$\therefore\left[\begin{array}{rr}1 & 2 \\ -1 & 3\end{array}\right] X=\left[\begin{array}{ll}0 & 1 \\ 2 & 4\end{array}\right]$
By $R_2+R_1$, we get, $\left[\begin{array}{ll}1 & 2 \\ 0 & 5\end{array}\right] X=\left[\begin{array}{ll}0 & 1 \\ 2 & 5\end{array}\right]$
By $\left(\frac{1}{5}\right) R_2$, we get, $\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] X=\left[\begin{array}{ll}0 & 1 \\ \frac{2}{5} & 1\end{array}\right]$
By $R_1-2 R_2$, we get, $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] X=\left[\begin{array}{rr}-\frac{4}{5} & -1 \\ \frac{2}{5} & 1\end{array}\right]$
$\therefore X=\left[\begin{array}{rr}-\frac{4}{5} & -1 \\ \frac{2}{5} & 1\end{array}\right]$
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