FP1 January 2010 Q9
9. \[\mathbf{M} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\[4pt] \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}\]
(a) Describe fully the geometrical transformation represented by the matrix \(\mathbf{M}\). (2)
The transformation represented by \(\mathbf{M}\) maps the point \(A\) with coordinates \((p,\ q)\) onto the point \(B\) with coordinates \((3\sqrt{2},\ 4\sqrt{2})\).
(b) Find the value of \(p\) and the value of \(q\). (4)
(c) Find, in its simplest surd form, the length \(OA\), where \(O\) is the origin. (2)
(d) Find \(\mathbf{M}^2\). (2)
The point \(B\) is mapped onto the point \(C\) by the transformation represented by \(\mathbf{M}^2\).
(e) Find the coordinates of \(C\). (2)
| Scheme | Marks |
|---|---|
| \(45^\circ\) or \(\dfrac{\pi}{4}\) rotation (anticlockwise), about the origin | B1, B1 |
| (2) |
Notes
Order of matrix multiplication needs to be correct to award Ms
(a) More than one transformation 0/2
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\[4pt] \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}\begin{pmatrix} p \\ q \end{pmatrix} = \begin{pmatrix} 3\sqrt{2} \\ 4\sqrt{2} \end{pmatrix}\) | M1 |
| \(p - q = 6\) and \(p + q = 8\) or equivalent | M1 A1 |
| \(p = 7\) and \(q = 1\) both correct | A1 |
| (4) |
Notes
(b) Second M1 for correct matrix multiplication to give two equations
Alternative
| Scheme | Marks |
|---|---|
| (b) \(\mathbf{M}^{-1} = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\[4pt] -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}\) | First M1 A1 |
| \(\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\[4pt] -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}\begin{pmatrix} 3\sqrt{2} \\ 4\sqrt{2} \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \end{pmatrix}\) | Second M1 A1 |
| Scheme | Marks |
|---|---|
| Length of \(OA\) (= length of \(OB\)) \(= \sqrt{7^2 + 1^2},\ = \sqrt{50} = 5\sqrt{2}\) | M1, A1 |
| (2) |
Notes
(c) Correct use of their \(p\) and their \(q\) award M1
| Scheme | Marks |
|---|---|
| \(\mathbf{M}^2 = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\[4pt] \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}\begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\[4pt] \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 3\sqrt{2} \\ 4\sqrt{2} \end{pmatrix}\) so coordinates are \((-4\sqrt{2},\ 3\sqrt{2})\) | M1 A1 |
| (2) | |
| [12] |
Notes
(e) Accept column vector for final A1.