tion
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the matrices $A=\begin{array}{cc}-3 & 1 \\ 4 & 1\end{array}, B=\begin{array}{cc}-3 & 3 \\ 2 & 0\end{array}$, and $C=\begin{array}{cc}-2 & 0 \\ -5 & 5\end{array}$, for the matrix $X$ in the equation $X(A-3 B)=C$.
Attempt 1 out of 2

Step 1 ：\(A - 3B = \left[\begin{array}{cc}-3 & 1 \\ 4 & 1\end{array}\right] - 3\left[\begin{array}{cc}-3 & 3 \\ 2 & 0\end{array}\right] = \left[\begin{array}{cc}-3+9 & 1-9 \\ 4-6 & 1-0\end{array}\right] = \left[\begin{array}{cc}6 & -8 \\ -2 & 1\end{array}\right]\)

Step 2 ：\(X = C / (A - 3B) = \left[\begin{array}{cc}-2 & 0 \\ -5 & 5\end{array}\right] / \left[\begin{array}{cc}6 & -8 \\ -2 & 1\end{array}\right]\)

Step 3 ：\((A - 3B)^{-1} = \frac{1}{6*1-(-8)*(-2)}\left[\begin{array}{cc}1 & 8 \\ 2 & 6\end{array}\right] = \frac{1}{-10}\left[\begin{array}{cc}1 & 8 \\ 2 & 6\end{array}\right] = \left[\begin{array}{cc}-0.1 & -0.8 \\ -0.2 & -0.6\end{array}\right]\)

Step 4 ：\(X = C * (A - 3B)^{-1} = \left[\begin{array}{cc}-2 & 0 \\ -5 & 5\end{array}\right] * \left[\begin{array}{cc}-0.1 & -0.8 \\ -0.2 & -0.6\end{array}\right] = \left[\begin{array}{cc}0.2 & 1.6 \\ -1 & -5\end{array}\right]\)

Step 5 ：\(\boxed{X = \left[\begin{array}{cc}0.2 & 1.6 \\ -1 & -5\end{array}\right]}\)