# Question 11:

## Recurrence relation (2 marks):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha$ is a root $\Rightarrow \alpha^4-3\alpha^2+5\alpha-2=0$ | | |
| $\Rightarrow \alpha^{n+4}-3\alpha^{n+2}+5\alpha^{n+1}-2\alpha^n=0$ | M1 | Substitutes $\alpha$ into equation, multiplies by $\alpha^n$ |
| Repeat for $\beta,\gamma,\delta$ and sum $\Rightarrow S_{n+4}-3S_{n+2}+5S_{n+1}-2S_n=0$ (AG) | A1 | 2 | |

## Part (i) — Finding $S_4$ (3 marks):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_2=0-2\times(-3)=6$ | B1 | Uses $\sum\alpha^2=\left(\sum\alpha\right)^2-2\sum\alpha\beta$ |
| $S_4=3\times6-5\times0+2\times4=26$ | M1A1 | 3 | Finds $S_4$ from formula |

## Part (ii) — Finding $\sum\alpha^2\beta^3$ (6 marks):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_{-1}=\frac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta}=\frac{-5}{-2}=\frac{5}{2}$ | M1A1 | Uses $S_{-1}=\frac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta}$ |
| $S_3=3\times0-5\times4+2\times\frac{5}{2}=-15$ | M1A1 | Finds $S_3$ from formula |
| $S_5=3\times(-15)-5\times6+2\times0=-75$ | M1A1 | 6 | Finds $S_5$ from formula |

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum\alpha^2\beta^3=S_2S_3-S_5$ | M1 | |
| $=6\times(-15)-(-75)=-15$ | M1A1 | 3 | **[14]** |

# Question 11 (OR):

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Reduces **M** to echelon form: $\begin{pmatrix} 2 & 1 & -1 & 4 \\ 3 & 4 & 6 & 1 \\ -1 & 2 & 8 & -7 \end{pmatrix} \rightarrow \ldots \rightarrow \begin{pmatrix} 2 & 1 & -1 & 4 \\ 0 & 1 & 3 & -2 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ | M1A1 | |
| $\text{Dim}(\mathbf{M}) = 4 - 2 = 2$ | A1 | Uses dimension theorem |

## Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Basis for $R$ is $\left\{ \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} \right\}$ | B1 | OE |
| $x = 2\lambda + \mu$, $y = 3\lambda + 4\mu$, $z = -\lambda + 2\mu \Rightarrow 2x - y + z = 0$ | M1A1 | Finds Cartesian equation for $R$ |

## Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $2x + y - z + 4t = 0$, $y + 3z - 2t = 0$, $t = \lambda$ and $z = \mu$, $\Rightarrow y = 2\lambda - 3\mu$ and $x = -3\lambda + 2\mu$ | M1 | Finds basis for null space |
| $\Rightarrow$ Basis of null space is $\left\{ \begin{pmatrix} -3 \\ 2 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ -3 \\ 1 \\ 0 \end{pmatrix} \right\}$ | A1A1 | OE |
| $2 \times 8 - 7 + k = 0 \Rightarrow k = -9$ | B1 | Evaluates $k$ |
| $5\begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix} - 2\begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ 7 \\ -9 \end{pmatrix}$ | M1A1 | OE, via equations — finds particular solution |
| $\mathbf{x} = \begin{pmatrix} 5 \\ -2 \\ 0 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} -3 \\ 2 \\ 0 \\ 1 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ -3 \\ 1 \\ 0 \end{pmatrix}$ | M1A1 | Finds general solution |

**Total: [14]**