1. **State the problem:** Find matrix $X$ such that $$A \cdot X \cdot C + 2 \cdot B = C$$ where $$A = \begin{bmatrix} 2 & 1 \\ -1 & 3 \end{bmatrix}, B = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}, C = \begin{bmatrix} -1 & 3 \\ 2 & 1 \end{bmatrix}.$$ 2. **Rewrite the equation:** $$A X C = C - 2 B.$$ Calculate the right side: $$2 B = 2 \times \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ 2 & 2 \end{bmatrix}.$$ Then, $$C - 2 B = \begin{bmatrix} -1 & 3 \\ 2 & 1 \end{bmatrix} - \begin{bmatrix} 4 & -2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} -5 & 5 \\ 0 & -1 \end{bmatrix}.$$ 3. **Isolate $X$:** Multiply both sides on the left by $A^{-1}$ and on the right by $C^{-1}$: $$X = A^{-1} (C - 2 B) C^{-1}.$$ 4. **Find $A^{-1}$:** Determinant of $A$: $$\det(A) = 2 \times 3 - (-1) \times 1 = 6 + 1 = 7.$$ Inverse: $$A^{-1} = \frac{1}{7} \begin{bmatrix} 3 & -1 \\ 1 & 2 \end{bmatrix}.$$ 5. **Find $C^{-1}$:** Determinant of $C$: $$\det(C) = (-1) \times 1 - 3 \times 2 = -1 - 6 = -7.$$ Inverse: $$C^{-1} = \frac{1}{-7} \begin{bmatrix} 1 & -3 \\ -2 & -1 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} -1 & 3 \\ 2 & 1 \end{bmatrix}.$$ 6. **Calculate $A^{-1} (C - 2 B)$:** $$A^{-1} (C - 2 B) = \frac{1}{7} \begin{bmatrix} 3 & -1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} -5 & 5 \\ 0 & -1 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 3 \times (-5) + (-1) \times 0 & 3 \times 5 + (-1) \times (-1) \\ 1 \times (-5) + 2 \times 0 & 1 \times 5 + 2 \times (-1) \end{bmatrix} = \frac{1}{7} \begin{bmatrix} -15 & 16 \\ -5 & 3 \end{bmatrix}.$$ 7. **Calculate $X$ by multiplying the above with $C^{-1}$:** $$X = \frac{1}{7} \begin{bmatrix} -15 & 16 \\ -5 & 3 \end{bmatrix} \times \frac{1}{7} \begin{bmatrix} -1 & 3 \\ 2 & 1 \end{bmatrix} = \frac{1}{49} \begin{bmatrix} (-15)(-1) + 16 \times 2 & (-15)(3) + 16 \times 1 \\ (-5)(-1) + 3 \times 2 & (-5)(3) + 3 \times 1 \end{bmatrix} = \frac{1}{49} \begin{bmatrix} 15 + 32 & -45 + 16 \\ 5 + 6 & -15 + 3 \end{bmatrix} = \frac{1}{49} \begin{bmatrix} 47 & -29 \\ 11 & -12 \end{bmatrix}.$$ **Final answer:** $$X = \frac{1}{49} \begin{bmatrix} 47 & -29 \\ 11 & -12 \end{bmatrix}.$$