FP1 June 2016 Q6
6. \[\mathbf{P} = \begin{pmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}\]
The transformation \(U\) maps the point \(A\), with coordinates \((p, q)\), onto the point \(B\), with coordinates \((6\sqrt{2}, 3\sqrt{2})\).
The transformation \(V\), represented by the \(2 \times 2\) matrix \(\mathbf{Q}\), is a reflection in the line with equation \(y = x\).
The transformation \(U\) followed by the transformation \(V\) is the transformation \(T\). The transformation \(T\) is represented by the matrix \(\mathbf{R}\).
| Scheme | Marks |
|---|---|
| Rotation, 135 degrees or \(\dfrac{3\pi}{4}\) radians (anticlockwise) about \(O\) or 225 degrees or \(\dfrac{5\pi}{4}\) clockwise about \(O\). | M1, A1 |
| (2) |
Notes
M1: Rotation only A1: 135 degrees about \(O\)
SC: 135 degrees about \(O\) only award M1A0.
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} -\tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \\ \tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \end{pmatrix}\begin{pmatrix} p \\ q \end{pmatrix} = \begin{pmatrix} 6\sqrt{2} \\ 3\sqrt{2} \end{pmatrix}\) | |
| \(-p - q = 12\) and \(p - q = 6\) or equivalent | M1A1 |
| \(p = -3\) and \(q = -9\) or \(\begin{pmatrix} -3 \\ -9 \end{pmatrix}\) | B1 cso |
| ALT Uses Inverse matrix \(\mathbf{P}^{-1}\) with vector \(= \begin{pmatrix} -\tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} \\ -\tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \end{pmatrix}\begin{pmatrix} 6\sqrt{2} \\ 3\sqrt{2} \end{pmatrix}\) | M1A1 |
| \(p = -3\) and \(q = -9\) or \(\begin{pmatrix} -3 \\ -9 \end{pmatrix}\) | B1 cso |
| (3) |
Notes
M1: Multiplies matrices in correct order to obtain two equations in \(p\) and \(q\).
A1: Two correct equations
B1 cso: \(p\) and \(q\) both correct, may be in vector form. No errors seen in solution.
ALT (b) M1: Attempt to find Inverse Matrix and pre-multiply A1: Correct Inverse Matrix used
B1 cso: \(p\) and \(q\) both correct, may be in vector form. No errors seen in solution.
| Scheme | Marks |
|---|---|
| \(\mathbf{Q} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| Accept \(\mathbf{T}\) if used instead of \(\mathbf{R}\) | |
| \(\mathbf{R} = \text{"}\mathbf{Q}\text{"}\mathbf{P} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} -\tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \\ \tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} \tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \\ -\tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \end{pmatrix}\) | M1 A1 A1 |
| (3) |
Notes
M1: Sets matrix product correct way round and obtains one correct term for their \(\mathbf{Q}\)
A1: Two correct terms from a correct \(\mathbf{Q}\). \(\mathbf{Q}\) incorrect award A0 here. A1: Completely correct matrix
| Scheme | Marks |
|---|---|
| \(\mathbf{R}^{-1} = \dfrac{1}{-1}\begin{pmatrix} -\tfrac{1}{\sqrt{2}} & +\tfrac{1}{\sqrt{2}} \\ +\tfrac{1}{\sqrt{2}} & +\tfrac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} \tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \\ -\tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \end{pmatrix}\) \(= \mathbf{R}\) (so matrix is self inverse and so transformation is self inverse) | B1 |
| ALT 1 (e) \(\mathbf{RR} = \begin{pmatrix} \tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \\ -\tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \end{pmatrix}\begin{pmatrix} \tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \\ -\tfrac{1}{\sqrt{2}} & -\tfrac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) (so \(\mathbf{R}\) is self inverse and so transformation is self inverse) | B1 |
| ALT 2 (e) Matrix represents a reflection (so is self inverse) | B1 |
| (1) | |
| (10 marks) |
Notes
B1: Calculates \(\mathbf{R}^{-1}\) and indicates that \(\mathbf{R}^{-1} = \mathbf{R}\) or calculates \(\mathbf{R}^2\) and indicates that \(\mathbf{R}^2 = \mathbf{I}\) or states that \(\mathbf{R}\) represents a reflection.