The given system of equation can be written in the form of $A X=B$, where

$A = \left[ {\begin{array}{*{20}{c}}
  2&1&1 \\ 
  1&{ – 2}&{ – 1} \\ 
  0&3&{ – 5} 
\end{array}} \right],X = \left[ {\begin{array}{*{20}{l}}
  x \\ 
  y \\ 
  z 
\end{array}} \right]{\text{ and }}B = \left[ {\begin{array}{*{20}{c}}
  1 \\ 
  {\frac{3}{2}} \\ 
  9 
\end{array}} \right]$

Now,

$|A|=2(10+3)-1(-5-3)+0=2(13)-1(-8)=26+8=34 \neq 0$

Thus $A$ is non-singular. Therefore, its inverse exists.

Now,

$A_{11}=13, A_{12}=5, A_{13}=3$

$A_{21}=8, A_{22}=-10, A_{23}=-6$

$A_{31}=1, A_{32}=3, A_{33}=-5$

$\therefore A^{-1}=\frac{1}{|A|}(a d j A)=\frac{1}{34}\left[\begin{array}{ccc}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -16 & -5\end{array}\right]$

$\therefore X=A^{-1} B=\frac{1}{34}\left[\begin{array}{ccc}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5\end{array}\right]\left[\begin{array}{c}1 \\ \frac{3}{2} \\ 9\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{34}\left[\begin{array}{l}13+12+9 \\ 5-15+27 \\ 3-9-45\end{array}\right]$

$=\frac{1}{34}\left[\begin{array}{c}34 \\ 17 \\ -51\end{array}\right]=\left[\begin{array}{c}1 \\ \frac{1}{2} \\ -\frac{3}{2}\end{array}\right]$

Hence, $x=1, y=\frac{1}{2},$ and $z=-\frac{3}{2}$