If the matrix $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1 \end{bmatrix}$$ satisfies the equation $$A^{20} + \alpha A^{19} + \beta A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ for some real numbers $$\alpha$$ and $$\beta$$, then $$\beta - \alpha$$ is equal to ______.

We find the eigenvalues of $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1 \end{bmatrix}$$. The characteristic polynomial is $$\det(A - \lambda I) = (2 - \lambda)[(1 - \lambda)(-1 - \lambda)] = -(2 - \lambda)(\lambda^2 - 1)$$, giving eigenvalues $$\lambda_1 = 1$$, $$\lambda_2 = 2$$, and $$\lambda_3 = -1$$.

Since all three eigenvalues are distinct, $$A$$ is diagonalizable, and for any polynomial $$f$$, the eigenvalues of $$f(A)$$ are $$f(\lambda_1)$$, $$f(\lambda_2)$$, and $$f(\lambda_3)$$.

Let $$f(\lambda) = \lambda^{20} + \alpha \lambda^{19} + \beta \lambda$$. The right-hand side matrix has eigenvalues $$1, 4, 1$$. We match each eigenvalue of $$f(A)$$ to the corresponding eigenvalue of the RHS.

From $$f(1) = 1$$: we get $$1 + \alpha + \beta = 1$$, so $$\alpha + \beta = 0$$, meaning $$\beta = -\alpha$$.

From $$f(-1) = 1$$: we get $$1 - \alpha - \beta = 1$$, which gives $$\alpha + \beta = 0$$. This is consistent with the equation above.

From $$f(2) = 4$$: we get $$2^{20} + \alpha \cdot 2^{19} + 2\beta = 4$$. Substituting $$\beta = -\alpha$$ gives $$2^{20} + 2^{19}\alpha - 2\alpha = 4$$, so $$\alpha(2^{19} - 2) = 4 - 2^{20}$$.

Factoring the right side: $$4 - 2^{20} = -4(2^{18} - 1)$$, and the left coefficient: $$2^{19} - 2 = 2(2^{18} - 1)$$. Dividing gives $$\alpha = \frac{-4(2^{18} - 1)}{2(2^{18} - 1)} = -2$$.

Therefore $$\beta = -\alpha = 2$$. We verify: $$f(2) = 2^{20} - 2 \cdot 2^{19} + 2 \cdot 2 = 2^{20} - 2^{20} + 4 = 4$$, which confirms the result.

The answer is $$\beta - \alpha = 2 - (-2) = 4$$.