The correct answer is $\boxed{\text{A}.}$
To find the eigenvalues of a matrix, we can use the following formula:
$$\lambda = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
where $a$, $b$, and $c$ are the coefficients of the characteristic polynomial of the matrix.
The characteristic polynomial of the matrix $A$ is given by:
$$p(x) = |xI – A| = x^3 – 20x^2 + 128x – 64$$
We can find the coefficients of $p(x)$ by expanding it:
$$p(x) = x^3 – 20x^2 + 128x – 64 = x(x^2 – 20x + 128) – 64 = x(x – 8)(x – 16)$$
Therefore, the coefficients of $p(x)$ are $a = 1$, $b = -20$, and $c = 128$.
Substituting these values into the formula for the eigenvalues, we get:
$$\lambda = \frac{-(-20) \pm \sqrt{(-20)^2 – 4 \cdot 1 \cdot 128}}{2 \cdot 1}$$
$$\lambda = \frac{20 \pm \sqrt{400 – 496}}{2}$$
$$\lambda = \frac{20 \pm \sqrt{-96}}{2}$$
$$\lambda = \frac{20 \pm 8i}{2}$$
$$\lambda = 10 \pm 4i$$
Therefore, the eigenvalues of the matrix $A$ are $\boxed{10 \pm 4i}$.
We can also find the eigenvalues of a matrix by using the power method. The power method is a iterative method that starts with an initial guess for an eigenvector and then repeatedly multiplies the matrix by the eigenvector to get closer to the actual eigenvector.
To use the power method, we start with an initial guess for an eigenvector, say $\mathbf{x}_0$. We then repeatedly multiply $\mathbf{x}_0$ by $A$ to get:
$$\mathbf{x}_1 = A\mathbf{x}_0$$
$$\mathbf{x}_2 = A\mathbf{x}_1 = A^2\mathbf{x}_0$$
$$\mathbf{x}_3 = A\mathbf{x}_2 = A^3\mathbf{x}_0$$
and so on.
The hope is that as we iterate, $\mathbf{x}_n$ will get closer and closer to an eigenvector of $A$.
To find the eigenvalue corresponding to the eigenvector $\mathbf{x}_n$, we can take the dot product of $\mathbf{x}_n$ with $A\mathbf{x}_n$:
$$\langle \mathbf{x}_n, A\mathbf{x}_n \rangle = \langle A\mathbf{x}_n, \mathbf{x}_n \rangle = \lambda_n \|\mathbf{x}_n\|^2$$
where $\lambda_n$ is the eigenvalue corresponding to $\mathbf{x}_n$.
We can use the power method to find the eigenvalues of the matrix $A$. We start with an initial guess for an eigenvector, say $\mathbf{x}_0 = (2, 1, 4)^T$. We then repeatedly multiply $\mathbf{x}_0$ by $A$ to get:
$$\mathbf{x}_1 = A\mathbf{x}_0 = \left[ \begin{array}{ccc} 4 & 8 & 4 \\ 8 & 16 & -4 \\ 4 & -4 & 15 \end{array} \right] \left[ \begin{array}{c} 2 \\ 1 \\ 4 \end{array} \right] = \left[ \begin{array}{c} 32 \\ 16 \\ 64 \end{array} \right]$$
$$\mathbf{x}_2 = A\mathbf{x}_1 = \left[ \begin{array}{ccc} 4 & 8 &