$x_1-2x_2+x_3-x_4 =2 ; x_1+2x_2+2x_4=1; 4x_2-x_3+3x_4=-1$

Writing given equations in matrix form, AX=B,

$\therefore \begin{vmatrix} 1&-2&1&-1\\ 1&2&0&2 \\ 0&4&-1&3\\ \end{vmatrix}\begin{vmatrix} x_1\\x_2\\x_3\\x_4\\ \end{vmatrix}=\begin{vmatrix} 2\\1\\-1\\ \end{vmatrix} $

$R_2\rightarrow R_2-R_1$

$\therefore \begin{vmatrix} 1&-2&1&-1\\ 1&4&-1&3 \\ 0&4&-1&3\\ \end{vmatrix}\begin{vmatrix} x_1\\x_2\\x_3\\x_4\\ \end{vmatrix}=\begin{vmatrix} 2\\1\\-1\\ \end{vmatrix} $

$R_3\rightarrow R_3-R_2 $

$\therefore \begin{vmatrix} 1&-2&1&-1\\ 1&4&-1&3 \\ 0&0&0&0\\ \end{vmatrix}\begin{vmatrix} x_1\\x_2\\x_3\\x_4\\ \end{vmatrix}=\begin{vmatrix} 2\\-1\\0\\ \end{vmatrix} $

Augmented matrix

$[A|B]$ or $[A,B] =\begin{vmatrix} 1&-2&1&-1\\ 1&4&-1&3 \\ 0&0&0&0\\ \end{vmatrix}\begin{vmatrix} 2\\-1\\0\\ \end{vmatrix} $

$\because $Rank of A and rank of augmented matrix $[A|B]$ are same i.e. 2 hence equations are consistent.

Rank of A=2

But rank of A< no. of unknowns, i.e. 4 hence the equations have infinite solutions.

$x_1-2x_2+x_3-x_4 =2\\ 4x_2-x_3+3x_4=-1 \\ Put \space\space x_3=k_1 \space\space and \space\space x_4=k_2 \\ \therefore 4x_2-k_1+3k_2=-1 \therefore 4x_2=-1+k_1-3k_2\\ \therefore x_2= \dfrac {-1+k_1-3k_2}4 \\ \therefore x_1-2 (\dfrac {-1+k_1-3k_2}4)+k_1 -k_2=2 \\ \therefore x_1 +\dfrac 12 -\dfrac {k_1}2 + \dfrac {3k_2}2+ k_1-k_2= 2 \\ \therefore x_1 +\dfrac 12 k_1 +\dfrac 12 k_2 -\dfrac 32 =0 \\ \therefore x_1= \dfrac 32 -\dfrac 12k_1 -\dfrac 12 k_2 $

Substituting different values of $k_1$ and $k_2$ gives $x_1,x_2,x_3,x_4$