AS October 2021 Paper 1 Q5
5 Matrices \(\mathbf{A}\) and \(\mathbf{B}\) are given by \(\mathbf{A} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} \dfrac{5}{13} & -\dfrac{12}{13} \\[6pt] \dfrac{12}{13} & \dfrac{5}{13} \end{pmatrix}\).
Matrix \(\mathbf{B}\) represents the transformation \(\mathrm{T_B}\).
Matrix \(\mathbf{C}\) is given by \(\mathbf{C} = \begin{pmatrix} 1 & 0 \\ 0 & -3 \end{pmatrix}\) and represents the transformation \(\mathrm{T_C}\).
The transformation \(\mathrm{T_{BC}}\) is transformation \(\mathrm{T_C}\) followed by transformation \(\mathrm{T_B}\).
An object shape of area 5 is transformed by \(\mathrm{T_{BC}}\) to an image shape \(N\).
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{AB} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \dfrac{5}{13} & -\dfrac{12}{13} \\[6pt] \dfrac{12}{13} & \dfrac{5}{13} \end{pmatrix} = \begin{pmatrix} -\dfrac{5}{13} & \dfrac{12}{13} \\[6pt] \dfrac{12}{13} & \dfrac{5}{13} \end{pmatrix}\) | M1 | 2.1 |
| \(\mathbf{BA} = \begin{pmatrix} \dfrac{5}{13} & -\dfrac{12}{13} \\[6pt] \dfrac{12}{13} & \dfrac{5}{13} \end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -\dfrac{5}{13} & -\dfrac{12}{13} \\[6pt] -\dfrac{12}{13} & \dfrac{5}{13} \end{pmatrix} \ne \mathbf{AB}\) so matrix multiplication is not commutative | A1 | 2.2a |
| [2] |
Notes
M1: BC. \(\mathbf{AB}\) or \(\mathbf{BA}\) correct.
Could see \(\dfrac{1}{13}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 5 & -12 \\ 12 & 5 \end{pmatrix} = \dfrac{1}{13}\begin{pmatrix} -5 & 12 \\ 12 & 5 \end{pmatrix}\)
A1: BC. Other multiplication correct and conclusion
| Scheme | Marks | AO |
|---|---|---|
| Rotation about \(O\) | M1 | 1.2 |
| \(67.4^\circ\) anticlockwise | A1 | 1.1 |
| [2] |
Notes
A1: or 1.18 rads
| Scheme | Marks | AO |
|---|---|---|
| \((\mathrm{T_B})^{-1}\) is a rotation about \(O\) by \(-67.4^\circ\) anticlockwise (or \(67.4^\circ\) clockwise) | M1 | 3.1a |
| So \(\mathbf{B}^{-1} = \begin{pmatrix} \cos(-67.4^\circ) & -\sin(-67.4^\circ) \\ \sin(-67.4^\circ) & \cos(-67.4^\circ) \end{pmatrix}\) \(= \begin{pmatrix} \dfrac{5}{13} & \dfrac{12}{13} \\[6pt] -\dfrac{12}{13} & \dfrac{5}{13} \end{pmatrix}\) | A1 | 1.1 |
| [2] |
Notes
M1: Correct inverse of their rotation \(\mathrm{T_B}\).
Could also be rotation of \(292.6^\circ\) anticlockwise
A1: or \(\mathbf{B}^{-1} = \begin{pmatrix} 0.385 & 0.923 \\ -0.923 & 0.385 \end{pmatrix}\) (allow 0.384 for 0.385)
NB: Question states “by considering the inverse transformation”.
SC1 For correct inverse by other method.
| Scheme | Marks | AO |
|---|---|---|
| \(\det\mathbf{B} = 1\) and \(\det\mathbf{C} = -3\) | M1 | 3.1a |
| So area of \(N = |1 \times -3| \times 5 = 15\) | A1 | 3.2a |
| [2] |
Notes
M1: Could find \(\mathbf{BC}\) and then find \(\det(\mathbf{BC}) = -3\)
A1: Area must be 15, do not allow \(-15\) or \(\pm 15\)