We have

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

By  $ R2-R1 \begin{bmatrix}1 &2 &3 &2 \\0 &4 &-1&3 \\0 &4 &-1 &3 \end{bmatrix} \begin{bmatrix}x_1\\x_2 \\x_3 \\x_4 \end{bmatrix} = \begin{bmatrix}2\\-1 \\-1 \end{bmatrix} $

By $ by R3-R2 \begin{bmatrix}1 &2 &3 &2 \\0 &4 &-1 &3 \\0 &0 &0 &0 \end{bmatrix} \begin{bmatrix} x_1\\x_2 \\x_3\\x_4\end{bmatrix} = \begin{bmatrix} 2\\-1\\0\end{bmatrix} \ \cdots (1)$

The rank of coefficient matrix  $\begin{bmatrix} 1 &2 &3 &2 \\0 &4 &-1 &3 \\0 &0 &0 &0 \end{bmatrix}$ is 2

and rank of augmented matrix $\begin{bmatrix} 1 &2 &3 &2 &2 \\0 &4 &-1 &3 &-1 \\0 &0 &0 &0 &0 \end{bmatrix}$ is also 2

Hence the equations are consistent.

But r=2<4.

Hence, the number of parameters =4-2=2

Hence equation have study doubly infinite solutions.

Now, from (1)

$x_1+2x_2+3x_3 +2x_4=2,$

$4x_2 - x_3 + 3x_4 = -1$

Put $x_3 = t_2$ and $x_4 = t_2$

Hence, $4x_2 = t_1 - 3t_2$ and $x_1 = 2 \big [ (t_1-3t_2)/4 \big ]+3t_1 + 2t_2$

Hence,

$x_2 = \dfrac {(t_1 - 3t_2)}{4}$ and

$x_1 = \left [ \dfrac {(t_1 - 3t_2)}{2} \right ] + 3t_1 + 2t_2$