Examine whether the following system of equation is consistent or unconsistent. If consistent find the complete solution.

$x+y+z=6,$

$x-y+z=2,$

$2x-y+3z=9$

Given system of equations can be written as $AX=B\Rightarrow \left[\begin{array}{ccc}1& 1& 1\\ 1& -1& 1\\ 2& -1& 3\end{array}\right]\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}6\\ 2\\ 9\end{array}\right]$

Augmented matrix $\left[AB\right]=\left[\begin{array}{cccc}1& 1& 1& 6\\ 1& -1& 1& 2\\ 2& -1& 3& 9\end{array}\right]$

$\left|A\right|=\left|\begin{array}{ccc}1& 1& 1\\ 1& -1& 1\\ 2& -1& 3\end{array}\right|=1(-3+1)-1(3-2)+1(-1+2)$

$=1(-2)-1\left(1\right)+1\left(1\right)$

$=-2-1+1=-2$

$\therefore \text{ Rank of }\left(A\right)=\mathrm{Rank}\text{ of }\left[AB\right]=3$

The system of equations are consistent and has a unique solution

$\left[AB\right]=\left[\begin{array}{cccc}1& 1& 1& 6\\ 1& -1& 1& 2\\ 2& -1& 3& 9\end{array}\right]$

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

$\left[\begin{array}{cccc}1& 1& 1& 6\\ 0& -2& 0& -4\\ 0& -3& 1& -3\end{array}\right]$

$R_2\to R_2/-2$

$\left[\begin{array}{cccc}1& 1& 1& 6\\ 0& 1& 0& 2\\ 0& -3& 1& -3\end{array}\right]$

$R_1\to R_1-R_2,R_3\to R_3+2R_2$

$\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\end{array}\right]R_1\to R_1-R_3$

$\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\end{array}\right]$

$\therefore x=1,y=2,z=3$ in the solution