(\mathbf{P} = \begin{pmatrix}1&-3&-3\\0&-2&5\\3&1&1\end{pmatrix}\) | B1 | Use of eigenvectors (ft) as columns |
| \(\mathbf{D} = \begin{pmatrix}1&0&0\\0&3^n&0\\0&0&(-4)^n\end{pmatrix}\) | M1 | Use of \(1, 3, -4\) (ft) in correct order; \(n\) not required for M1 |
| A1 | Power \(n\); \(-4^n\) A0 | |
| [3] |
# Question 3:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{M} - \lambda\mathbf{I} = \begin{pmatrix} 1-\lambda & 3 & 0 \\ 3 & -2-\lambda & -1 \\ 0 & -1 & 1-\lambda \end{pmatrix}$ | | |
| $\det(\mathbf{M}-\lambda\mathbf{I})$ | M1 | Forming $\det(\mathbf{M}-\lambda\mathbf{I})$ |
| $= (1-\lambda)[(-2-\lambda)(1-\lambda)-1] - 3[3(1-\lambda)]$ | A1 | Any correct form; Sarrus: $(1-\lambda)^2(-2-\lambda)-10(1-\lambda)$ or e.g. $\lambda - 1 + (1-\lambda)(\lambda^2+\lambda-11)$ |
| $= (1-\lambda)(\lambda^2+\lambda-3) - 9(1-\lambda)$ | | |
| $\Rightarrow \lambda^3 - 13\lambda + 12 = 0$ | A1(ag) | Condone omission of $= 0$ |
| **[3]** | | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\lambda-1)(\lambda^2+\lambda-12) = 0$ | M1 | Factorising as far as quadratic; allow one error |
| $\Rightarrow (\lambda-1)(\lambda-3)(\lambda+4) = 0$ | A1 | |
| $\Rightarrow$ eigenvalues are $1, 3, -4$ | A1 | |
| $\lambda=1$: $\begin{pmatrix}0&3&0\\3&-3&-1\\0&-1&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$ | M2 | For any one of $\lambda = 1, 3, -4$: obtaining two independent equations; from which an eigenvector could be found |
| | M1 | Obtaining a non-zero eigenvector |
| $\Rightarrow y=0, 3x-z=0$ | A1 | o.e.; allow e.g. $3y=0, 3x-3y-z=0$ |
| $\Rightarrow$ eigenvector is $\begin{pmatrix}1\\0\\3\end{pmatrix}$ | A1 | |
| $\lambda=3$: $\begin{pmatrix}-2&3&0\\3&-5&-1\\0&-1&-2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$ | | |
| $\Rightarrow -2x+3y=0, -y-2z=0$ | A1 | o.e. |
| $\Rightarrow y=-2z, x=-3z$ | | |
| $\Rightarrow$ eigenvector is $\begin{pmatrix}-3\\-2\\1\end{pmatrix}$ | A1 | |
| $\lambda=-4$: $\begin{pmatrix}5&3&0\\3&2&-1\\0&-1&5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$ | | |
| $\Rightarrow 5x+3y=0, -y+5z=0$ | A1 | o.e. |
| $\Rightarrow y=5z, x=-3z$ | | |
| $\Rightarrow$ eigenvector is $\begin{pmatrix}-3\\5\\1\end{pmatrix}$ | A1 | |
| **[12]** | | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. $\mathbf{P} = \begin{pmatrix}1&-3&-3\\0&-2&5\\3&1&1\end{pmatrix}$ | B1 | Use of eigenvectors (ft) as columns |
| $\mathbf{D} = \begin{pmatrix}1&0&0\\0&3^n&0\\0&0&(-4)^n\end{pmatrix}$ | M1 | Use of $1, 3, -4$ (ft) in correct order; $n$ not required for M1 |
| | A1 | Power $n$; $-4^n$ A0 |
| **[3]** | | |3 You are given the matrix $\mathbf { M } = \left( \begin{array} { r r r } 1 & 3 & 0 \\ 3 & - 2 & - 1 \\ 0 & - 1 & 1 \end{array} \right)$.\\
(i) Show that the characteristic equation of $\mathbf { M }$ is
$$\lambda ^ { 3 } - 13 \lambda + 12 = 0 .$$
(ii) Find the eigenvalues and corresponding eigenvectors of $\mathbf { M }$.\\
(iii) Write down a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that
$$\mathbf { M } ^ { n } = \mathbf { P D P } ^ { - 1 } .$$
(You are not required to calculate $\mathbf { P } ^ { - 1 }$.)
\hfill \mbox{\textit{OCR MEI FP2 2013 Q3 [18]}}