Answer
See below
Work Step by Step
$\begin{bmatrix}
2 & -1 &3|14\\3&1&-2|-1 \\7&2&-3|3\\5&&-1&-2|5
\end{bmatrix}\approx \begin{bmatrix}
7&2&-3|3\\3&1&-2|-1 \\2 & -1 &3|14\\5&-1&-2|5
\end{bmatrix}\approx \begin{bmatrix}
7&2&-3|3\\ 0& \frac{1}{7}&-\frac{5}{7}|-\frac{16}{7} \\0 & -\frac{11}{7} &\frac{27}{7}| \frac{92}{7}\\ 0&-\frac{17}{7}&\frac{1}{7}| \frac{20}{7}
\end{bmatrix} \approx \begin{bmatrix}
7&2&-3|3\\ 0&-\frac{17}{7}&\frac{1}{7}| \frac{20}{7}\\0 & 0 &\frac{64}{7}| \frac{192}{17}\\ 0& 0 &-\frac{12}{17}|-\frac{36}{7}
\end{bmatrix}\approx \begin{bmatrix}
7&2&-3|3\\ 0&-\frac{17}{7}&\frac{1}{7}| \frac{20}{7}\\0 & 0 &\frac{64}{7}| \frac{192}{17}\\ 0& 0 & 0|0
\end{bmatrix}$
$1.P_{13}\\
2.A_{12}(-\frac{3}{7}),A_{13}(-\frac{2}{7}),A_{14}(-\frac{5}{7})\\
3.P_{24}\\
4.A_{23}(-\frac{11}{17}),A_{24}(\frac{1}{17})\\
5.A_{34}(\frac{3}{16})$
The unique solution $(2,-1,3)$
