Answer
See below
Work Step by Step
$\begin{bmatrix}
2 &-1 &4|0\\5& 1 & 2 |1 \\1 & -1& 3|0
\end{bmatrix} \approx \begin{bmatrix}
1 & -1& 3|0\\5& 1 & 2 |1\\2 &-1 &4|0
\end{bmatrix} \approx \begin{bmatrix}
1 & -1& 3|0\\0&6 &-13|1\\0& 1 & -2 |0
\end{bmatrix} \approx \begin{bmatrix}
1 & -1& 3|0\\0&1&-\frac{13}{6}|\frac{1}{6}\\0& 1 & -2 |0
\end{bmatrix} \approx \begin{bmatrix}
1 & -1& 3|0\\0&1&-\frac{13}{6}|\frac{1}{6}\\0& 0& \frac{1}{6}|-\frac{1}{6}
\end{bmatrix} \approx \begin{bmatrix}
1 & -1& 3|0\\0&1&-\frac{13}{6}|\frac{1}{6}\\0& 0& 1|-1
\end{bmatrix}$
$1.P_{13}\\
2.A_{12}(-5),A_{13}(-2)\\
3.M_2(\frac{1}{6}), A_{23}(-1)\\
4.M_3(6)$
From the third row,
$x_3=-1\\
x_2-\frac{13}{6}x_3=\frac{1}{6}\rightarrow x_2=-2\\
x_1-x_2+3x_3=0 \rightarrow x_1=1$
Thus, $A^{-1}=\begin{bmatrix}
1\\-2\\-1
\end{bmatrix}$
