## Question 10(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}6&-8&7\\7&-9&7\\6&-6&5\end{pmatrix}\begin{pmatrix}1\\1\\0\end{pmatrix} = \begin{pmatrix}-2\\-2\\0\end{pmatrix} \Rightarrow \lambda_1 = -2$ | M1A1 | |

## Question 10(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda_2 = -1 \Rightarrow \mathbf{e}_2 = \begin{pmatrix}1\\0\\-1\end{pmatrix}$ | M1A1 | OE |

## Question 10(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\det \mathbf{A} = 10 \Rightarrow \lambda_3 = 5 \Rightarrow \mathbf{e}_3 = \begin{pmatrix}1\\1\\1\end{pmatrix}$ | B1B1B1 | OE |

## Question 10(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{P} = \begin{pmatrix}1&1&1\\1&0&1\\0&-1&1\end{pmatrix}$ | B1FT | |
| $\mathbf{D} = \begin{pmatrix}-2&0&0\\0&-1&0\\0&0&5\end{pmatrix}$ (Check for consistency.) | B1FT | |
| $\det\mathbf{P} = -1$, $\text{Adj}\,\mathbf{P} = \begin{pmatrix}1&-2&1\\-1&1&0\\-1&1&-1\end{pmatrix} \Rightarrow \mathbf{P}^{-1} = \begin{pmatrix}-1&2&-1\\1&-1&0\\1&-1&1\end{pmatrix}$ | M1A1 | Alternative Method: Reduced row echelon form M1, A1 |
| $\mathbf{A}^n = \begin{pmatrix}1&1&1\\1&0&1\\0&-1&1\end{pmatrix}\begin{pmatrix}(-2)^n&0&0\\0&(-1)^n&0\\0&0&5^n\end{pmatrix}\begin{pmatrix}-1&2&-1\\1&-1&0\\1&-1&1\end{pmatrix}$ | M1 | Allow $\mathbf{A}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}$ |
| $= \begin{pmatrix}[5^n+(-1)^n-(-2)^n] & [2\cdot(-2)^n+(-1)^{n+1}-5^n] & [5^n-(-2)^n]\\ [5^n-(-2)^n] & [2\cdot(-2)^n-5^n] & [5^n-(-2)^n]\\ [5^n-(-1)^n] & [(-1)^n-5^n] & [5^n]\end{pmatrix}$ | A1 | |

## Question 11E(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = 2a\!\left(\cos2\theta\cos\tfrac{1}{2}\pi - \sin2\theta\sin\tfrac{1}{2}\pi\right) = 2a(0 - \sin2\theta\cdot1) = -2a\sin2\theta$ | M1A1 | AG |
| Sketch showing loops in 2nd and 4th quadrant | B1 | Loops in 2nd and 4th quadrant |
| Symmetry and shape correct | B1 | Symmetry and shape correct |

## Question 11E(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = -4a\sin\theta\cos\theta = -4a\cdot\dfrac{y}{r}\cdot\dfrac{x}{r} \Rightarrow r^3 = -4axy \Rightarrow (x^2+y^2)^{\frac{3}{2}} = -4axy$ | M1A1 | AG |

## Question 11E(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Area of one loop is $\dfrac{1}{2}\displaystyle\int_{\frac{1}{2}\pi}^{\pi} 4a^2\sin^2 2\theta\, \mathrm{d}\theta = a^2\displaystyle\int_{\frac{1}{2}\pi}^{\pi} 2\sin^2 2\theta\, \mathrm{d}\theta$ | M1 | OE |
| $= a^2\displaystyle\int_{\frac{1}{2}\pi}^{\pi}(1-\cos4\theta)\,\mathrm{d}\theta$ | M1A1 | |
| $= a^2\!\left[\theta - \tfrac{1}{4}\sin4\theta\right]_{\frac{1}{2}\pi}^{\pi} = \dfrac{1}{2}\pi a^2$ | M1A1 | |

## Question 11E(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = r\sin\theta = -2a\sin2\theta\sin\theta$ | B1 | |
| $\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = -2a(2\cos2\theta\sin\theta + \sin2\theta\cos\theta) = 0$ | M1 | |
| $2\cos2\theta\sin\theta = -\sin2\theta\cos\theta \;\; (\Rightarrow 2\tan\theta = -\tan2\theta)$ | A1 | AG |

## Question 11O(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}1&-1&0&2\\3&-1&4&0\\5&-8&-6&19\\-2&3&2&-7\end{pmatrix} \Rightarrow\cdots\Rightarrow \begin{pmatrix}1&-1&0&2\\0&1&2&-3\\0&0&0&0\\0&0&0&0\end{pmatrix}$ | M1A1 | |
| $r(\mathbf{A}) = 4 - 2 = 2$ | A1FT | |
| $x - y + 2t = 0 \qquad y + 2z - 3t = 0$ | M1 | |
| $z = \lambda',\; t = \mu' \Rightarrow x = -2\lambda' + \mu',\; y = -2\lambda' + 3\mu'$; Basis of null space is $\left\{\begin{pmatrix}2\\2\\-1\\0\end{pmatrix}, \begin{pmatrix}1\\3\\0\\1\end{pmatrix}\right\}$ | A1A1 | AG |

## Question 11O(ii):

$\mathbf{A}\left\{\mathbf{x} - \begin{pmatrix} p \\ q \\ 0 \\ 0 \end{pmatrix}\right\} = 0 \Rightarrow \mathbf{x} = \begin{pmatrix} p \\ q \\ 0 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 2 \\ -1 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ 3 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} p+2\lambda+\mu \\ q+2\lambda+3\mu \\ -\lambda \\ \mu \end{pmatrix}$ | M1A1 |

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## Question 11O(iii):

$p\begin{pmatrix} 1 \\ 3 \\ 5 \\ -2 \end{pmatrix} + q\begin{pmatrix} -1 \\ -1 \\ -8 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 7 \\ 18 \\ -7 \end{pmatrix} \Rightarrow p - q = 3$ and $3p - q = 7 \Rightarrow p = 2,\ q = -1$ | M1A1 A1 |

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## Question 11O(iv):

$2 + 2\lambda + \mu = 4 \Rightarrow 2\lambda + \mu = 2$

$-1 + 2\lambda + 3\mu = 9 \Rightarrow 2\lambda + 3\mu = 10$ | M1 |

$\Rightarrow \lambda = -1,\ \mu = 4 \Rightarrow$ Solution of $\mathbf{Ax} = \begin{pmatrix} 3 \\ 7 \\ 18 \\ -7 \end{pmatrix}$ is $\mathbf{x} = \begin{pmatrix} 4 \\ 9 \\ 1 \\ 4 \end{pmatrix}$ | A1A1 |