AS June 2023 Paper 1 Q9
9 Matrix \(\mathbf{R}\) is given by \(\mathbf{R} = \begin{pmatrix} a & 0 & -b \\ 0 & 1 & 0 \\ b & 0 & a \end{pmatrix}\) where \(a\) and \(b\) are constants.
The constants \(a\) and \(b\) are given by \(a = \dfrac{\sqrt{2}}{4}(\sqrt{3} + 1)\) and \(b = \dfrac{\sqrt{2}}{4}(\sqrt{3} - 1)\).
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} a & 0 & -b \\ 0 & 1 & 0 \\ b & 0 & a \end{pmatrix}\begin{pmatrix} a & 0 & -b \\ 0 & 1 & 0 \\ b & 0 & a \end{pmatrix}\) \(= \begin{pmatrix} a \times a + -b \times b & 0 & -ab - ab \\ 0 & 0 + 1 \times 1 + 0 & 0 \\ ab + ab & 0 & -b \times b + a \times a \end{pmatrix}\) | M1 | 1.1 |
| \(= \begin{pmatrix} a^2 - b^2 & 0 & -2ab \\ 0 & 1 & 0 \\ 2ab & 0 & a^2 - b^2 \end{pmatrix}\) | A1 | 1.1 |
| [2] |
Notes
M1: Some indication of knowledge of how to multiply 3 by 3 matrices
Can be implied by 3 non-zero correct terms
A1: Final solutions must have all terms simplified.
| Scheme | Marks | AO |
|---|---|---|
| \(ab = \dfrac{\sqrt{2}}{4}\left(\sqrt{3} + 1\right)\dfrac{\sqrt{2}}{4}\left(\sqrt{3} - 1\right) = \dfrac{2}{16}(3 - 1) = \dfrac{1}{4}\) or \(a^2 - b^2 = (a - b)(a + b) = \dfrac{2\sqrt{2}}{4} \times \dfrac{2\sqrt{2}\sqrt{3}}{4}\) \(= \dfrac{1}{2}\sqrt{3}\) | B1 | 3.1a |
| \(R^2 = \begin{pmatrix} \frac{1}{2}\sqrt{3} & 0 & -2 \times \frac{1}{4} \\ 0 & 1 & 0 \\ 2 \times \frac{1}{4} & 0 & \frac{1}{2}\sqrt{3} \end{pmatrix}\) \(= \begin{pmatrix} \frac{1}{2}\sqrt{3} & 0 & -\frac{1}{2} \\ 0 & \frac{1}{2} \times 2 & 0 \\ \frac{1}{2} & 0 & \frac{1}{2}\sqrt{3} \end{pmatrix} = \frac{1}{2}\begin{pmatrix} \sqrt{3} & 0 & -1 \\ 0 & 2 & 0 \\ 1 & 0 & \sqrt{3} \end{pmatrix}\) so \(k = \tfrac{1}{2}\) | B1 | 2.2a |
| [2] |
Notes
B1: (1st) Explicitly finding an expression for either \(ab\) (or \(2ab\)) or \(a^2 - b^2\)
B1: (2nd) AG. Explicitly finding the expression for the other substituting into expression for \(\mathbf{R}^2\) (or carrying out the matrix multiplication again). \(k\) can be embedded.
If \(k\) embedded need to see \(\frac{1}{2} \times 2\) or \(\frac{2}{2}\).
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{R}^4 = \begin{pmatrix} \frac{1}{2}\sqrt{3} & 0 & -\frac{1}{2} \\ 0 & 1 & 0 \\ \frac{1}{2} & 0 & \frac{1}{2}\sqrt{3} \end{pmatrix}\begin{pmatrix} \frac{1}{2}\sqrt{3} & 0 & -\frac{1}{2} \\ 0 & 1 & 0 \\ \frac{1}{2} & 0 & \frac{1}{2}\sqrt{3} \end{pmatrix}\) \(= \begin{pmatrix} \frac{1}{2} & 0 & -\frac{1}{2}\sqrt{3} \\ 0 & 1 & 0 \\ \frac{1}{2}\sqrt{3} & 0 & \frac{1}{2} \end{pmatrix}\) | ||
| \(\mathbf{R}^6 = \begin{pmatrix} \frac{1}{2}\sqrt{3} & 0 & -\frac{1}{2} \\ 0 & 1 & 0 \\ \frac{1}{2} & 0 & \frac{1}{2}\sqrt{3} \end{pmatrix}\begin{pmatrix} \frac{1}{2} & 0 & -\frac{1}{2}\sqrt{3} \\ 0 & 1 & 0 \\ \frac{1}{2}\sqrt{3} & 0 & \frac{1}{2} \end{pmatrix}\) \(= \begin{pmatrix} 0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}\) | B1 | 1.1 |
| \(\mathbf{R}^{12} = \begin{pmatrix} 0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}\) \(= \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}\) | B1 | 1.1 |
| \(\mathbf{R}^{24} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}\) \(= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) | B1 | 1.1 |
| [3] |
Notes
\(\mathbf{R}^4\): For reference – does not need to be seen in working.
SC – if answers left in terms of k then award B1 if one or two correct and B2 if all three correct
B1: (1st) For correct \(\mathbf{R}^6\)
By calculator expected, so intermediate steps might not be shown.
B1: (2nd) For correct \(\mathbf{R}^{12}\)
B1: (3rd) For correct \(\mathbf{R}^{24}\)
| Scheme | Marks | AO |
|---|---|---|
| Rotation | B1 | 3.1a |
| \((360^\circ/24 =)\ 15^\circ\) | B1 | 2.2a |
| Clockwise about the \(y\)-axis. | B1 | 3.2a |
| [3] |
Notes
B1: (2nd) Also allow \(345^\circ\) (Rotation in opposite sense)
Also allow in radians \(\frac{\pi}{12}\)
B1: (3rd) Both sense and axis must be correct.
Could be a rotation of \(345^\circ\) anticlockwise about \(y\)-axis
If correct transformation is combined with an incorrect one (such as correct rotation combined with a reflection) then maximum mark is B2.