I have the following matrix:
$$A=\begin{pmatrix}1&2&1&6\\-2&3&5&6\\ 3&6&3&6\\-4&-4&0&23\end{pmatrix}$$
The question on my problem sheet asks me to find all solutions to $Ax=(2,0,-6,31)^T$ by using gaussian elimination.
(I don't know how to write augmented matrices but my "$b$" will be represented by the red column)
$$\begin{pmatrix}1&2&1&6&\color{red}{2}\\-2&3&5&6 &\color{red}{0}\\ 3&6&3&6& \color{red}{-6}\\-4&-4&0&23& \color{red}{31}\end{pmatrix} \stackrel{R_2+2R_1, R_3-3R_1, R4+4R_1}{\longrightarrow} \begin{pmatrix}1&2&1&6&\color{red}{2}\\0&7&7&18 &\color{red}{4}\\ 0&0&0&-12& \color{red}{-12}\\0&4&4&47& \color{red}{39}\end{pmatrix} \stackrel{R_3 \iff R_4,R_2/7, R_3/4 }{\longrightarrow} \begin{pmatrix}1&2&1&6&\color{red}{2}\\0&1&1&\frac{18}{7} &\color{red}{\frac{4}{17}}\\ 0&1&1&\frac{47}{4}& \color{red}{\frac{39}{4}}\\0&0&0&1& \color{red}{1}\end{pmatrix} $$
$$ \stackrel{R_3-R_2}{\longrightarrow} \begin{pmatrix}1&2&1&6&\color{red}{2}\\0&1&1&\frac{18}{7} &\color{red}{\frac{4}{17}}\\ 0&0&0&\frac{257}{28}& \color{red}{\frac{647}{68}}\\0&0&0&1& \color{red}{1}\end{pmatrix} \stackrel{R_3-\frac{257}{28}R_4}{\longrightarrow} \begin{pmatrix}1&2&1&6&\color{red}{2}\\0&1&1&\frac{18}{7} &\color{red}{\frac{4}{17}}\\ 0&0&0&0& \color{red}{\frac{40}{119}}\\0&0&0&1& \color{red}{1}\end{pmatrix}$$
Since $$0x_1+0x_2+0x_3+0x_4\not=\frac{40}{119}$$
am I right in saying this system of equation has no solutions? However, the answer given in the problem sheet is: $$x_1=x_3$$
$$x_2=-2-x_3$$
$$x_4=1$$
$$x_3=\text{free}$$
What am I doing wrong?