Question: Exercises 9–14 require techniques from 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 Exercises 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 variableis involved.)

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

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

It is given that\(A = \left( {\begin{array}{*{20}{c}}2&7\\7&2\end{array}} \right)\)and\(I = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\)is identity matrix. Find the matrix\(\left( {A - \lambda I} \right)\)as shown below:

\(\begin{array}{c}A - \lambda I = \left( {\begin{array}{*{20}{c}}1&0&{ - 1}\\2&3&{ - 1}\\0&6&0\end{array}} \right) - \lambda \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{1 - \lambda }&0&{ - 1}\\2&{3 - \lambda }&{ - 1}\\0&6&{ - \lambda }\end{array}} \right)\end{array}\)

For \(n \ge 2\) the determinant of an \(n \times n\) matrix \(A = [{a_{ij}}]\) is the sum of \(n\)terms of the form \( \pm {a_{1j}}\det {A_{1j}},\) with plus and minus signs alternating, where the entries \({a_{11}},{a_{12}}, \ldots ,{a_{1n}}\) are from the first row of \(A\). In symbols,

\(\begin{gathered} {\text{det}}A = {a_{11}}\det {A_{11}} - {a_{12}}\det {A_{12}} + \ldots + {\left( { - 1} \right)^{1 + n}}{a_{1n}}\det {A_{1n}} \\ = \mathop \sum \limits_{j = 1}^n {\left( { - 1} \right)^{1 + j}}{a_{1j}}\det {A_{1j}} \\ \end{gathered} \)

With the help of above defined formula, the\(\det A\)is calculated as follows:

\(\begin{array}{c}det\left( {A - \lambda I} \right) = det\left( {\begin{array}{*{20}{c}}{1 - \lambda }&0&{ - 1}\\2&{3 - \lambda }&{ - 1}\\0&6&{ - \lambda }\end{array}} \right)\\ = \left( {1 - \lambda } \right)\left| {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 1}\\6&{ - \lambda }\end{array}} \right| - 0\left| {\begin{array}{*{20}{c}}2&{ - 1}\\0&{ - \lambda }\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}2&{3 - \lambda }\\0&6\end{array}} \right|\\ = \left( {1 - \lambda } \right)\left( {\left( {3 - \lambda } \right)\left( { - \lambda } \right) - \left( 6 \right)\left( { - 1} \right)} \right) - 0\left( {\left( 2 \right)\left( { - \lambda } \right) - 0\left( { - 1} \right)} \right) - 1\left( {\left( 2 \right)\left( 6 \right) - 0} \right)\\ = - 3\lambda + 3{\lambda ^2} + {\lambda ^2} - {\lambda ^3} + 6 - 6\lambda \\ = - {\lambda ^3} + 4{\lambda ^2} - 9\lambda + 6\end{array}\)

Thus, the characteristic polynomial is \( - {\lambda ^3} + 4{\lambda ^2} - 9\lambda + 6\).