Question

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

Short answer

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

Step-by-step solution

Definition of the characteristic polynomial

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 \(n\).

Determine the characteristic polynomial of the matrix

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 \( - {\lambda ^3} + 4{\lambda ^2} + 25\lambda - 28\).