Solve the following system of equations by Gauss-Jordan method.

$x+y+z=3,$

$2x+2y-z=3,$

$x+y-z=1$

The matrix equation is $Ax=D$

where $A=\left[\begin{array}{ccc}1& 1& 1\\ 2& 2& -1\\ 1& 1& -1\end{array}\right]x=\left[\begin{array}{l}x\\ y\\ z\end{array}\right]$ and $D=\left[\begin{array}{l}3\\ 3\\ 1\end{array}\right]$

The augmented of matrix $\left[\begin{array}{l}A\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}D\end{array}\right]=\left[\begin{array}{cccc}1& 1& 1& 3\\ 2& 2& -1& 3\\ 1& 1& -1& 1\end{array}\right]$

an applying $R_2\to R_2-2R_1,$ $R_3\to R_3-R_1$

$~\left[\begin{array}{cccc}1& 1& 1& 3\\ 0& 0& -3& -3\\ 0& 0& -2& -2\end{array}\right]$

an applying $R_3\to R_2-\left(\frac{2}{3}\right)R_2$

we get $~\left[\begin{array}{cccc}1& 1& 1& 3\\ 0& 0& -3& -3\\ 0& 0& 0& 0\end{array}\right]$

Hence the following is the system of equations equilent to the given system of equations $x+y+z=3, -3z=-3$

Hence $z=1, x+y=2$

The solution set is $x=k,y=2-k,z=1$where $k\in R$