AS June 2018 Paper 1 Q5
5.
\[\mathbf{A} = \begin{pmatrix}-\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2}\\[8pt] \dfrac{\sqrt{3}}{2} & -\dfrac{1}{2}\end{pmatrix}\]The transformation \(V\), represented by the \(2 \times 2\) matrix \(\mathbf{B}\), is a reflection in the line \(y = -x\)
Given that \(U\) followed by \(V\) is the transformation \(T\), which is represented by the matrix \(\mathbf{C}\),
| Scheme | Marks | AO |
|---|---|---|
| Rotation | B1 | 1.1b |
| 120 degrees (anticlockwise) or \(\dfrac{2\pi}{3}\) radians (anticlockwise) Or 240 degrees clockwise or \(\dfrac{4\pi}{3}\) radians clockwise | B1 | 2.5 |
| About (from) the origin. Allow (0, 0) or \(O\) for origin. | B1 | 1.2 |
| (3) |
Notes
B1: Identifies the transformation as a rotation
B1: Correct angle. Allow equivalents in degrees or radians.
B1: Identifies the origin as the centre of rotation
These marks can only be awarded as the elements of a single transformation
| Scheme | Marks | AO |
|---|---|---|
| \[\begin{pmatrix}0 & -1\\ -1 & 0\end{pmatrix}\] | B1 | 1.1b |
| (1) |
Notes
B1: Shows the correct matrix in the correct form
| Scheme | Marks | AO |
|---|---|---|
| \[\begin{pmatrix}0 & -1\\ -1 & 0\end{pmatrix}\begin{pmatrix}-\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2}\\[8pt] \dfrac{\sqrt{3}}{2} & -\dfrac{1}{2}\end{pmatrix} = \begin{pmatrix}\ldots & \ldots\\ \ldots & \ldots\end{pmatrix}\] | M1 | 1.1b |
| \[\begin{pmatrix}0 & -1\\ -1 & 0\end{pmatrix}\begin{pmatrix}-\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2}\\[8pt] \dfrac{\sqrt{3}}{2} & -\dfrac{1}{2}\end{pmatrix} = \begin{pmatrix}-\dfrac{\sqrt{3}}{2} & \dfrac{1}{2}\\[8pt] \dfrac{1}{2} & \dfrac{\sqrt{3}}{2}\end{pmatrix}\] | A1ft | 1.1b |
| (2) |
Notes
M1: Multiplies the matrices in the correct order (evidence of multiplication can be taken from 3 correct or 3 correct ft elements)
A1ft: Correct matrix (follow through their matrix from part (b))
A correct matrix or a correct follow through matrix implies both marks.
| Scheme | Marks | AO |
|---|---|---|
| \[\begin{pmatrix}-\dfrac{\sqrt{3}}{2} & \dfrac{1}{2}\\[8pt] \dfrac{1}{2} & \dfrac{\sqrt{3}}{2}\end{pmatrix}\begin{pmatrix}1\\ k\end{pmatrix} = \begin{pmatrix}1\\ k\end{pmatrix} = \ldots \text{ or } \begin{pmatrix}-\dfrac{\sqrt{3}}{2} & \dfrac{1}{2}\\[8pt] \dfrac{1}{2} & \dfrac{\sqrt{3}}{2}\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = \begin{pmatrix}x\\ y\end{pmatrix} = \ldots\]Note: \(\begin{pmatrix}-\dfrac{\sqrt{3}}{2} & \dfrac{1}{2}\\[8pt] \dfrac{1}{2} & \dfrac{\sqrt{3}}{2}\end{pmatrix}\begin{pmatrix}1\\ k\end{pmatrix} = \begin{pmatrix}-\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}k\\[8pt] \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}k\end{pmatrix} = \begin{pmatrix}1\\ k\end{pmatrix}\) can score M1 (for the matrix equation) but needs an equation to be “extracted” to score the next A1 | M1 | 3.1a |
| \[-\frac{\sqrt{3}}{2} + \frac{1}{2}k = 1 \text{ or } \frac{1}{2} + \frac{\sqrt{3}}{2}k = k\]or\[x = -\frac{\sqrt{3}}{2}x + \frac{1}{2}y \text{ or } y = \frac{1}{2}x + \frac{\sqrt{3}}{2}y\](Note that candidates may then substitute \(x = 1\) which is acceptable) | A1ft | 1.1b |
| \[-\frac{\sqrt{3}}{2} + \frac{1}{2}k = 1 \text{ or } x = -\frac{\sqrt{3}}{2}x + \frac{1}{2}y \Rightarrow k = 2 + \sqrt{3}\left(\text{or } \frac{1}{2 - \sqrt{3}}\right)\] | A1 | 1.1b |
| \[\frac{1}{2} + \frac{\sqrt{3}}{2}k = k \text{ or } y = \frac{1}{2}x + \frac{\sqrt{3}}{2}y \Rightarrow k = 2 + \sqrt{3}\left(\text{or } \frac{1}{2 - \sqrt{3}}\right)\] | B1 | 1.1b |
| (4) | ||
| (10 marks) |
Notes
M1: Translates the problem into a matrix multiplication to obtain at least one equation in \(k\) or in \(x\) and \(y\)
A1ft: Obtains one correct equation (follow through their matrix from part (c))
A1: Correct value for \(k\) in any form
B1: Checks their answer by independently solving both equations correctly to obtain \(2 + \sqrt{3}\) both times or substitutes \(2 + \sqrt{3}\) into the other equation to confirm its validity