Solution:

$A=\begin{bmatrix} 4 & -2 & 2 \\ 6 & -3 & 4 \\ 3 & -2 & 3 \end{bmatrix}$

|A|=2

$|A-\lambda I|=0$

$A=\left |\begin{matrix} 4-\lambda & -2 & 2 \\ 6 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right |= 0$

$\lambda^3-4\lambda^2+(-12-9+12+12-6+8)\lambda-2=0$

$\lambda^3-4\lambda^2+5\lambda-2=0$

$\lambda =2,1,1$

Eigen values are repeated therefore given matrix A is derogatory.

To find Minimal polynomial of A

$(\lambda-2)(\lambda -1) = 0$

$\lambda^2-3\lambda+2=0$ Annhilates A

$A^2-3A+2I$------(1)

$A=\begin{bmatrix} 4 & -2 & 2 \\ 6 & -3 & 4 \\ 3 & -2 & 3 \end{bmatrix}$

$A^2$=$A.A$=$\begin{bmatrix} 4 & -2 & 2 \\ 6 & -3 & 4 \\ 3 & -2 & 3 \end{bmatrix}$ . $\begin{bmatrix} 4 & -2 & 2 \\ 6 & -3 & 4 \\ 3 & -2 & 3 \end{bmatrix}$

$A^2=\begin{bmatrix} 10& -6 & 6 \\ 18 & -11 & 12 \\ 9 & -6 & 7\end{bmatrix}$

$I=\begin{bmatrix} 1& 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$

$I=\begin{bmatrix} 10& -6 & 6 \\ 18 & -11 & 12 \\ 9 & -6 & 7\end{bmatrix}$-3$\begin{bmatrix} 4& -2 & 2 \\ 6 & -3 & 4 \\ 3 & -2 & 3\end{bmatrix}$+2$\begin{bmatrix} 1& 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$

$\begin{bmatrix} 0& 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$

Hence A is derogatory.

Minimal polynomial is $(x^2-3x+2)$