Answer
See below
Work Step by Step
1. Find eigenvalues:
$\left (\lambda -2\right )^2 ( \lambda +1)=0$
$\lambda_1=\lambda_2=2, \lambda_3=-1$
We just need to test with $\lambda_1,\lambda_2$
For $\lambda=2$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
-1-\lambda & -3 & 1 \\
-1 & -1-\lambda & 1\\
-1 & -3 & 3-\lambda
\end{bmatrix}=\begin{bmatrix}
-3& -3 & 1\\
-1 & -3 & 1\\
-1 & -3 & 1
\end{bmatrix}$
We obtain reduced row echelon form:
$\begin{bmatrix}
-3 & -3 & 1| 0 \\
-1 & -3 & 1| 0 \\
-1 & -3 & 1
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & 0 | 0 \\
0 & 3 & 1 | 0\\
0 & 0 & 0
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Let $r,s$ be free variables.
$\vec{V}=r(-3,1,0)+s(1,0,1) \\
E_2=\{(-3,1,0) +(1,0,1)\}
\rightarrow dim(E_2)=2$
Hence, matrix $A$ is non-defective.
