| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| 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 diagonalization question requiring eigenvalue/eigenvector calculation, forming P and D matrices, and using the result to find M^n. While it involves multiple steps and the final part requires understanding that M^n = PD^nP^(-1), these are routine Further Maths techniques with no novel insight required. The calculations are straightforward for a 2×2 matrix, making this slightly easier than average for FP2. |
| Spec | 4.03h Determinant 2x2: calculation4.03l Singular/non-singular matrices4.03m det(AB) = det(A)*det(B) |
| Answer | Marks | Guidance |
|---|---|---|
| Characteristic equation is \((7-\lambda)(-1-\lambda)+12=0\), \(\lambda^2-6\lambda+5=0\), \(\lambda=1,5\) | M1, A1A1 | |
| When \(\lambda=1\): \(\begin{pmatrix}7&3\\-4&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x\\y\end{pmatrix}\) | M1 | or \(\begin{pmatrix}6&3\\-4&-2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\) can be awarded for either eigenvalue; Equation relating \(x\) and \(y\) |
| \(7x+3y=x\), \(-4x-y=y\) | M1 | or any (non-zero) multiple |
| \(y=-2x\), eigenvector is \(\begin{pmatrix}1\\-2\end{pmatrix}\) | A1 | |
| When \(\lambda=5\): \(\begin{pmatrix}7&3\\-4&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=5\begin{pmatrix}x\\y\end{pmatrix}\) | M1 | |
| \(7x+3y=5x\), \(-4x-y=5y\) | M1 | |
| \(y=\frac{3}{2}x\), eigenvector is \(\begin{pmatrix}3\\2\end{pmatrix}\) | A1 | |
| 8 | SR \((M-\lambda I)v=\lambda v\) can earn M1A1A1M0M1A0M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P=\begin{pmatrix}1&3\\-2&-2\end{pmatrix}\), \(D=\begin{pmatrix}1&0\\0&5\end{pmatrix}\) | B1 ft, B1 ft | B0 if \(P\) is singular; For B2, the order must be consistent |
| Answer | Marks | Guidance |
|---|---|---|
| \(M=PDP^{-1}\) | M1 | May be implied |
| \(M^n=PD^nP^{-1} = P\begin{pmatrix}1&0\\0&5^n\end{pmatrix}P^{-1}\) | M1 | |
| \(= \begin{pmatrix}1&3\\-2&-2\end{pmatrix}\begin{pmatrix}1&0\\0&5^n\end{pmatrix}\begin{pmatrix}-\frac{1}{2}&\frac{3}{4}\\frac{1}{2}&-\frac{1}{4}\end{pmatrix}\) | A1 ft | Dependent on M1M1 |
| \(= \begin{pmatrix}1&3\times 5^n\\-2&-2\times 5^n\end{pmatrix}\begin{pmatrix}-\frac{1}{2}&\frac{3}{4}\\frac{1}{2}&-\frac{1}{4}\end{pmatrix} = \frac{1}{4}\begin{pmatrix}-2+6\times 5^n&-3+3\times 5^n\\4-4\times 5^n&6-2\times 5^n\end{pmatrix}\) | M1 | Obtaining at least one element in a product of three matrices |
| \(a=-\frac{1}{2}+\frac{3}{2}\times 5^n\), \(b=-\frac{3}{4}+\frac{3}{4}\times 5^n\), \(c=1-5^n\), \(d=\frac{3}{2}-\frac{1}{2}\times 5^n\) | A1 ag, A2 | Give A1 for one of \(b,c,d\) correct; SR If \(M^n=P^{-1}D^nP\) is used, max marks are M0M1A0B1M1A0A1; SR If their \(P\) is singular, max marks are M1M1A1B0M0 |
| 8 |
## 3(i)
| Characteristic equation is $(7-\lambda)(-1-\lambda)+12=0$, $\lambda^2-6\lambda+5=0$, $\lambda=1,5$ | M1, A1A1 | |
|---|---|---|
| When $\lambda=1$: $\begin{pmatrix}7&3\\-4&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x\\y\end{pmatrix}$ | M1 | or $\begin{pmatrix}6&3\\-4&-2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$ can be awarded for either eigenvalue; Equation relating $x$ and $y$ |
|---|---|---|
| $7x+3y=x$, $-4x-y=y$ | M1 | or any (non-zero) multiple |
|---|---|---|
| $y=-2x$, eigenvector is $\begin{pmatrix}1\\-2\end{pmatrix}$ | A1 | |
|---|---|---|
| When $\lambda=5$: $\begin{pmatrix}7&3\\-4&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=5\begin{pmatrix}x\\y\end{pmatrix}$ | M1 | |
|---|---|---|
| $7x+3y=5x$, $-4x-y=5y$ | M1 | |
|---|---|---|
| $y=\frac{3}{2}x$, eigenvector is $\begin{pmatrix}3\\2\end{pmatrix}$ | A1 | |
|---|---|---|
| | 8 | SR $(M-\lambda I)v=\lambda v$ can earn M1A1A1M0M1A0M1A0 |
## 3(ii)
| $P=\begin{pmatrix}1&3\\-2&-2\end{pmatrix}$, $D=\begin{pmatrix}1&0\\0&5\end{pmatrix}$ | B1 ft, B1 ft | B0 if $P$ is singular; For B2, the order must be consistent |
|---|---|---|
## 3(iii)
| $M=PDP^{-1}$ | M1 | May be implied |
|---|---|---|
| $M^n=PD^nP^{-1} = P\begin{pmatrix}1&0\\0&5^n\end{pmatrix}P^{-1}$ | M1 | |
|---|---|---|
| $= \begin{pmatrix}1&3\\-2&-2\end{pmatrix}\begin{pmatrix}1&0\\0&5^n\end{pmatrix}\begin{pmatrix}-\frac{1}{2}&\frac{3}{4}\\frac{1}{2}&-\frac{1}{4}\end{pmatrix}$ | A1 ft | Dependent on M1M1 |
|---|---|---|
| $= \begin{pmatrix}1&3\times 5^n\\-2&-2\times 5^n\end{pmatrix}\begin{pmatrix}-\frac{1}{2}&\frac{3}{4}\\frac{1}{2}&-\frac{1}{4}\end{pmatrix} = \frac{1}{4}\begin{pmatrix}-2+6\times 5^n&-3+3\times 5^n\\4-4\times 5^n&6-2\times 5^n\end{pmatrix}$ | M1 | Obtaining at least one element in a product of three matrices |
|---|---|---|
| $a=-\frac{1}{2}+\frac{3}{2}\times 5^n$, $b=-\frac{3}{4}+\frac{3}{4}\times 5^n$, $c=1-5^n$, $d=\frac{3}{2}-\frac{1}{2}\times 5^n$ | A1 ag, A2 | Give A1 for one of $b,c,d$ correct; SR If $M^n=P^{-1}D^nP$ is used, max marks are M0M1A0B1M1A0A1; SR If their $P$ is singular, max marks are M1M1A1B0M0 |
|---|---|---|
| | 8 | |3 You are given the matrix $\mathbf { M } = \left( \begin{array} { r r } 7 & 3 \\ - 4 & - 1 \end{array} \right)$.\\
(i) Find the eigenvalues, and corresponding eigenvectors, of the matrix $\mathbf { M }$.\\
(ii) Write down a matrix $\mathbf { P }$ and a diagonal matrix $\mathbf { D }$ such that $\mathbf { P } ^ { - 1 } \mathbf { M P } = \mathbf { D }$.\\
(iii) Given that $\mathbf { M } ^ { n } = \left( \begin{array} { l l } a & b \\ c & d \end{array} \right)$, show that $a = - \frac { 1 } { 2 } + \frac { 3 } { 2 } \times 5 ^ { n }$, and find similar expressions for $b , c$ and $d$.
Section B (18 marks)
\hfill \mbox{\textit{OCR MEI FP2 2008 Q3 [18]}}