$\begin{aligned} & \text { By }(1)-(2), 3 Y=A-B, \therefore Y=\frac{1}{3}(A-B) \\ & \therefore Y=\frac{1}{3}\left\{\left[\begin{array}{cc}2 & -1 \\ 1 & 3 \\ -3 & -2\end{array}\right]-\left[\begin{array}{cc}-2 & 1 \\ 3 & -1 \\ 4 & -2\end{array}\right]\right\}=\frac{1}{3}\left[\begin{array}{cc}4 & -2 \\ -2 & 4 \\ -7 & 0\end{array}\right] \\ & =\left[\begin{array}{cc}\frac{4}{3} & -\frac{2}{3} \\ -\frac{2}{3} & \frac{4}{3} \\ -\frac{7}{3} & 0\end{array}\right] \\ & \text { From (1) } \mathrm{X}+\mathrm{Y}=\mathrm{A}, \quad \therefore \mathrm{X}=\mathrm{A}-\mathrm{Y} \text {, } \\ & \therefore X=\left[\begin{array}{cc}2 & -1 \\ 1 & 3 \\ -3 & -2\end{array}\right]-\left[\begin{array}{cc}\frac{4}{3} & -\frac{2}{3} \\ -\frac{2}{3} & \frac{4}{3} \\ -\frac{7}{3} & 0\end{array}\right] \\ & X=\left[\begin{array}{cc}2-\frac{4}{3} & -1+\frac{2}{3} \\ 1+\frac{2}{3} & 3-\frac{4}{3} \\ -3+\frac{7}{3} & -2+0\end{array}\right]=\left[\begin{array}{cc}\frac{2}{3} & -\frac{1}{3} \\ \frac{5}{3} & \frac{5}{3} \\ -\frac{2}{3} & -2\end{array}\right] \\ & \end{aligned}$
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