Answer
See below
Work Step by Step
$\begin{bmatrix}
2 & -1 & -1|2\\4&3&-2|-1 \\1&4&1|4
\end{bmatrix}\approx \begin{bmatrix}
4&3&-2|-1 \\2 & -1 & -1|2\\1&4&1|4
\end{bmatrix}\approx \begin{bmatrix}
4&3&-2|-1 \\ 0& \frac{-5}{2}&0 |\frac{5}{2}\\0 & \frac{13}{4} &\frac{3}{2}| \frac{17}{4}
\end{bmatrix} \approx \begin{bmatrix}
4&3&-2|1\\0 & \frac{13}{4} &\frac{3}{2}| \frac{17}{4}\\0& \frac{-5}{2}&0 |\frac{5}{2}
\end{bmatrix}\approx\begin{bmatrix}
4&3&-2|-1 \\ 0 & \frac{13}{4} &\frac{3}{2}| \frac{17}{4}\\0&0&\frac{15}{13}|\frac{75}{13} \end{bmatrix}$
$1.P_{12}\\
2.A_{12}(-\frac{1}{2}),A_{13}(-\frac{1}{4})\\
3.P_{23}\\
4.A_{23}(\frac{10}{13})$
The unique solution $(3,-1,5)$
