# Question 11:

## EITHER (First Method)

### Part (i)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{P} = \begin{pmatrix} 0 & -1 & 1 \\ 1 & 0 & 1 \\ -1 & 1 & 0 \end{pmatrix}$, $\mathbf{D} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}$ | B1B1 | Columns can be in any order but must match |
| $\det \mathbf{P} = 2$ | B1 | |
| $\mathbf{P}^{-1} = \frac{1}{2}\begin{pmatrix} -1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ | M1A1 | Adj $\div$ Det; no working for $\frac{1}{3}$; row operations M1A1A1 (3 errors) |
| $\mathbf{A} = \mathbf{PDP}^{-1}$ | M1 | |
| $\mathbf{A} = \begin{pmatrix} 0 & -1 & 2 \\ -1 & 0 & 2 \\ 1 & 1 & 0 \end{pmatrix} \cdot \frac{1}{2}\begin{pmatrix} -1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ | M1A1 | |
| $= \begin{pmatrix} 1.5 & 0.5 & 0.5 \\ 1.5 & 0.5 & 1.5 \\ -1 & 1 & 0 \end{pmatrix}$ | A1 | **Part mark: 9** |

### Part (ii)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{A}^{2n} = \mathbf{PD}^{2n}\mathbf{P}^{-1}$ | M1 | |
| $= \frac{1}{2}\begin{pmatrix} 0 & -1 & 1 \\ 1 & 0 & 1 \\ -1 & 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2^{2n} \end{pmatrix}\begin{pmatrix} -1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ | A1 | |
| $= \frac{1}{2}\begin{pmatrix} 0 & -1 & 2^{2n} \\ 1 & 0 & 2^{2n} \\ -1 & 1 & 0 \end{pmatrix}\begin{pmatrix} -1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ | M1A1 | |
| $= \frac{1}{2}\begin{pmatrix} 2^{2n}+1 & 2^{2n}-1 & 2^{2n}-1 \\ 2^{2n}-1 & 2^{2n}+1 & 2^{2n}-1 \\ 0 & 0 & 2 \end{pmatrix}$ | A1 | **Part mark: 5** |

**Total: [14]**

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## EITHER (Alternative Method)

### Part (i)

| Answer/Working | Marks | Guidance |
|---|---|---|
| Using $\mathbf{Ae} = \lambda\mathbf{e}$ three times to form 3 sets of linear equations | M1A1 | |
| $b-c=0$, $e-f=-1$, $h-j=1$ | | |
| $-a+c=-1$, $-d+f=0$, $-g+j=1$ | | |
| $a+b=2$, $d+e=2$, $g+h=0$ | | |
| Solves one set, then other two sets | M1 | |
| $\mathbf{A} = \begin{pmatrix} 1.5 & 0.5 & 0.5 \\ 1.5 & 0.5 & 1.5 \\ -1 & 1 & 0 \end{pmatrix}$ | A1 | **Part mark: 4** |

### Part (ii)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{P}$ and $\mathbf{D}$ as above | B1B1 | |
| $\mathbf{P}^{-1} = \frac{1}{2}\begin{pmatrix} -1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ | B1, M1A1 | **Part mark: 5** |
| $\mathbf{A}^{2n} = \mathbf{PD}^{2n}\mathbf{P}^{-1}$ | M1 | |
| Full multiplication as above | A1, M1A1 | |
| $= \frac{1}{2}\begin{pmatrix} 2^{2n}+1 & 2^{2n}-1 & 2^{2n}-1 \\ 2^{2n}-1 & 2^{2n}+1 & 2^{2n}-1 \\ 0 & 0 & 2 \end{pmatrix}$ | A1 | **Part mark: 5** |

**Total: [14]**

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## OR (Third Method)

### Part 1

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix} 1 & -1 & -1 & 1 \\ 2 & -1 & -4 & 3 \\ 3 & -3 & -2 & 2 \\ 5 & -4 & -6 & 5 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & -1 & -1 & 1 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ | M1A1 | Reduces to echelon form |
| $r(\mathbf{A}) = 4 - 1 = 3$ | A1 | **Part mark: 3** |
| $x - y - z + t = 0$, $y - 2z + t = 0$, $z - t = 0$ | M1 | |
| $\Rightarrow t = \lambda,\ z = \lambda,\ y = \lambda,\ x = \lambda$ | | |
| Basis of null space is $\left\{\begin{pmatrix}1\\1\\1\\1\end{pmatrix}\right\}$ | A1 | |
| $\left\{\mathbf{Ax} = \begin{pmatrix}p\\q\\r\\0\end{pmatrix}\right\} \Rightarrow \mathbf{x} = \begin{pmatrix}p\\q\\r\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\1\\1\end{pmatrix}$ (AG) | M1A1 | **Part mark: 4** |
| $p - q - r = 3$, $2p - q - 4r = 7$, $3p - 3q - 2r = 8$ | M1 | |
| $p = 1,\ q = -1,\ r = -1$ | A1, A1 | B2 all correct; B1 two correct with no working. **Part mark: 3** |
| $\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = \frac{11}{4} \Rightarrow 4\lambda^2 - 2\lambda + \frac{1}{4} = 0$ | M1A1 | |
| $\Rightarrow \left(2\lambda - \frac{1}{2}\right)^2 = 0 \Rightarrow \lambda = \frac{1}{4} \Rightarrow \mathbf{x} = \begin{pmatrix}1.25\\-0.75\\-0.75\\0.25\end{pmatrix}$ | M1A1 | **Part mark: 4** |

**Total: [14]**

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## OR (Alternative for 2nd part)

| Answer/Working | Marks | Guidance |
|---|---|---|
| Writes $\mathbf{x} = \begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}$ and forms equations from $\mathbf{Ax} = p\begin{pmatrix}1\\2\\3\\5\end{pmatrix} + q\begin{pmatrix}-1\\-1\\-3\\-4\end{pmatrix} + r\begin{pmatrix}-1\\-4\\-2\\-6\end{pmatrix}$ | M1A1 | |
| $x_1 - x_2 - x_3 + x_4 = p - q - r$ etc. | M1 | |
| Obtains $x_1 = x_4 + p$, $x_2 = x_4 + q$, $x_3 = x_4 + r$ | A1 | |
| Sets $x_4 = \lambda$: $\mathbf{x} = \begin{pmatrix}p+\lambda\\q+\lambda\\r+\lambda\\\lambda\end{pmatrix}$ | | Those who verify only get M1A1 (2/4) |