Show that the linear transformation represented by M transforms any point on the line $y = x$ to a point on the same line.

Explain what happens to points on the line $4y + x = 0$ when they are transformed by M.

State the two eigenvalues of M.

State two eigenvectors of M which correspond to the two eigenvalues.

$\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \end{array}\,\,\,\begin{array}{*{20}{c}} { - 4} \\ { - 1} \end{array}} \right)\left( {\begin{array}{*{20}{c}} k \\ k \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 2k} \\ { - 2k} \end{array}} \right)\left( { = - 2\left( {\begin{array}{*{20}{c}} k \\ k \end{array}} \right)} \right)$      M1A1

hence still on the line $y = x$     AG

[2 marks]

consider $\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \end{array}\,\,\,\begin{array}{*{20}{c}} { - 4} \\ { - 1} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {4k} \\ { - k} \end{array}} \right)$      M1

$= \left( {\begin{array}{*{20}{c}} {12k} \\ { - 3k} \end{array}} \right)\left( { = 3\left( {\begin{array}{*{20}{c}} {4k} \\ { - k} \end{array}} \right)} \right)$      A1

hence the line is invariant      A1

[3 marks]

hence the eigenvalues are −2 and 3      A1A1

[2 marks]

$\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)$ and $\left( {\begin{array}{*{20}{c}} 4 \\ { - 1} \end{array}} \right)$ or equivalent      A1A1

[2 marks]