Given $A = \begin{bmatrix} 5 & -6 & -6 \\ -1 & 4 & 2 \\ 3 & -6 & -4 \end{bmatrix}$

The characteristic equation of matrix A is

$$|A - \lambda I| = 0 \\ (-1)^3 \lambda^3 + (-)^2 s_{1} \lambda^2 + (-1) s_{2} \lambda + |A| = 0 ........(1)$$

where $$s_{1} = trace(A) = (5 + 4 - 4) = 5 \\ \therefore s_{1} = 5 \\ s_{2} = \begin{vmatrix} 4 & 2 \\ -6 & -4 \end{vmatrix} + \begin{vmatrix} 5 & -6 \\ 3 & -4 \end{vmatrix} + \begin{vmatrix} 5 & -6 \\ -1 & 4 \end{vmatrix} \\ = -4 -2 + 14 = 8 \\ \therefore s_{@} = 8 \\ |A| = 5(-4) + 6(-2) - 6(-6) = 4 \\ \therefore |A| = 4 $$

Equation (1) becomes,

Hence $$\phi(\lambda) = -\lambda^3 + 5\lambda^2 - 8\lambda + 4 = 0 \\ \phi(\lambda) = \lambda^3 - 5\lambda^2 + 8\lambda - 4 = 0 \\ \therefore \phi(\lambda) = (\lambda - 2) (\lambda^2 - 3\lambda + 2) = 0 \\ \therefore \phi(\lambda) = (\lambda - 2)(\lambda - 2) (\lambda - 1) = 0$$

$\therefore$ The eigen values are $\lambda$ = 1, 2, 2

To find the minimal polynomial of A:

Since each eigen value of A is also a root of its minimal polynomial, so if $f(\lambda)$ is the minimal polynomial of A, then both $(\lambda - 1)$ and $(\lambda - 2)$ are factors of $m(\lambda)$.

To check whether the polynomial $f(\lambda) = (\lambda - 1) (\lambda - 2) = \lambda^2 - 3\lambda + 2$ annihilates A or not.

i.e. to verify $A^2 - 3A + 2I = 0 \\ A^2 = \begin{bmatrix} 5 & - 6 & - 6 \\ -1 & 4 & 2 \\ 3 & -6 & -4 \end{bmatrix}\begin{bmatrix} 5 & - 6 & - 6 \\ -1 & 4 & 2 \\ 3 & -6 & -4 \end{bmatrix} \\ A^2 = \begin{bmatrix} 13 & -18 & -18 \\ -3 & 10 & 6 \\ 9 & -18 & -14 \end{bmatrix} \\ A^2 - 3A + 2I = \begin{bmatrix} 13 & -18 & -18 \\ -3 & 10 & 6 \\ 9 & -18 & -14 \end{bmatrix} - 3 \begin{bmatrix} 5 & - 6 & - 6 \\ -1 & 4 & 2 \\ 3 & -6 & -4 \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ A62 - 3A + 2I = O = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$\therefore f(\lambda)$ annihilates A. Thus $f(\lambda)$ is the monic polynomial of lowest degree which annihilates A. Hence $f(\lambda)$ is the minimal polynomial of A. Since its degree is less than the order of A(2 < 3). Therefore A is derogatory.