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.)

11. \(\left( {\begin{array}{*{20}{c}}4&0&0\\5&3&2\\{ - 2}&0&2\end{array}} \right)\)

Short answer

The characteristic polynomial of the matrix is \({\lambda ^3} + 9{\lambda ^2} - 26\lambda + 24\).

Step-by-step solution

The Characteristic equation

Aneigenvalue of a\(n \times n\)matrix\(A\)is ascalar \(\lambda \)such that if\(\lambda \)satisfies the characteristic equation\(\det \left( {A - \lambda I} \right) = 0\).

Determine the characteristic polynomial of the matrix

The unique arrangement of zeros in \(A\) is to create a cofactor expansion on the first row.

Determine the characteristic polynomial of the matrix as shown below:

\(\begin{array}{c}\det \left( {A - \lambda I} \right) = \det \left( {\begin{array}{*{20}{c}}{4 - \lambda }&0&0\\5&{3 - \lambda }&2\\{ - 2}&0&{2 - \lambda }\end{array}} \right)\\ = \left( {4 - \lambda } \right)\det \left( {\begin{array}{*{20}{c}}{3 - \lambda }&2\\0&{2 - \lambda }\end{array}} \right)\\ = \left( {4 - \lambda } \right)\left( {3 - \lambda } \right)\left( {2 - \lambda } \right)\\ = \left( {4 - \lambda } \right)\left( {{\lambda ^2} - 5\lambda + 6} \right)\\ = {\lambda ^3} + 9{\lambda ^2} - 26\lambda + 24\end{array}\)

It would not be necessary to express the characteristic polynomial in the expanded form if only the eigenvalues were needed.

Thus, the characteristic polynomial of the matrix is \({\lambda ^3} + 9{\lambda ^2} - 26\lambda + 24\).