## Question 5(a):

| $-20 + 3k = 0$ | M1 | 1 mark |
| $k = \frac{20}{3}$ | A1 | 1 mark |
| | | **2** |

## Question 5(b):

| Consider $\begin{pmatrix}5 & 6\\-3 & -4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}5x+6y\\-3x-4y\end{pmatrix} = \begin{pmatrix}X\\Y\end{pmatrix}$ | M1 | 1 mark |
| Line through $O$, so $Y = mX$ | M1 | 1 mark |
| Invariant line, so $Y = mX$: $-3x - 4mx = m(5x + 6mx)$ | M1A1 | 2 marks |
| $6m^2 + 9m + 3 = 0$ | A1 | 1 mark |
| $m = -1, -\frac{1}{2}$ | A1 | 1 mark |
| Invariant lines $y = -x$, $2y + x = 0$ | A1 | 1 mark |
| | | **6** |

## Question 5(c)(i):

| $\det\mathbf{A} = -20 + 18 = -2$ | M1 | 1 mark |
| Area $= 2\times10 = 20\text{ cm}^2$ | A1 | 1 mark |
| | | **2** |

## Question 5(c)(ii):

| $\mathbf{A}^{-1} = \frac{1}{-2}\begin{pmatrix}-4 & -6\\3 & 5\end{pmatrix} = \begin{pmatrix}2 & 3\\-\frac{3}{2} & -\frac{5}{2}\end{pmatrix}$ | M1A1 | **2** |

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