Let $λ$ be Eigen value and x be corresponding Eigen vector of matrix A. $$\text {Characteristic Equation is } |A- λI|=0$$

$$\therefore \begin{vmatrix} -9-\lambda & 4 & 4 \\ -8 & 3-\lambda & 4 \\ -16 & 8 & 7-\lambda \\ \end{vmatrix}=0$$

On solving we get

$$\lambda^3-(-9+3+7)\lambda^2+(-11-1+5)\lambda-3=0$$

$\therefore \lambda^3-\lambda^2-5\lambda -3=0\\ \therefore \text { Eigen value $(\lambda)$ are } 3,-1,-1$

Case 1: $\lambda=-1\\ \therefore [A-\lambda I]x=0$

$$\therefore \begin{bmatrix} -8 & 4 & 4 \\ -8 & 4 & 4 \\ -16 & 8 & 8 \\ \end{bmatrix}.\begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix}=\begin{bmatrix}0\\ 0\\0\end{bmatrix}$$

$$R_2-R_1 ;R_3-2R_1;1/4R;$$

$$\therefore \begin{bmatrix} -2 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}.\begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix}=\begin{bmatrix}0\\ 0\\0\end{bmatrix}.... (1)$$

Number of unknown $(n)=3$

Rank (r) = Number of non-zero rows=1

Here, Algebraic multiplicity (A.M.)= No. of times $(x=-1)$ is repeated=2

Geometric multiplicity $(G.M.)=n-r=3-1=2$

$\therefore$ AM=GM for $\lambda=-1$

Expanding (1)

$$-2X_1+x_2+x_3=0..... (2)$$

$$\therefore -2t+s+x_3=0$$

$$\therefore s_3=2t-s$$

$\therefore $ Eigen vector $$\begin{bmatrix} x_1\\x_2\\x_3\end {bmatrix}= \begin{bmatrix} t\\s\\2t-s\end {bmatrix}=\begin{bmatrix} 1t+0s\\0t+1s\\2t-s\end {bmatrix}$$

$$=\begin{bmatrix} 1\\0\\2\end {bmatrix}t+ \begin{bmatrix} 0\\1\\-1\end {bmatrix}s$$

$\therefore $ Eigen Vector $x_1=[1\space \space0 \space \space2] $

$\therefore $ Eigen Vector $x_2=[0\space \space1 \space \space-1] $

Case 2: $\lambda=3\\ \therefore [A-\lambda I]X=0$

$$\therefore \begin{bmatrix} -12 & 4 & 4 \\ -8 & 0 & 4 \\ -16 & 4 & 4 \\ \end{bmatrix}.\begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix}=\begin{bmatrix}0\\ 0\\0\end{bmatrix}$$

$$R_2-2/3R_1 ;R_3-4/3R_1;1/4R_1;$$

$$\therefore \begin{bmatrix} -3 & 1 & 1 \\ 0 & -8/3 & 4/3 \\ 0 & 8/3 & 4/3 \\ \end{bmatrix}.\begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix}=\begin{bmatrix}0\\ 0\\0\end{bmatrix}....(3)$$

Here , $n=3$ and $r=1$

A.M.= No. of times $λ =3$ is repeated =1

G.M.= $n-r=3-2=1$

$\therefore$ A.M=G.M for $\lambda =3$

Expanding (3),

$$-3x_1+x_2+x_3=0.... (4)$$

$$-2x_2+x_3=0.....(5)$$

From (5) $x_3=2x_2$

From (4) $-3x_1+x_2+2x_2=0\\ \therefore 3x_2=3x_1\\ \therefore x_2=x_1$

$\therefore $ Eigen vector

$$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} =\begin{bmatrix} x_2 \\x_2 \\2x_2\\ \end{bmatrix} = \begin{bmatrix} 1\\1\\2\\ \end{bmatrix}x^2$$

$\therefore $ Eigen vector $x_3 =[1\space\space1\space\space2 ]$

Since, A.M. = G.M. for all Eigen values, matrix A is diagonalizable.

$$\therefore M^{-1}AM=D$$

So, the given matrix a is diagonalizable to diagonal matrix D by the transforming matrix M, where

$$D= \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 3 \\ \end{bmatrix}$$ and $$M= \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ -1 & 2 & 2 \\ \end{bmatrix}$$