It is given that \(\lambda\) is an eigenvalue of the non-singular square matrix \(\mathbf { A }\), with corresponding eigenvector \(\mathbf { e }\).
Show that \(\lambda ^ { - 1 }\) is an eigenvalue of \(\mathbf { A } ^ { - 1 }\) for which \(\mathbf { e }\) is a corresponding eigenvector.
The matrix \(\mathbf { A }\) is given by
$$\mathbf { A } = \left( \begin{array} { r r r }
2 & 0 & 3
15 & - 4 & 3
3 & 0 & 2
\end{array} \right)$$
Given that - 1 is an eigenvalue of \(\mathbf { A }\), find a corresponding eigenvector.
It is also given that \(\left( \begin{array} { l } 0 1 0 \end{array} \right)\) and \(\left( \begin{array} { l } 1 2 1 \end{array} \right)\) are eigenvectors of \(\mathbf { A }\). Find the corresponding eigenvalues.
Hence find a matrix \(\mathbf { P }\) and a diagonal matrix \(\mathbf { D }\) such that \(\mathbf { A } ^ { - 1 } = \mathbf { P D P } ^ { - 1 }\).
Use the characteristic equation of \(\mathbf { A }\) to show that \(\mathbf { A } ^ { - 1 } = p \mathbf { A } ^ { 2 } + q l\), where \(p\) and \(q\) are rational numbers to be determined.