91Ó°ÊÓ

The eigenvalue of an \(n \times n\) matrix \(A\) is a scalar \(\lambda \) such that \(\lambda \) satisfies the characteristic equation \(\det \left( {A - \lambda I} \right) = 0\).

When \(A\) is an \(n \times n\) matrix, \(\det \left( {A - \lambda I} \right)\) is the characteristic polynomial of \(A\), which is the polynomial of degree \(n\).

Use the cofactor expression along the third row to obtain the characteristic polynomial of the matrix, as shown below.

\[\begin{array}\det \left( {A - \lambda I} \right) = \det \left[ {\begin{array}{*{20}{c}}{ - 1 - \lambda }&0&1\\{ - 3}&{4 - \lambda }&1\\0&0&{2 - \lambda }\end{array}} \right]\\ = \left( {2 - \lambda } \right)\det \left[ {\begin{array}{*{20}{c}}{ - 1 - \lambda }&0\\{ - 3}&{4 - \lambda }\end{array}} \right]\\ = \left( {2 - \lambda } \right)\left( { - 1 - \lambda } \right)\left( {4 - \lambda } \right)\\ = \left( {2 - \lambda } \right)\left( { - 4 - 3\lambda + {\lambda ^2}} \right)\\ = - {\lambda ^3} + 5{\lambda ^2} - 2\lambda - 8\end{array}\]

Thus, the characteristic polynomial of the matrix is \( - {\lambda ^3} + 5{\lambda ^2} - 2\lambda - 8\).