Answer
$\rightarrow \lambda_1=-4; \lambda_2=\lambda_3=7$
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
$\det(A−\lambda)=\det \begin{bmatrix} 6-\lambda & 0 & -2\\ 0& 7-\lambda & 0 \\ -5 & 0 & -3-\lambda \end{bmatrix}=0$
$(6- \lambda).(7- \lambda)(-3-\lambda)-2(7-\lambda)(5)=0$
$(6- \lambda).(-21- 4\lambda+\lambda^2)-10(7-\lambda)=0$
$-\lambda^3+10\lambda^2+7\lambda-196=0$
$(7-\lambda)(\lambda^2-3\lambda-28)$
$(7-\lambda)(\lambda-7)(\lambda+4)$
$\rightarrow \lambda_1=-4; \lambda_2=\lambda_3=7$
The eigenvalues of the given matrix A are :
$\rightarrow \lambda_1=-4; \lambda_2=\lambda_3=7$
