- In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
$$\mathbf { M } = \left( \begin{array} { r r r }
2 & 0 & 3 \\
0 & - 4 & - 3 \\
0 & - 4 & 0
\end{array} \right)$$
Given that \(\mathbf { M }\) has exactly two distinct eigenvalues \(\lambda _ { 1 }\) and \(\lambda _ { 2 }\) where \(\lambda _ { 1 } < \lambda _ { 2 }\)
- determine a normalised eigenvector corresponding to the eigenvalue \(\lambda _ { 1 }\)
The line \(l _ { 1 }\) has equation \(\mathbf { r } = \left( \begin{array} { r } 4 \\ - 1 \\ 0 \end{array} \right) + \mu \left( \begin{array} { r } 2 \\ 0 \\ - 1 \end{array} \right)\), where \(\mu\) is a scalar parameter.
The transformation \(T\) is represented by \(\mathbf { M }\).
The line \(l _ { 1 }\) is transformed by \(T\) to the line \(l _ { 2 }\) - Determine a vector equation for \(l _ { 2 }\), giving your answer in the form \(\mathbf { r } \times \mathbf { b } = \mathbf { c }\) where \(\mathbf { b }\) and \(\mathbf { c }\) are constant vectors.