# Question 8:

## Part (a)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}5&-2&5\\0&3&p\\-6&6&-4\end{pmatrix}\begin{pmatrix}2\\1\\-2\end{pmatrix} = \begin{pmatrix}-2\\3-2p\\2\end{pmatrix} = -1\times\begin{pmatrix}2\\2p-3\\-2\end{pmatrix}$; corresponding eigenvalue is $-1$ | B1 | Correct eigenvalue of $-1$ |

## Part (a)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2p-3=1 \Rightarrow p=\ldots$ | M1 | Correct equation with eigenvalue set up; need only see middle equation |
| $p=2$ | A1* | Correct proof (full matrix calculation not necessary) |

## Part (a)(iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\det\begin{pmatrix}5-\lambda&-2&5\\0&3-\lambda&p\\-6&6&-4-\lambda\end{pmatrix}=0$ | M1 | Applies $\det(\mathbf{A}-\lambda\mathbf{I})=0$ to achieve a cubic in $\lambda$; allow with $p$ instead of 2 |
| $\Rightarrow \lambda^3-4\lambda^2+\lambda+6=0$ | A1 | Correct simplified cubic |
| $(\lambda+1)(\lambda^2-5\lambda+6)=0 \Rightarrow (\lambda+1)(\lambda-2)(\lambda-3)=0$; eigenvalues are $-1$, $2$ and $3$ | A1 | Correct eigenvalues |
| For $\lambda=2$: $3x-2y+5z=0$, $y+2z=0$, $-6x+6y-6z=0 \Rightarrow x/y/z=\ldots$ | M1 | Forms and solves eigenvector equations for at least one (other than $-1$) eigenvalue |
| Either $k\begin{pmatrix}3\\2\\-1\end{pmatrix}$ (for $\lambda=2$) or $m\begin{pmatrix}1\\1\\0\end{pmatrix}$ (for $\lambda=3$) | A1 | One correct (other) eigenvector |
| Both eigenvector systems attempted | M1 | Both eigenvectors attempted |
| Both $k\begin{pmatrix}3\\2\\-1\end{pmatrix}$ (for $\lambda=2$) and $m\begin{pmatrix}1\\1\\0\end{pmatrix}$ (for $\lambda=3$) | A1 | Both (other) eigenvectors correct |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{P}=\begin{pmatrix}2&3&1\\1&2&1\\-2&-1&0\end{pmatrix}$ and $\mathbf{D}=\begin{pmatrix}-1&0&0\\0&2&0\\0&0&3\end{pmatrix}$ | B1ft | Correct corresponding $\mathbf{P}$ and $\mathbf{D}$; follow through on (a); columns consistent between $\mathbf{P}$ and $\mathbf{D}$ |

## Part (c)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dot{u}=ku \Rightarrow \int\frac{1}{u}\,du = k\int dt \Rightarrow \ln u = kt (+c)$ | M1 | Separates variables and attempts integration (constant not required) |
| So $u=Ae^{kt}$ or $u=e^{kt+c}$ | A1 | Correct answer for $u$, either form, including constant of integration |

## Part (c)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}\dot{x}\\\dot{y}\\\dot{z}\end{pmatrix}=\mathbf{PDP}^{-1}\begin{pmatrix}x\\y\\z\end{pmatrix} \Rightarrow \begin{pmatrix}\dot{u}\\\dot{v}\\\dot{w}\end{pmatrix}=\mathbf{P}^{-1}\begin{pmatrix}\dot{x}\\\dot{y}\\\dot{z}\end{pmatrix}=\mathbf{D}\begin{pmatrix}u\\v\\w\end{pmatrix}=\begin{pmatrix}-u\\2v\\3w\end{pmatrix}$ | M1 | Correct use of substitution to decouple the system |
| $\Rightarrow \begin{pmatrix}u\\v\\w\end{pmatrix}=\begin{pmatrix}Ae^{-t}\\Be^{2t}\\Ce^{3t}\end{pmatrix}$ | M1 | Correct solutions for $u$, $v$, $w$ |
| $\Rightarrow \begin{pmatrix}x\\y\\z\end{pmatrix}=\mathbf{P}\begin{pmatrix}Ae^{-t}\\Be^{2t}\\Ce^{3t}\end{pmatrix}=\ldots$ | M1 | Multiplies by $\mathbf{P}$ to recover $x,y,z$ |
| $\Rightarrow \begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2Ae^{-t}+3Be^{2t}+Ce^{3t}\\Ae^{-t}+2Be^{2t}+Ce^{3t}\\-2Ae^{-t}-Be^{2t}\end{pmatrix}$ | A1 | Correct final answer |

## Question (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses **P** and **D** to transform system into equation in $u$, $v$ and $w$ | M1 | May be implied |
| Forms the solution for $u$, $v$ and $w$ using their eigenvalues | M1 | |
| Reverses the substitution (multiplies by their **P**) to get solution for $x$, $y$ and $z$ | M1 | |
| Correct answer, in matrix form or as separate equations | A1 | Award when first seen and isw |

**NB:** Different orderings of the columns of **P** and **D** will give the terms in different orders.