Start from forming a new matrix by subtracting $\lambda$ from the diagonal entries of the given matrix: $$ \left[\begin{array}{ccc} 8-\lambda & -8 & -2 \\ 4 & -\lambda-3 & -2 \\ 3 & -4 & 1-\lambda \end{array}\right] $$ The determinant of the obtained matrix is $-(\lambda-3)(\lambda-2)(\lambda-1)$ (for steps, see determinant calculator). Solve the equation $-(\lambda-3)(\lambda-2)(\lambda-1)=0$. The roots are $\lambda_{1}=3, \lambda_{2}=2, \lambda_{3}=1$

These are the eigenvalues.

Next, find the eigenvectors. - $\lambda=3$ $$ \left[\begin{array}{ccc} 8-\lambda & -8 & -2 \\ 4 & -\lambda-3 & -2 \\ 3 & -4 & 1-\lambda \end{array}\right]=\left[\begin{array}{ccc} 5 & -8 & -2 \\ 4 & -6 & -2 \\ 3 & -4 & -2 \end{array}\right] $$

The null space of this matrix is $\left\{\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]\right\}$ This is the eigenvector. - $\lambda=2$ $$ \left[\begin{array}{ccc} 8-\lambda & -8 & -2 \\ 4 & -\lambda-3 & -2 \\ 3 & -4 & 1-\lambda \end{array}\right]=\left[\begin{array}{ccc} 6 & -8 & -2 \\ 4 & -5 & -2 \\ 3 & -4 & -1 \end{array}\right] $$ The null space of this matrix is $\left\{\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]\right\}$ This is the eigenvector. - $\lambda=1$ $$ \left[\begin{array}{ccc} 8-\lambda & -8 & -2 \\ 4 & -\lambda-3 & -2 \\ 3 & -4 & 1-\lambda \end{array}\right]=\left[\begin{array}{ccc} 7 & -8 & -2 \\ 4 & -4 & -2 \\ 3 & -4 & 0 \end{array}\right] $$

The null space of this matrix is $\left\{\left[\begin{array}{c}2 \\ \frac{3}{2} \\ 1\end{array}\right]\right\}$

This is the eigenvector.

ANSWER

Eigenvalue: 3 A, multiplicity: 1 A , eigenvector: $\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]$ A.

Eigenvalue: 2 A, multiplicity: 1 A, eigenvector: $\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]$ A.

Eigenvalue: $1 \mathrm{~A}$, multiplicity: $1 \mathrm{~A}$, eigenvector: $\left[\begin{array}{c}2 \\ \frac{3}{2} \\ 1\end{array}\right]=\left[\begin{array}{c}2 \\ 1.5 \\ 1\end{array}\right] \mathrm{A} .$