FP1 January 2011 Q8
8. \[\mathbf{A} = \begin{pmatrix} 2 & -2 \\ -1 & 3 \end{pmatrix}\]
(a) Find \(\det\mathbf{A}\). (1)
(b) Find \(\mathbf{A}^{-1}\). (2)
The triangle \(R\) is transformed to the triangle \(S\) by the matrix \(\mathbf{A}\).
Given that the area of triangle \(S\) is 72 square units,
(c) find the area of triangle \(R\). (2)
The triangle \(S\) has vertices at the points \((0,\ 4)\), \((8,\ 16)\) and \((12,\ 4)\).
(d) Find the coordinates of the vertices of \(R\). (4)
| Scheme | Marks |
|---|---|
| \(\mathbf{A} = \begin{pmatrix} 2 & -2 \\ -1 & 3 \end{pmatrix}\) | |
| \(\det\mathbf{A} = 2(3) - (-1)(-2) = 6 - 2 = \underline{4}\) 4 | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\mathbf{A}^{-1} = \dfrac{1}{4}\begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix}\) \(\dfrac{1}{\det\mathbf{A}}\begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix}\) | M1 |
| \(\dfrac{1}{4}\begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix}\) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| Area\((R) = \dfrac{72}{4} = \underline{18}\ (\text{units})^2\) \(\dfrac{72}{\text{their }\det\mathbf{A}}\) or \(72(\text{their }\det\mathbf{A})\) | M1 |
| 18 or ft answer. | A1ft |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathbf{AR} = \mathbf{S} \Rightarrow \mathbf{A}^{-1}\mathbf{AR} = \mathbf{A}^{-1}\mathbf{S} \Rightarrow \mathbf{R} = \mathbf{A}^{-1}\mathbf{S}\) | |
| \(\mathbf{R} = \dfrac{1}{4}\begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 0 & 8 & 12 \\ 4 & 16 & 4 \end{pmatrix}\) At least one attempt to apply \(\mathbf{A}^{-1}\) by any of the three vertices in \(\mathbf{S}\). | M1 |
| \(= \dfrac{1}{4}\begin{pmatrix} 8 & 56 & 44 \\ 8 & 40 & 20 \end{pmatrix}\) | |
| \(= \begin{pmatrix} 2 & 14 & 11 \\ 2 & 10 & 5 \end{pmatrix}\) At least one correct column o.e. | A1ft |
| At least two correct columns o.e. | A1 |
| Vertices are \((2,\ 2)\), \((14,\ 10)\) and \((11,\ 5)\). All three coordinates correct. | A1 |
| (4) | |
| [9] |