Answer
See below
Work Step by Step
$\begin{bmatrix}
-1 & -2 & 3|0\\-1& 1 & 1 |0 \\-1 & -2& -1|1
\end{bmatrix} \approx \begin{bmatrix}
1 & 2 & -3|0\\-1& 1 & 1 |0 \\-1 & -2& -1|1
\end{bmatrix} \approx \begin{bmatrix}
1 & 2 & -3|0\\0& 3 & -2|0 \\0 & 0& -4|1
\end{bmatrix} \approx \begin{bmatrix}
1 & 2 & -3|0\\0& 1& -\frac{2}{3}|0 \\0 & 0& 1| -\frac{1}{4}
\end{bmatrix} $
$1.M_1(-1)\\
2.A_{12}(1),A_{13}(1)\\
3.M_3(\frac{-1}{4}), M_2(\frac{1}{3})$
From the third row,
$x_3=-\frac{1}{4}\\
x_2-\frac{2}{3}x_3=0 \rightarrow x_2=-\frac{1}{6}\\
x_1+2x_2-3x_3=0 \rightarrow x_1=-\frac{5}{12}$
Thus, $A^{-1}=\begin{bmatrix}
-\frac{5}{12}\\
-\frac{1}{6}\\
-\frac{1}{4}
\end{bmatrix}$
