By considering the points \((1,{\text{ }}0)\) and \((0,{\text{ }}1)\) determine the \(2 \times 2\) matrix which represents

(i)     an anticlockwise rotation of \(\theta \) about the origin;

(ii)     a reflection in the line \(y = (\tan \theta )x\).

Determine the matrix \(A\) which represents a rotation from the direction \(\left( {\begin{array}{*{20}{c}} 1 \\ 0 \end{array}} \right)\) to the direction \(\left( {\begin{array}{*{20}{c}} 1 \\ 3 \end{array}} \right)\).

A triangle whose vertices have coordinates \((0,{\text{ }}0)\), \((3,{\text{ }}1)\) and \((1,{\text{ }}5)\) undergoes a transformation represented by the matrix \({A^{ - 1}}XA\), where \(X\) is the matrix representing a reflection in the \(x\)-axis. Find the coordinates of the vertices of the transformed triangle.

The matrix \(B = {A^{ - 1}}XA\) represents a reflection in the line \(y = mx\). Find the value of \(m\).

(i)     under an anti-clockwise rotation of \(\theta \)

\((1,{\text{ }}0) \to (\cos \theta ,{\text{ }}\sin \theta )\)

\((0,{\text{ }}1) \to ( - \sin \theta ,{\text{ }}\cos \theta )\)     M1

rotation matrix is \(\left( {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta } \\  {\sin \theta }&{\cos \theta } \end{array}} \right)\)     A1

(ii)          M1

under a reflection in the line \(y = (\tan \theta )x\)

\((1,{\text{ }}0) \to (\cos 2\theta ,{\text{ }}\sin 2\theta )\)

\((0,{\text{ }}1) \to (\sin 2\theta ,{\text{ }} - \cos 2\theta )\)     M1

matrix for reflection in the line \(y = (\tan \theta )x: \left( {\begin{array}{*{20}{c}} {\cos 2\theta }&{\sin 2\theta } \\ {\sin 2\theta }&{ - \cos 2\theta } \end{array}} \right)\)     A1

in this case \(\tan \theta  = 3\)     (M1)

\(\Rightarrow \sin \theta  = \frac{3}{{\sqrt {10} }}\)

hence rotation matrix is \(\left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt {10} }}}&{ - \frac{3}{{\sqrt {10} }}} \\  {\frac{3}{{\sqrt {10} }}}&{\frac{1}{{\sqrt {10} }}} \end{array}} \right)\)     A1

\({A^{ - 1}} = \left( {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt {10} }}}&{\frac{3}{{\sqrt {10} }}} \\ { - \frac{3}{{\sqrt {10} }}}&{\frac{1}{{\sqrt {10} }}} \end{array}} \right)\)     (A1)

\(X = \left( {\begin{array}{*{20}{c}} 1&0 \\ 0&{ - 1} \end{array}} \right)\)     (A1)

\( \Rightarrow {A^{ - 1}}XA = \left( {\begin{array}{*{20}{c}} { - \frac{8}{{10}}}&{ - \frac{6}{{10}}} \\ { - \frac{6}{{10}}}&{\frac{8}{{10}}} \end{array}} \right)\)     (M1)A1

\( \Rightarrow {A^{ - 1}}XA(G) = \left( {\begin{array}{*{20}{c}} { - \frac{8}{{10}}}&{ - \frac{6}{{10}}} \\ { - \frac{6}{{10}}}&{\frac{8}{{10}}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 0&3&1 \\ 0&1&5 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0&{ - 3}&{ - \frac{{19}}{5}} \\ 0&{ - 1}&{\frac{{17}}{5}} \end{array}} \right)\)     (M1)

hence coordinates are \((0,{\text{ }}0)\), \(( - 3,{\text{ }} - 1)\) and \(\left( { - \frac{{19}}{5},{\text{ }}\frac{{17}}{5}} \right)\)     A1

\(B = \left( {\begin{array}{*{20}{c}} { - \frac{8}{{10}}}&{ - \frac{6}{{10}}} \\ { - \frac{6}{{10}}}&{\frac{8}{{10}}} \end{array}} \right)\)

the matrix for the reflection in the line \(y = (\tan \theta )x\) is \(\left( {\begin{array}{*{20}{c}} {\cos 2\theta }&{\sin 2\theta } \\ {\sin 2\theta }&{ - \cos 2\theta } \end{array}} \right)\)

\(\cos 2\theta  =  - \frac{4}{5},{\text{ }}\sin 2\theta  =  - \frac{3}{5}\)     (A1)(A1)

\(\cos 2\theta  = 2{\cos ^2}\theta  - 1\)     (M1)

\( \Rightarrow 2{\cos ^2}\theta  = \frac{2}{{10}}\)

\( \Rightarrow \cos \theta  =  \pm \frac{1}{{\sqrt {10} }}\)     (A1)

\( \Rightarrow \cos \theta  =  - \frac{1}{{\sqrt {10} }}\) and \(\sin \theta  = \frac{3}{{\sqrt {10} }}\)     (A1)

\( \Rightarrow \tan \theta  =  - 3\)

\( \Rightarrow m =  - 3\)     A1

This proved to be a more challenging question for many candidates. In part a) many candidates appeared to not know how to find the matrices and for those who attempted to find them, arithmetic errors were common. A number of wholly correct solutions to parts b), c) and d) were seen, but many candidates seemed unfamiliar with this style of question and made errors or simply gave up part way through the process.

This proved to be a more challenging question for many candidates. In part a) many candidates appeared to not know how to find the matrices and for those who attempted to find them, arithmetic errors were common. A number of wholly correct solutions to parts b), c) and d) were seen, but many candidates seemed unfamiliar with this style of question and made errors or simply gave up part way through the process.

This proved to be a more challenging question for many candidates. In part a) many candidates appeared to not know how to find the matrices and for those who attempted to find them, arithmetic errors were common. A number of wholly correct solutions to parts b), c) and d) were seen, but many candidates seemed unfamiliar with this style of question and made errors or simply gave up part way through the process.

This proved to be a more challenging question for many candidates. In part a) many candidates appeared to not know how to find the matrices and for those who attempted to find them, arithmetic errors were common. A number of wholly correct solutions to parts b), c) and d) were seen, but many candidates seemed unfamiliar with this style of question and made errors or simply gave up part way through the process.