Let $$A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix}$$, $$\alpha > 0$$, such that $$\det(A) = 0$$ and $$\alpha + \beta = 1$$. If I denotes the $$2 \times 2$$ identity matrix, then the matrix $$(1 + A)^8$$ is:

We are given $$A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix}$$ with $$\alpha \gt 0$$, $$\det(A) = 0$$, and $$\alpha + \beta = 1$$.

From $$\det(A) = 0$$: $$\alpha\beta + 6 = 0$$, so $$\alpha\beta = -6$$.

From $$\alpha + \beta = 1$$: $$\beta = 1 - \alpha$$. Substituting: $$\alpha(1 - \alpha) = -6$$, giving $$\alpha - \alpha^2 = -6$$, so $$\alpha^2 - \alpha - 6 = 0$$.

Factoring: $$(\alpha - 3)(\alpha + 2) = 0$$. Since $$\alpha \gt 0$$, we get $$\alpha = 3$$ and $$\beta = -2$$.

So $$A = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix}$$ and $$I + A = \begin{bmatrix} 4 & -1 \\ 6 & -1 \end{bmatrix}$$.

Let $$B = I + A$$. We compute $$B^2$$: $$B^2 = \begin{bmatrix} 4 & -1 \\ 6 & -1 \end{bmatrix}\begin{bmatrix} 4 & -1 \\ 6 & -1 \end{bmatrix} = \begin{bmatrix} 16-6 & -4+1 \\ 24-6 & -6+1 \end{bmatrix} = \begin{bmatrix} 10 & -3 \\ 18 & -5 \end{bmatrix}$$.

Note that $$\text{tr}(B) = 3$$ and $$\det(B) = -4 + 6 = 2$$. By Cayley-Hamilton, $$B^2 = 3B - 2I$$, i.e., $$B^2 - 3B + 2I = 0$$.

We can express higher powers using the recurrence $$B^n = 3B^{n-1} - 2B^{n-2}$$. The characteristic equation $$\lambda^2 - 3\lambda + 2 = 0$$ has roots $$\lambda = 1$$ and $$\lambda = 2$$.

So $$B^n = \alpha_0 \cdot 1^n \cdot I + \alpha_1$$ ... Let us write $$B^n = c_1 \cdot 2^n I + c_2 \cdot 1^n I$$ — actually since the eigenvalues are 1 and 2, we write $$B^n = (2^n - 1)(B - I) + (2 \cdot 1^n - 1 \cdot 2^n)(- I) + ...$$ Let us use the direct formula.

Since $$B$$ has eigenvalues 1 and 2, we can write $$B = P \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} P^{-1}$$, so $$B^n = P \begin{bmatrix} 1 & 0 \\ 0 & 2^n \end{bmatrix} P^{-1}$$.

Using the formula $$B^n = \frac{2^n(B - I) - 1^n(B - 2I)}{2 - 1} = 2^n(B - I) - (B - 2I) = (2^n - 1)B - (2^n - 2)I$$.

For $$n = 8$$: $$B^8 = (256 - 1)B - (256 - 2)I = 255B - 254I$$.

$$B^8 = 255\begin{bmatrix} 4 & -1 \\ 6 & -1 \end{bmatrix} - 254\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1020 - 254 & -255 \\ 1530 & -255 - 254 \end{bmatrix} = \begin{bmatrix} 766 & -255 \\ 1530 & -509 \end{bmatrix}$$.

Hence, the correct answer is Option D.