Answer
See below
Work Step by Step
$\begin{bmatrix}
4 &2&3|12\\
1&-1&1|2\\
3&1&-1|2
\end{bmatrix} \approx\begin{bmatrix}
1&-1&1|2\\
4 &2&3|12\\
3&1&-1|2
\end{bmatrix} \approx \begin{bmatrix}
1&-1&1|2\\
0 & 6& -1|4\\
0&4&-4|-4
\end{bmatrix}\approx \begin{bmatrix}
1&-1&1|2\\
0 &4& -4|-4\\
0& 6&-1|4
\end{bmatrix} \approx \begin{bmatrix}
1&-1&1|2\\
0 &1& -1|-1\\
0& 6&-1|4
\end{bmatrix}\approx \begin{bmatrix}
1& 0& 0|1\\
0 &1& -1|-1\\
0& 0 & 5| 10
\end{bmatrix} \approx \begin{bmatrix}
1& 0& 0|1\\
0 &1& -1|-1\\
0& 0 & 1| 2
\end{bmatrix}\approx \begin{bmatrix}
1& 0& 0|1\\
0 &1& 0|1\\
0& 0 & 1| 2
\end{bmatrix}$
$1.P_{12}\\
2.A_{12}(-4),A_{13}(-3)\\
3.P_{23}\\
4.M_2(\frac{1}{4})\\
5.A_{21}(1),A_{23}(-6)\\
6.M_2(\frac{1}{5})\\
7.A_{32}(1)$
By taking real solutions, we have
$x^3_1=1\\
x_2^2=1\\
x_3=2$
Hence, $x_1=1\\
x_2=\pm 1\\
x_3=2$
The original system of equations has two solutions: $(1,1,2)$ and $(1,-1,2)$
We can see that there is no contradiction of Theorem 2.5.9, this system is not linear in $x_1,x_2$ and $x_3$
