Use Exercise 12 to find the eigenvalues of the matrices in Exercises 13 and 14.

14. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{5}}&{{\bf{ - 6}}}&{{\bf{ - 7}}}\\{\bf{2}}&{\bf{4}}&{\bf{5}}&{\bf{2}}\\{\bf{0}}&{\bf{0}}&{{\bf{ - 7}}}&{{\bf{ - 4}}}\\{\bf{0}}&{\bf{0}}&{\bf{3}}&{\bf{1}}\end{array}} \right)\)

Assume \(G = \left( {\begin{array}{*{20}{c}}U&X\\0&V\end{array}} \right)\).

Then we have,

\(\begin{aligned}{c}A &= \left( {\begin{aligned}{*{20}{c}}1&5&{ - 6}&{ - 7}\\2&4&5&2\\0&0&{ - 7}&{ - 4}\\0&0&3&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}U&X\\0&V\end{aligned}} \right)\end{aligned}\)

On comparison we get,

\(U = \left( {\begin{aligned}{*{20}{c}}1&5\\2&4\end{aligned}} \right)\),\(V = \left( {\begin{aligned}{*{20}{c}}{ - 7}&{ - 4}\\3&1\end{aligned}} \right)\)

Now characteristic polynomial of \(U\) are as shown below:

\(\begin{aligned}{c}\det \left( {\begin{aligned}{*{20}{c}}{1 - \lambda }&5\\2&{4 - \lambda }\end{aligned}} \right) &= \left( {1 - \lambda } \right)\left( {4 - \lambda } \right) - 10\\ &= {\lambda ^2} - 5\lambda - 6\\ &= \left( {\lambda + 1} \right)\left( {\lambda - 6} \right)\end{aligned}\)

\(\begin{aligned}{c}\det \left( {\begin{aligned}{*{20}{c}}{ - 7 - \lambda }&{ - 4}\\3&{1 - \lambda }\end{aligned}} \right) &= \left( { - 7 - \lambda } \right)\left( {1 - \lambda } \right) + 12\\ &= {\lambda ^2} + 6\lambda + 5\\ &= \left( {\lambda + 1} \right)\left( {\lambda + 5} \right)\end{aligned}\)

On multiplication we get the characteristic polynomial of \(A\).

\(\begin{aligned}{c}A &= \left( {\lambda + 1} \right)\left( {\lambda - 6} \right)\left( {\lambda + 1} \right)\left( {\lambda + 5} \right)\\ &= {\left( {\lambda + 1} \right)^2}\left( {\lambda + 5} \right)\left( {\lambda - 6} \right)\end{aligned}\)

Thus, the eigenvalues of \(A\) are \( - 1\), \( - 1\), \( - 5\), \(6\).