$x + y + z = 6$

$y + 3z = 11$

$x + z = 2y{\text{ or }}x - 2y + z = 0$

This system can be written as $A X = B$, where

$A=\left[\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 3 \\
1 & 2 & 1
\end{array}\right], X=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] \text { and } B=\left[\begin{array}{c}
6 \\
11 \\
0
\end{array}\right]$

Here $|\mathrm{A}|=1(1+6)-(0-3)+(0-1)=9 \neq 0 .$ Now we find adj $\mathrm{A}$

$A_{11}=1(1+6)=7, \quad A_{12}=-(0-3)=3, \quad A_{13}=-1$
$A_{21}=-(1+2)=-3, \quad A_{22}=0, \quad A_{23}=-(-2-1)=3$
$\mathrm{A}_{31}=(3-1)=2, \quad \mathrm{A}_{32}=-(3-0)=-3, \quad \mathrm{A}_{33}=(1-0)=1$

Hence         $adj$ $\mathrm{A}=\left[\begin{array}{ccc}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]$

Thus     $\mathrm{A}^{-1}=\frac{1}{|\mathrm{A}|}$ adj $(\mathrm{A})=\frac{1}{9}\left[\begin{array}{ccc}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]$

since      $\mathrm{X}=\mathrm{A}^{-1} \mathrm{B}$

$X=\frac{1}{9}\left[\begin{array}{ccc}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]\left[\begin{array}{c}6 \\ 11 \\ 0\end{array}\right]$

$or$ $\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{9}\left[\begin{array}{l}42-33+0 \\ 18+0+0 \\ -6+33+0\end{array}\right]=\frac{1}{9}\left[\begin{array}{l}9 \\ 18 \\ 27\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$

Thus $\quad x=1, y=2, z=3$