Question

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

13. \(A = \left( {\begin{array}{*{20}{c}}3&{ - 2}&8\\0&5&{ - 2}\\0&{ - 4}&3\end{array}} \right)\)

Short answer

The eigenvalues of \(A\) are \(1\), \(7\), \(3\).

Step-by-step solution

Step 1: Write the echelon form

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

Then we have,

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

Step 2: Find the characteristic polynomial of \(V\)

Find the determinant of\(V - \lambda I\).

\(\begin{aligned}{c}\det \left( {\begin{aligned}{*{20}{c}}{5 - \lambda }&{ - 2}\\{ - 4}&{3 - \lambda }\end{aligned}} \right) &= \left( {5 - \lambda } \right)\left( {3 - \lambda } \right) - 8\\ &= 15 - 5\lambda - 3\lambda + {\lambda ^2} - 8\\ &= {\lambda ^2} - 8\lambda + 7\\ &= \left( {\lambda - 1} \right)\left( {\lambda - 7} \right)\end{aligned}\)

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

\(A = \left( {\lambda - 1} \right)\left( {\lambda - 7} \right)\left( {\lambda - 3} \right)\)

Thus, the eigenvalues of \(A\) are \(1\), \(7\), \(3\).