Solve the given system of equations, or else show that

Step 1
Consider the following system:
${\displaystyle x_1+2x_2-x_3=1}$
${\displaystyle 2x_1+x_2+x_3=1}$
${\displaystyle x_1-x_2+2x_3=1}$
Solve the system
Write augmented matrix
${\displaystyle \left[\begin{array}{cccc}1& 2& -1& 1\\ 2& 1& 1& 1\\ 1& -1& 2& 1\end{array}\right]}$
${\displaystyle {\overrightarrow{R_3-R_1}}^{R_2-2R_1}}$
${\displaystyle \left[\begin{array}{cccc}1& 2& -1& 1\\ 0& -3& 3& -1\\ 0& -3& 3& 0\end{array}\right]}$
$\underrightarrow{R_3-R_2}$
${\displaystyle \left[\begin{array}{cccc}1& 2& -1& 1\\ 0& -3& 3& -1\\ 0& 0& 0& 1\end{array}\right]}$
From the last row, we got ${\displaystyle 0=1}$ which is absqrt.
Hence the system has no solution
Step 2
Consider the following syste:
${\displaystyle x_1+2x_2-x_3=-2}$
${\displaystyle -2x_1-4x_2+2x_3=4}$
${\displaystyle 2x_1+4x_2-2x_3=-4}$
Solve the system
Write the augmented matrix
${\displaystyle \left[\begin{array}{cccc}1& 2& -1& -2\\ -2& -4& 2& 4\\ 2& 4& -2& -4\end{array}\right]}$
$$\begin{array}{c}{R}_2+2R_1\\ \to \\ R_3-2R_1\end{array}$$
${\displaystyle \left[\begin{array}{cccc}1& 2& -1& -2\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]}$
From this, we get ${\displaystyle x_1+2x_2-x_3=-2}$
Here, ${\displaystyle x_2,x_3}$ are free variables.
So choose ${\displaystyle x_2=s,x_3=t}$, where s, t are any parameters.
Then ${\displaystyle x_1=-2-2s+t}$
Hence the solution to the given system is
${\displaystyle \left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=\left(\begin{array}{c}-2-2s+t\\ s\\ t\end{array}\right)}$, where s, t are any parameters