Answer
The eigenvalues of the given matrix A are $\rightarrow \lambda_1=-6; \lambda_2=7, \lambda_3=2$
Work Step by Step
We are given:
$\det (A- \lambda I)=0$
Using cofactor expansion along the first column of $A−\lambda I$, we obtain
$\det(A−\lambda)=\det \begin{bmatrix}
2-\lambda & 0 & 0\\
-1 & -6-\lambda & 0 \\
3 & 3 & 7-\lambda
\end{bmatrix}$
$=(2- \lambda).(-6- \lambda).(7-\lambda)$
$=(-12+4\lambda+\lambda^2)(7-\lambda)$
$=-84+12\lambda + 28\lambda -4\lambda^2+7\lambda^2 -\lambda^3$
$=-\lambda^3+3\lambda^2+40\lambda-84$
$\rightarrow \lambda_1=-6; \lambda_2=7, \lambda_3=2$
The eigenvalues of the given matrix A are $\rightarrow \lambda_1=-6; \lambda_2=7, \lambda_3=2$
