Question: Exercises 9-14 require techniques section 3.1. Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\) determinants described prior to Exercise 15-18 in Section 3.1. [Note: Finding the characteristic polynomial of a \(3 \times 3\) matrix is not easy to do with just row operations, because the variable \(\lambda \) is involved.

14. \(\left[ {\begin{array}{*{20}{c}}5&- 2&3\\0&1&0\\6&7&- 2\end{array}} \right]\)

The eigen value 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 second row to obtain the characteristic polynomial of the matrix, as shown below.

\[\begin{array}\det \left( {A - \lambda I} \right) = \det \left[ {\begin{array}{*{20}{c}}{5 - \lambda }&{ - 2}&3\\0&{1 - \lambda }&0\\6&7&{ - 2 - \lambda }\end{array}} \right]\\ = \left( {1 - \lambda } \right)\det \left[ {\begin{array}{*{20}{c}}{5 - \lambda }&3\\6&{ - 2 - \lambda }\end{array}} \right]\\ = \left( {1 - \lambda } \right)\left[ {\left( {5 - \lambda } \right)\left( { - 2 - \lambda } \right) - \left( 3 \right)\left( 6 \right)} \right]\\ = \left( {1 - \lambda } \right)\left( {{\lambda ^2} - 3\lambda - 28} \right)\\ = - {\lambda ^3} + 4{\lambda ^2} + 25\lambda - 28\,\\ = \left( {1 - \lambda } \right)\left( {\lambda - 7} \right)\left( {\lambda + 4} \right)\end{array}\]

Thus, the characteristic polynomial of the matrix is \( - {\lambda ^3} + 4{\lambda ^2} + 25\lambda - 28\).