By considering the image of $(0, 0)$, find $\mathit{q}$.

By considering the image of $(1, 0)$ and $(0, 1)$, show that

$\mathit{P}=\left(\begin{array}{cc}\frac{\sqrt{3}}{4} & \frac{1}{4}\\ -\frac{1}{4} & \frac{\sqrt{3}}{4}\end{array}\right)$.

Write down the matrix $\mathit{S}$.

Use $\mathit{P}=\mathit{R}\mathit{S}$ to find the matrix $\mathit{R}$.

Hence find the angle and direction of the rotation represented by $\mathit{R}$.

Write down an equation satisfied by $\left(\begin{array}{c}a\\ b\end{array}\right)$.

Find the value of $a$ and the value of $b$.

$\mathit{P}\left(\begin{array}{c}0\\ 0\end{array}\right)+\mathit{q}=\left(\begin{array}{c}0\\ 1\end{array}\right)$        (M1)

$\mathit{q}=\left(\begin{array}{c}0\\ 1\end{array}\right)$          A1

[2 marks]

EITHER

$\mathit{P}\left(\begin{array}{c}1\\ 0\end{array}\right)+\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}\frac{\sqrt{3}}{4}\\ \frac{3}{4}\end{array}\right)$          M1

hence $\mathit{P}\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}\frac{\sqrt{3}}{4}\\ -\frac{1}{4}\end{array}\right)$          A1

$\mathit{P}\left(\begin{array}{c}0\\ 1\end{array}\right)+\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}\frac{1}{4}\\ 1+\frac{\sqrt{3}}{4}\end{array}\right)$          M1

hence $\mathit{P}\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}\frac{1}{4}\\ \frac{\sqrt{3}}{4}\end{array}\right)$          A1

OR

$\left(\begin{array}{cc}a & b\\ c & d\end{array}\right)\left(\begin{array}{c}1\\ 0\end{array}\right)+\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}\frac{\sqrt{3}}{4}\\ \frac{3}{4}\end{array}\right)$          M1

hence $\left(\begin{array}{cc}a & b\\ c & d\end{array}\right)\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}\frac{\sqrt{3}}{4}\\ -\frac{1}{4}\end{array}\right)$          A1

$\left(\begin{array}{c}a\\ c\end{array}\right)=\left(\begin{array}{c}\frac{\sqrt{3}}{4}\\ -\frac{1}{4}\end{array}\right)$

$\left(\begin{array}{cc}a & b\\ c & d\end{array}\right)\left(\begin{array}{c}0\\ 1\end{array}\right)+\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}\frac{1}{4}\\ 1+\frac{\sqrt{3}}{4}\end{array}\right)$          M1

$\left(\begin{array}{cc}a & b\\ c & d\end{array}\right)\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}\frac{1}{4}\\ \frac{\sqrt{3}}{4}\end{array}\right)$          A1

$\left(\begin{array}{c}b\\ d\end{array}\right)=\left(\begin{array}{c}\frac{1}{4}\\ \frac{\sqrt{3}}{4}\end{array}\right)$

THEN

$\Rightarrow \mathit{P}=\left(\begin{array}{cc}\frac{\sqrt{3}}{4} & \frac{1}{4}\\ -\frac{1}{4} & \frac{\sqrt{3}}{4}\end{array}\right)$          AG

[4 marks]

$\left(\begin{array}{cc}\frac{1}{2} & 0\\ 0 & \frac{1}{2}\end{array}\right)$          A1

[1 mark]

EITHER

${\mathit{S}}^{-1}=\left(\begin{array}{cc}2 & 0\\ 0 & 2\end{array}\right)$         (A1)

$\mathit{R}=\mathit{P}{\mathit{S}}^{-1}$         (M1)

Note: The M1 is for an attempt at rearranging the matrix equation. Award even if the order of the product is reversed.

$\mathit{R}=\left(\begin{array}{cc}\frac{\sqrt{3}}{4} & \frac{1}{4}\\ -\frac{1}{4} & \frac{\sqrt{3}}{4}\end{array}\right)\left(\begin{array}{cc}2 & 0\\ 0 & 2\end{array}\right)$         (A1)

OR

$\left(\begin{array}{cc}\frac{\sqrt{3}}{4} & \frac{1}{4}\\ -\frac{1}{4} & \frac{\sqrt{3}}{4}\end{array}\right)=\mathit{R}\left(\begin{array}{cc}0.5 & 0\\ 0 & 0.5\end{array}\right)$

let $\mathit{R}=\left(\begin{array}{cc}a & b\\ c & d\end{array}\right)$

attempt to solve a system of equations         M1

$\frac{\sqrt{3}}{4}=0.5a, \frac{1}{4}=0.5b$

$-\frac{1}{4}=0.5c, \frac{\sqrt{3}}{4}=0.5d$          A2

Note: Award A1 for two correct equations, A2 for all four equations correct.

THEN

$\mathit{R}=\left(\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2}\\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right)$  OR  $\left(\begin{array}{cc}0.866 & 0.5\\ -0.5 & 0.866\end{array}\right)$  OR  $\left(\left(\begin{array}{cc}0.866025\dots & 0.5\\ -0.5 & 0.866025\dots \end{array}\right)\right)$          A1

Note: The correct answer can be obtained from reversing the matrices, so do not award if incorrect product seen. If the given answer is obtained from the product $\mathit{R}={\mathit{S}}^{-1}\mathit{P}$, award (A1)(M1)(A0)A0.

[4 marks]

clockwise         A1

arccosine or arcsine of value in matrix seen         (M1)

$30°$         A1

Note: Both A1 marks are dependent on the answer to part (c)(i) and should only be awarded for a valid rotation matrix.

[3 marks]

METHOD 1

$\left(\begin{array}{c}a\\ b\end{array}\right)=\mathit{P}\left(\begin{array}{c}a\\ b\end{array}\right)+\mathit{q}$         A1

METHOD 2

$\left(\begin{array}{c}x\text{'}\\ y\text{'}\end{array}\right)=\mathit{P}\left(\begin{array}{c}x-a\\ y-b\end{array}\right)+\left(\begin{array}{c}a\\ b\end{array}\right)$         A1

Note: Accept substitution of $x$ and $y$ (and $x\text{'}$ and $y\text{'}$) with particular points given in the question.

[1 mark]

METHOD 1

solving $\left(\begin{array}{c}a\\ b\end{array}\right)=\mathit{P}\left(\begin{array}{c}a\\ b\end{array}\right)+\mathit{q}$ using simultaneous equations or $\mathit{a}={\left(\mathit{I}-\mathit{P}\right)}^{-1}\mathit{q}$         (M1)

$a=0.651 \left(0.651084\dots \right), b=1.48 \left(1.47662\dots \right)$        A1A1

$\left(a=\frac{5+2\sqrt{3}}{13}, b=\frac{14+3\sqrt{3}}{13}\right)$

METHOD 2

$\left(\begin{array}{c}0\\ 1\end{array}\right)=\mathit{P}\left(\begin{array}{c}0-a\\ 0-b\end{array}\right)+\left(\begin{array}{c}a\\ b\end{array}\right)$         (M1)

Note: This line, with any of the points substituted, may be seen in part (d)(i) and if so the M1 can be awarded there.

$\left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\mathit{I}-\mathit{P}\right)\left(\begin{array}{c}a\\ b\end{array}\right)$

$a=0.651084\dots , b=1.47662\dots$        A1A1

$\left(a=\frac{5+2\sqrt{3}}{13}, b=\frac{14+3\sqrt{3}}{13}\right)$

[3 marks]

Part (i) proved to be straightforward for most candidates. A common error in part (ii) was for candidates to begin with the matrix P and to show it successfully transformed the points to their images. This received no marks. For a ‘show that’ question it is expected that the work moves to rather than from the given answer.

(b), (c) These two parts dealt generally with more familiar aspects of matrix transformations and were well done.

(b), (c) These two parts dealt generally with more familiar aspects of matrix transformations and were well done.

The trick of recognizing that $(a,b)$ was invariant was generally not seen and as such the question could not be successfully answered.