## Question 4:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda = 3 \Rightarrow \|\mathbf{M} - 3\mathbf{I}\| = \begin{vmatrix} 3 & k & 2 \\ k & 2 & 0 \\ 2 & 0 & 4 \end{vmatrix} = 0 \Rightarrow 3(8) - k(4k) + 2(-4) = 0$ | M1 | Correct interpretation of 3 being an eigenvalue leading to formation of a quadratic equation in $k$ only. If method not clear look for at least 2 correct "components". NB rule of Sarrus gives $24 - 8 - 4k^2 = 0$ |
| $\Rightarrow 4k^2 = 16 \Rightarrow k = \ldots$ | dM1 | Solves quadratic. Depends on the first M. |
| $k = \pm 2$ | A1 | Correct values |

**Way 2:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix} 6 & k & 2 \\ k & 5 & 0 \\ 2 & 0 & 7 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = 3\begin{pmatrix} x \\ y \\ z \end{pmatrix} \Rightarrow \begin{cases} 6x + ky + 2z = 3x \\ kx + 5y = 3y \\ 2x + 7z = 3z \end{cases}$; $z = -\frac{1}{2}x,\ y = -\frac{1}{2}kx \Rightarrow 6x - \frac{k^2x}{2} - x = 3x \Rightarrow \frac{k^2}{2} = 2$ | M1 | Eliminates $z$ and $y$ and reaches a quadratic equation in $k$ only |
| $\frac{k^2}{2} = 2 \Rightarrow k = \ldots$ | dM1 | Solves quadratic. Depends on the first M. |
| $k = \pm 2$ | A1 | Correct values |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $k = -2 \Rightarrow \|\mathbf{M} - \lambda\mathbf{I}\| = \begin{vmatrix} 6-\lambda & -2 & 2 \\ -2 & 5-\lambda & 0 \\ 2 & 0 & 7-\lambda \end{vmatrix}$ $\Rightarrow (6-\lambda)(7-\lambda)(5-\lambda) + 2(2\lambda - 14) + 2(2\lambda - 10) = 0$ | M1 | Applies a value of $k$ from (a) and a recognisable attempt at the characteristic equation (the "= 0" is not needed here). If method not clear look for at least 2 correct "components". |
| $\Rightarrow \lambda^3 - 18\lambda^2 + 99\lambda - 162 = 0 \Rightarrow \lambda = \ldots$ | dM1 | Solves cubic. May use $\lambda = 3$ as a factor or calculator to solve. Depends on the first mark. Allow complex roots. |
| $\lambda = 6,\ 9\ (,3)$ | A1 | Correct values. Allow to come from $k = 2$ |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix} 6 & -2 & 2 \\ -2 & 5 & 0 \\ 2 & 0 & 7 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = 3\begin{pmatrix} x \\ y \\ z \end{pmatrix} \Rightarrow \begin{cases} 6x - 2y + 2z = 3x \\ -2x + 5y = 3y \\ 2x + 7z = 3z \end{cases} \Rightarrow \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \ldots$ or $\begin{pmatrix} 3 & -2 & 2 \\ -2 & 2 & 0 \\ 2 & 0 & 4 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ | M1 | Correct strategy for finding the eigenvector using a value of $k$ from (a). Note that the cross product of any 2 rows or columns of $\mathbf{M} - 3\mathbf{I}$ gives an eigenvector |
| $p\begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}$ | A1 | Any correct eigenvector |
| $\frac{1}{3}\begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}$ | A1 | Any correct normalised eigenvector |

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