1. **Problem statement:** Solve the matrix equation $$A \cdot X + B = X + C$$ for the matrix $$X$$, where $$A = \begin{pmatrix}4 & -2 \\ 0 & 4\end{pmatrix}, B = \begin{pmatrix}-4 & 3 \\ 4 & -4\end{pmatrix}, C = \begin{pmatrix}23 & 21 \\ -5 & 23\end{pmatrix}.$$ 2. **Rewrite the equation:** Move all terms involving $$X$$ to one side and constants to the other: $$A \cdot X - X = C - B.$$ 3. **Factor out $$X$$:** Since $$X$$ is a matrix, factor it as $$ (A - I) \cdot X = C - B,$$ where $$I$$ is the identity matrix $$\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}.$$ 4. **Calculate $$A - I$$:** $$A - I = \begin{pmatrix}4 & -2 \\ 0 & 4\end{pmatrix} - \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}3 & -2 \\ 0 & 3\end{pmatrix}.$$ 5. **Calculate $$C - B$$:** $$C - B = \begin{pmatrix}23 & 21 \\ -5 & 23\end{pmatrix} - \begin{pmatrix}-4 & 3 \\ 4 & -4\end{pmatrix} = \begin{pmatrix}23 - (-4) & 21 - 3 \\ -5 - 4 & 23 - (-4)\end{pmatrix} = \begin{pmatrix}27 & 18 \\ -9 & 27\end{pmatrix}.$$ 6. **Solve for $$X$$:** $$X = (A - I)^{-1} \cdot (C - B).$$ 7. **Find the inverse of $$A - I$$:** The matrix is $$M = \begin{pmatrix}3 & -2 \\ 0 & 3\end{pmatrix}.$$ The determinant is $$\det(M) = 3 \times 3 - 0 \times (-2) = 9.$$ The inverse is $$M^{-1} = \frac{1}{9} \begin{pmatrix}3 & 2 \\ 0 & 3\end{pmatrix} = \begin{pmatrix}\frac{1}{3} & \frac{2}{9} \\ 0 & \frac{1}{3}\end{pmatrix}.$$ 8. **Multiply $$M^{-1}$$ by $$C - B$$:** $$X = \begin{pmatrix}\frac{1}{3} & \frac{2}{9} \\ 0 & \frac{1}{3}\end{pmatrix} \cdot \begin{pmatrix}27 & 18 \\ -9 & 27\end{pmatrix}.$$ Calculate each element: - Top-left: $$\frac{1}{3} \times 27 + \frac{2}{9} \times (-9) = 9 - 2 = 7.$$ - Top-right: $$\frac{1}{3} \times 18 + \frac{2}{9} \times 27 = 6 + 6 = 12.$$ - Bottom-left: $$0 \times 27 + \frac{1}{3} \times (-9) = 0 - 3 = -3.$$ - Bottom-right: $$0 \times 18 + \frac{1}{3} \times 27 = 0 + 9 = 9.$$ 9. **Final solution:** $$X = \begin{pmatrix}7 & 12 \\ -3 & 9\end{pmatrix}.$$