Answer
$\rightarrow \lambda_1=2; \lambda_2=-5;\lambda_3=0$
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
$\det(A−\lambda)=\det \begin{bmatrix} -5-\lambda & -1 & 1\\ -2& -1-\lambda & 0 \\ -5 & 2 & 3-\lambda \end{bmatrix}=0$
$(3- \lambda).[(-5- \lambda)(-1-\lambda)-(-1)(-2)]+[(-2.2-(-1-\lambda).=0$
$(3- \lambda).(3+6 \lambda+\lambda^2)+(-9-5\lambda)=0$
$9+18\lambda+3\lambda^2-3\lambda-6\lambda^2-\lambda^3-9-5\lambda=0$
$-\lambda^3-3\lambda^2+10\lambda=0$
$(7-\lambda)(\lambda^2-3\lambda-28)$
$(\lambda-2)(\lambda+5)\lambda=0$
$\rightarrow \lambda_1=2; \lambda_2=-5;\lambda_3=0$
The eigenvalues of the given matrix A are :
$\rightarrow \lambda_1=2; \lambda_2=-5;\lambda_3=0$
