The characteristic eq. of A is

$\begin{vmatrix} 8 - \lambda & -8 & -2 \\ 4 & -3 - \lambda & -2 \\ 3 & -4 & 1 - \lambda \end{vmatrix} = 0$

Simplifying we get

$\therefore (8 - \lambda)[(-3 - \lambda) (1 - \lambda) - 8] + [4(1 - \lambda) + 6] - 2[-16 + 9 + 3\lambda] = 0 \\ \therefore (8 - \lambda) [-3 - \lambda + 3\lambda + \lambda^2 - 8] + 8[4 - 4\lambda + 6] - 2[-7 + 3\lambda] = 0 \\ \therefore (8 - \lambda) [\lambda^2 + 2\lambda - 11] + 80 - 32\lambda - 6\lambda + 14 = 0 \\ \therefore -\lambda^3 - 2\lambda^3 + 11\lambda + 8\lambda^2 + 16\lambda - 88 + 904 - 38\lambda = 0 \\ \therefore -\lambda^3 + 6\lambda^2 - 11\lambda + 6 = 0 \\ \therefore \lambda = 3,2,1$

For $\lambda = 1 \hspace{0.25cm} [A - \lambda I]X = 0 \\ \begin{bmatrix} 7 & -8 & -2 \\ 4 & -4 & -2 \\ 3 & -4 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ 7x_{1} - 8x_{2} - 2x_{3} = 0 \\ 4x_{1} - 4x_{2} - 2x_{3} = 0$

By cramer's rule

$\frac{x_{1}}{\begin{vmatrix} -8 & -2 \\ -4 & -2 \end{vmatrix}} = \frac{-x_{2}}{\begin{vmatrix} 7 & -2 \\ 4 & -2 \end{vmatrix}} = \frac{x_{3}}{\begin{vmatrix} 7 & -8 \\ 4 & 4 \end{vmatrix}} \\ \therefore \frac{x_{1}}{8} = \frac{x_{2}}{6} = \frac{x_{3}}{4} \\ \therefore \frac{x_{1}}{4} = \frac{x_{2}}{3} = \frac{x_{3}}{2}$

Corresponding to Eigen value 1 the eigen vector is [4, 3, 2]

2) For $\lambda = 2 \hspace{0.25cm} [A - \lambda_{2} I]X = 0$ given

$\begin{bmatrix} 6 & -8 & -2 \\ 4 & -5 & -2 \\ 3 & -4 & -1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ 6x_{1} - 8x_{2} - 2x_{3} = 0 \\ 4x_{1} - 5x_{2} - 2x_{3} = 0$

By Cramer's rule

$\frac{x_{1}}{\begin{vmatrix} -8 & -2 \\ -5 & -2 \end{vmatrix}} = \frac{-x_{2}}{\begin{vmatrix} 6 & -2 \\ 4 & -2 \end{vmatrix}} = \frac{x_{3}}{\begin{vmatrix} 6 & - 8 \\ 4 & -5 \end{vmatrix}} \\ \therefore \frac{x_{1}}{3} = \frac{x_{2}}{2} = \frac{x_{3}}{1}$

Corresponding to Eigen value 2 the Eigen vector is [3, 2, 1]

3) For $\lambda = 3, \hspace{0.25cm} [A - \lambda_{3} I]X = 0$ given

$\begin{bmatrix} 5 & -8 & -2 \\ 4 & -6 & -2 \\ 3 & -4 & -2 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ 5x_{1} - 8x_{2} - 2x_{3} = 0 \\ 4x_{1} - 6x_{2} - 2x_{3} = 0$

By cramer's rule

$\frac{x_{1}}{\begin{vmatrix} -8 & -2 \\ -6 & -2 \end{vmatrix}} = \frac{-x_{2}}{\begin{vmatrix} 5 & -2 \\ 4 & -2 \end{vmatrix}} = \frac{x_{3}}{\begin{vmatrix} 5 & -8 \\ 4 & -6 \end{vmatrix}} \\ \therefore \frac{x_{1}}{2} = \frac{x_{2}}{1} = \frac{x_{3}}{1}$

Corresponding to Eigen value 3 the eigen vector is [2, 1, 1]

Since $ M^{-1}AM = 0$ the matrix $A = \begin{bmatrix} 8 & -8 & 2 \\ 4 & -3 & -2 \\ 3 & -4 & 2 \end{bmatrix}$ will be diagonal matrix.

$D = \begin{bmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ by the transforming matrix