Answer
$\rightarrow \lambda_1=7; \lambda_2=-9;\lambda_3=5$
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
$\det(A−\lambda)=\det \begin{bmatrix} -5-\lambda & -5 & -2\\ -8& 1-\lambda & 0 \\ -5 & 3 & 7-\lambda \end{bmatrix}=0$
$(7- \lambda).[(-5- \lambda)(1-\lambda)-(-5)(-8)]=0$
$(7- \lambda).(-5+ 4\lambda+\lambda^2)-40=0$
$(7- \lambda).( \lambda^2+4\lambda-45)=0$
$(7-\lambda)(\lambda^2-3\lambda-28)$
$(7-\lambda)(\lambda+9)(\lambda-5)$
$\rightarrow \lambda_1=7; \lambda_2=-9;\lambda_3=5$
The eigenvalues of the given matrix A are :
$\rightarrow \lambda_1=7; \lambda_2=-9;\lambda_3=5$
