Answer
The eigenvalues of the given matrix A are $\lambda_1=3+i; \lambda_2=3-i$
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
$\det(A−\lambda)=\det \begin{bmatrix}
2-\lambda & -1\\
2 & 4-\lambda
\end{bmatrix}$
$=(2- \lambda).(4- \lambda)-(-1).2$
$=8-2\lambda -4\lambda +\lambda^2 + 2$
$=\lambda^2 -6\lambda +10$
$\rightarrow \lambda_1=3+i; \lambda_2=3-i$
The eigenvalues of the given matrix A are $\lambda_1=3+i; \lambda_2=3-i$
