FP1 June 2013 (R) Q6
6. \[\mathbf{A} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}\]
The transformation represented by \(\mathbf{B}\) followed by the transformation represented by \(\mathbf{A}\) is equivalent to the transformation represented by \(\mathbf{P}\).
Triangle \(T\) is transformed to the triangle \(T^{\prime}\) by the transformation represented by \(\mathbf{P}\).
Given that the area of triangle \(T^{\prime}\) is 24 square units,
Triangle \(T^{\prime}\) is transformed to the original triangle \(T\) by the matrix represented by \(\mathbf{Q}\).
| Scheme | Marks |
|---|---|
| \(\mathbf{P} = \mathbf{AB}\ \left\{= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}\right\}\) \(\mathbf{P} = \mathbf{AB}\), seen or implied. | M1 |
| \(\mathbf{P} = \begin{pmatrix} 1 & 4 \\ -2 & -3 \end{pmatrix}\) Correct answer. | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\det\mathbf{P} = 1(-3) - (4)(-2)\ \{= -3 + 8 = 5\}\) Applies “\(ad - bc\)”. | M1 |
| \(\text{Area}(T) = \underline{\dfrac{24}{5}}\ (\text{units})^2\) \(\dfrac{24}{\text{their } \det\mathbf{P}}\), dependent on previous M \(\underline{\dfrac{24}{5}}\) or \(\underline{4.8}\) | dM1 A1ft |
| (3) |
| Scheme | Marks |
|---|---|
| \(\mathbf{QP} = \mathbf{I} \Rightarrow \mathbf{QPP}^{-1} = \mathbf{IP}^{-1} \Rightarrow \mathbf{Q} = \mathbf{P}^{-1}\) | |
| \(\mathbf{Q} = \mathbf{P}^{-1} = \dfrac{1}{5}\begin{pmatrix} -3 & -4 \\ 2 & 1 \end{pmatrix}\) \(\mathbf{Q} = \mathbf{P}^{-1}\) stated or an attempt to find \(\mathbf{P}^{-1}\). Correct ft inverse matrix. | M1 A1ft |
| (2) | |
| (7 marks) |
Notes
Using \(\mathbf{BA}\), area is the same in (b) and inverse is \(\dfrac{1}{5}\begin{pmatrix} 1 & -2 \\ 4 & -3 \end{pmatrix}\) in (c) and could gain ft marks.