\( \left.\begin{array}{l}\text { Solve for } X \\ X=\left[\begin{array}{cc}8 & 6 \\ -3 & 2\end{array}\right] X+\left[\begin{array}{cc}-3 & 2 \\ -2 & 8\end{array}\right]=\left[\begin{array}{cc}-6 & 6 \\ 7 & -2\end{array}\right] X \\ \square\end{array}\right] \)

To solve for \( X \) in the equation \[ X = \begin{bmatrix} 8 & 6 \\ -3 & 2 \end{bmatrix} X + \begin{bmatrix} -3 & 2 \\ -2 & 8 \end{bmatrix} = \begin{bmatrix} -6 & 6 \\ 7 & -2 \end{bmatrix} X, \] we can rearrange the equation to isolate \( X \). First, let's rewrite the equation: \[ X - \begin{bmatrix} 8 & 6 \\ -3 & 2 \end{bmatrix} X - \begin{bmatrix} -6 & 6 \\ 7 & -2 \end{bmatrix} X = \begin{bmatrix} -3 & 2 \\ -2 & 8 \end{bmatrix}. \] This can be simplified to: \[ X - \left( \begin{bmatrix} 8 & 6 \\ -3 & 2 \end{bmatrix} + \begin{bmatrix} -6 & 6 \\ 7 & -2 \end{bmatrix} \right) X = \begin{bmatrix} -3 & 2 \\ -2 & 8 \end{bmatrix}. \] Next, we need to compute the sum of the two matrices: \[ \begin{bmatrix} 8 & 6 \\ -3 & 2 \end{bmatrix} + \begin{bmatrix} -6 & 6 \\ 7 & -2 \end{bmatrix} = \begin{bmatrix} 8 - 6 & 6 + 6 \\ -3 + 7 & 2 - 2 \end{bmatrix} = \begin{bmatrix} 2 & 12 \\ 4 & 0 \end{bmatrix}. \] Now, we can rewrite the equation as: \[ X - \begin{bmatrix} 2 & 12 \\ 4 & 0 \end{bmatrix} X = \begin{bmatrix} -3 & 2 \\ -2 & 8 \end{bmatrix}. \] Factoring out \( X \) gives us: \[ \left( I - \begin{bmatrix} 2 & 12 \\ 4 & 0 \end{bmatrix} \right) X = \begin{bmatrix} -3 & 2 \\ -2 & 8 \end{bmatrix}, \] where \( I \) is the identity matrix: \[ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \] Now we compute \( I - \begin{bmatrix} 2 & 12 \\ 4 & 0 \end{bmatrix} \): \[ I - \begin{bmatrix} 2 & 12 \\ 4 & 0 \end{bmatrix} = \begin{bmatrix} 1 - 2 & 0 - 12 \\ 0 - 4 & 1 - 0 \end{bmatrix} = \begin{bmatrix} -1 & -12 \\ -4 & 1 \end{bmatrix}. \] Thus, we have: \[ \begin{bmatrix} -1 & -12 \\ -4 & 1 \end{bmatrix} X = \begin{bmatrix} -3 & 2 \\ -2 & 8 \end{bmatrix}. \] To solve for \( X \), we need to multiply both sides by the inverse of \( \begin{bmatrix} -1 & -12 \\ -4 & 1 \end{bmatrix} \). First, we find the determinant of this matrix: \[ \text{det} = (-1)(1) - (-12)(-4) = -1 - 48 = -49. \] Now, the inverse of a \( 2 \times 2 \) matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by \[ \frac{1}{\text{det}} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}. \] Thus, the inverse of \( \begin{bmatrix} -1 & -12 \\ -4 & 1 \end{bmatrix} \) is: \[ \frac{1}{-49} \begin{bmatrix} 1 & 12 \\ 4 & -1 \end{bmatrix} = \begin{bmatrix} -\frac{1}{49} & -\frac