\(\begin{bmatrix}1&0&-1&-2&0\\0&1&2&3&0\end{bmatrix}\)

Since A is in reduced echelon form , we find the general solution

\(x_1-x_3-2x_4=0\)

\(x_2+2x_3+3x_4=0\)

Then

\((1) \Rightarrow x_1=x_3+2x_4\)

\((2) \Rightarrow x_2=-2x_3-3x_4\)

In vectors form , the general solution , we obtain

\(x= \begin{bmatrix}x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} =\begin{bmatrix}x_3+2x_4\\ -2x_3-3x_4\\ x_3\\ x_4 \end{bmatrix}=\begin{bmatrix}x_3\\ -2x_3\\ x_3\\ 0 \end{bmatrix}+ \begin{bmatrix}2x_4\\ -3x_4\\ 0\\ x_4 \end{bmatrix}=x_3\begin{bmatrix}1\\ -2\\1\\ 0 \end{bmatrix}+x_4\begin{bmatrix}2\\ -3\\ 0\\1 \end{bmatrix}\)

Hence ,

\(x=x_3\begin{bmatrix}1\\ -2\\1\\ 0 \end{bmatrix}+x_4\begin{bmatrix}2\\ -3\\ 0\\1 \end{bmatrix}\)