Answer
$\rightarrow \lambda_1=2; \lambda_2=14, \lambda_3=0$
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
$\det(A−\lambda)=\det \begin{bmatrix} 2-\lambda & 0 & 0\\ 7& 7-\lambda & 7 \\ 7 & 7 & 7-\lambda \end{bmatrix}=0$
$(2- \lambda).[(7- \lambda)^2-49]=0$
$(2- \lambda).(14- \lambda)(-\lambda)=0$
$\rightarrow \lambda_1=2; \lambda_2=14, \lambda_3=0$
The eigenvalues of the given matrix A are
$\rightarrow \lambda_1=2; \lambda_2=14, \lambda_3=0$
