Writing given equations in matrix form, AX=B,

$ \therefore \left[ \begin{array}{cccc} 1& -2& 1& -1\\ 1& 2& 0& 2\\ 0& 4& -1& 3 \end{array}\right] \left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right] \;= \; \left[ \begin{array}{c} 2 \\ 1 \\ -1 \end{array}\right] \\ \; \\ \; \\ \; \\ R_2 \rightarrow R_2 - R_1 \\ \; \\ \left[ \begin{array}{cccc} 1& -2& 1& -1\\ 0& 4& -1& 3\\ 0& 4& -1& 3 \end{array}\right] \left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right] \;= \; \left[ \begin{array}{c} 2 \\ -1 \\ -1 \end{array}\right] \\ \; \\ \; \\ \; \\ R_3 \rightarrow R_3 - R_2 \\ \; \\ \left[ \begin{array}{cccc} 1& -2& 1& -1\\ 0& 4& -1& 3\\ 0& 0& 0& 0 \end{array}\right] \left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right] \;= \; \left[ \begin{array}{c} 2 \\ -1 \\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ $

Augmented matrix

$ [A :B] or [A,B] \;=\; \left[ \begin{array}{cccc|c} 1& -2& 1& -1 & 2\\ 0& 4& -1& 3 & -1\\ 0& 0& 0& 0 & 0 \end{array}\right] \\ \; \\ $

$ \therefore $ 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 $ x_3=k_1 \; and \; 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 \Bigg( \dfrac{-1 +k_1-3k_2}{4} \Bigg) + k_1 - k_2 \;=\; 2 \\ \; \\ \; \\ \therefore x_1 + \dfrac{1}{2} - \dfrac{k_1}{2} + \dfrac{3k_2}{2} + k_1 - k_2 \;=\; 2 \\ \; \\ \; \\ \therefore x_1 + \dfrac{k_1}{2} + \dfrac{k_2}{2} - \dfrac{3}{2} \;=\; 0 \\ \; \\ \; \\ \therefore x_1 = \dfrac{3}{2} - \dfrac{k_1}{2} - \dfrac{k_2}{2} \\ \; \\ $

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