Answer
The eigenvalues of the given matrix A are $\lambda_1=1; \lambda_2=14$
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
$\det(A−\lambda)=\det \begin{bmatrix}
2-\lambda & 4\\
3 & 13-\lambda
\end{bmatrix}$
$=(2- \lambda).(13- \lambda)-4.3$
$=26-2\lambda -13\lambda +\lambda^2 -12$
$=\lambda^2 -15\lambda +14$
$\rightarrow \lambda_1=1; \lambda_2=14$
The eigenvalues of the given matrix A are $\lambda_1=1; \lambda_2=14$
