Let A, Band C be three $$2\times 2$$ matrices with real entries such that $$B=(I+A)^{-1}$$ and A+C=1. If $$BC=\begin{bmatrix}1 & -5 \\-1 & 2 \end{bmatrix}$$ and $$CB\begin{bmatrix}x_{1}\\ x_{2} \end{bmatrix}=\begin{bmatrix}12\\-6 \end{bmatrix}$$, then $$x_{1}+x_{2}$$ is

$$B = (I + A)^{-1}$$, $$A + C = I$$ (so $$C = I - A$$), $$BC = \begin{bmatrix} 1 & -5 \\ -1 & 2 \end{bmatrix}$$, and $$CB\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 12 \\ -6 \end{bmatrix}$$.

$$BC = (I+A)^{-1}(I-A)$$

$$CB = (I-A)(I+A)^{-1}$$

Since any matrix commutes with its own polynomial, $$A$$ commutes with $$I + A$$. Therefore, $$A$$ commutes with $$(I + A)^{-1}$$ (the inverse of a matrix that commutes with $$A$$ also commutes with $$A$$). This means:

$$(I + A)^{-1}(I - A) = (I - A)(I + A)^{-1}$$

Hence $$CB = BC$$.

Since $$CB = BC = \begin{bmatrix} 1 & -5 \\ -1 & 2 \end{bmatrix}$$, the equation becomes:

$$\begin{bmatrix} 1 & -5 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 12 \\ -6 \end{bmatrix}$$

This gives the system:

$$x_1 - 5x_2 = 12 \quad \cdots (1)$$

$$-x_1 + 2x_2 = -6 \quad \cdots (2)$$

Adding equations (1) and (2): $$-3x_2 = 6 \implies x_2 = -2$$.

Substituting into (1): $$x_1 - 5(-2) = 12 \implies x_1 + 10 = 12 \implies x_1 = 2$$.

$$x_1 + x_2 = 2 + (-2) = 0$$

The correct answer is Option (1): 0.