Given $A = \begin{bmatrix} 7 & 4 & -1 \\ 4 & 7 & -1 \\ -4 & -4 & 4 \end{bmatrix}$

The characteristic equation of matrix A is

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

Where $$s_{1} = trace(A) = (7 + 7 + 4) = 18 \\ \therefore s_{1} = 18 \\ s_{2} = \begin{vmatrix} 7 & -1 \\ -4 & 4 \end{vmatrix} + \begin{vmatrix} 7 & - 1 \\ -4 & 4 \end{vmatrix} + \begin{vmatrix} 7 & 4 \\ 4 & 7 \end{vmatrix} \\ = 24 + 24 + 33 = 81 \\ \therefore s_{2} = 81 \\ |A| = 7(24) - 4(12) - 1(12) = 108 \\ \therefore |A| = 108$$

Equation (1) becomes,

Hence $$\phi(\lambda) = -\lambda^3 + 18\lambda^2 - 81 \lambda + 108 = 0 \\ \phi(\lambda) = \lambda^3 - 18\lambda^2 + 81 \lambda - 108 = 0 \\ \therefore \phi(\lambda) = (\lambda - 3) (\lambda^2 - 15\lambda + 36) = 0 \\ \therefore \phi(\lambda) = (\lambda - 3) (\lambda - 3) (\lambda - 12) = 0$$

$\therefore$ The eigen values are $\lambda = 3,3,12$

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 $(\lambda - 3)$ and $(\lambda - 12)$ are factors of $m(\lambda)$.

To check whether the polynomial $f(\lambda) = (\lambda - 3)(\lambda - 12) = \lambda^2 - 15\lambda + 36$ annihilates A or not.

$A^2 = \begin{bmatrix} 7 & 4 & -1 \\ 4 & 7 & -1 \\ -4 & -4 & 4 \end{bmatrix}\begin{bmatrix} 7 & 4 & -1 \\ 4 & 7 & -1 \\ -4 & -4 & 4 \end{bmatrix} \\ A^2 = \begin{bmatrix} 69 & 60 & -15 \\ 60 & 69 & -15 \\ -60 & -60 & 24 \end{bmatrix} \\ A^2 - 15A + 36I = \begin{bmatrix} 69 & 60 & -15 \\ 60 & 69 & -15 \\ -60 & -60 & 24 \end{bmatrix} - 15 \begin{bmatrix} 7 & 4 & -1 \\ 4 & 7 & -1 \\ -4 & -4 & 4 \end{bmatrix} + 36 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ A^2 - 15A + 36I = O = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

$\therefore f(\lambda)$ 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. Therefore A is derogatory.