The matrix M is defined by M = \(\left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\).

The eigenvalues of M are denoted by \({\lambda _1},{\text{ }}{\lambda _2}\).

(a)     Show that \({\lambda _1} + {\lambda _2} = a + d\) and \({\lambda _1}{\lambda _2} = \det \)(M).

(b)     Given that \(a + b = c + d = 1\), show that 1 is an eigenvalue of M.

(c)     Find eigenvectors for the matrix \(\left( {\begin{array}{*{20}{c}}2&{ - 1}\\3&{ - 2}\end{array}} \right)\).

(a)     the eigenvalues satisfy \(\left| {\begin{array}{*{20}{c}}   {a - \lambda }&b \\   c&{d - \lambda } \end{array}} \right| = 0\)     (M1)

\({\lambda ^2} - (a + d)\lambda  + ad - bc = 0\)     A1

using the sum and product properties of the roots of a quadratic equation     R1

\({\lambda _1} + {\lambda _2} = a + d,{\text{ }}{\lambda _1}{\lambda _2} = ad - bc = \det \,\)(M)     AG

[3 marks]

(b)     let \(f(\lambda ) = {\lambda ^2} - (a + d)\lambda  + ad - bc\)

putting \(b = 1 - a\) and \(d = 1 - c\), consider     M1

\(f(1) = 1 - a - 1 + c + a - ac - c + ac = 0\)     A1

therefore \(\lambda  = 1\) is an eigenvalue     AG

[2 marks]

Note: Allow substitution for \(b\), \(c\) into the quadratic equation for \(\lambda \) followed by solution of this equation.

(c)     using any valid method     (M1)

the eigenvalues are 1 and –1     A1

an eigenvector corresponding to \(\lambda  = 1\) satisfies

\(\left( {\begin{array}{*{20}{c}}2&{ - 1}\\3&{ - 2}\end{array}} \right)\left( \begin{array}{l}x\\y\end{array} \right) = \left( \begin{array}{l}x\\y\end{array} \right)\) or \(\left( {\begin{array}{*{20}{c}}1&{ - 1}\\3&{ - 3}\end{array}} \right)\left( \begin{array}{l}x\\y\end{array} \right) = \left( \begin{array}{l}0\\0\end{array} \right)\)     M1A1

\(\left( \begin{array}{l}x\\y\end{array} \right) = \left( \begin{array}{l}1\\1\end{array} \right)\) or any multiple     A1

an eigenvector corresponding to \(\lambda  =  - 1\) satisfies

\(\left( {\begin{array}{*{20}{c}}2&{ - 1}\\3&{ - 2}\end{array}} \right)\left( \begin{array}{l}x\\y\end{array} \right) =  - \left( \begin{array}{l}x\\y\end{array} \right)\) or \(\left( {\begin{array}{*{20}{c}}3&{ - 1}\\3&{ - 1}\end{array}} \right)\left( \begin{array}{l}x\\y\end{array} \right) = \left( \begin{array}{l}0\\0\end{array} \right)\)     M1

\(\left( \begin{array}{l}x\\y\end{array} \right) = \left( \begin{array}{l}1\\3\end{array} \right)\) or any multiple     A1

Note: Award M1A1A1 for calculating the first eigenvector and M1A1 for the second irrespective of the order in which they are calculated.

[7 marks]