Solution:
(i) Given that $3 x+4 y+2 z=8$
$\begin{array}{l}
2 y-3 z=3 \\
x-2 y+6 z=-2
\end{array}$
We can write the given system in matrix form as:
$\left[\begin{array}{rrr}
2 & 1 & 1 \\
1 & 3 & -1 \\
3 & 1 & -2
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{Z}
\end{array}\right]=\left[\begin{array}{l}
2 \\
5 \\
6
\end{array}\right] \text { Or } \mathrm{A} \mathrm{X}=\mathrm{B}$
$A=\left[\begin{array}{ccc}2 & 1 & 1 \\ 1 & 3 & -1 \\ 3 & 1 & -2\end{array}\right], X=\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{l}2 \\ 5 \\ 6\end{array}\right]$
Now, $|\mathrm{A}|=2^{\left|\begin{array}{ll}3 & -1 \\ 1 & -2\end{array}\right|-1\left|\begin{array}{ll}1 & -1 \\ 3 & -2\end{array}\right|+1\left|\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right|}$
$\begin{array}{l}
=2(-6+1)-1(-2+3)+1(1-9) \\
=-10-1-8 \\
=-19
\end{array}$
Therefore, the above system of equations has a unique solution, given by
$X=A^{-1} B$
Cofactors of A are
$\begin{array}{l}
C_{11}=(-1)^{1+1}-6+1=-5 \\
C_{21}=(-1)^{2+1}(24+4)=-28 \\
C_{31}=(-1)^{3+1}-1-3=-4 \\
C_{12}=(-1)^{1+2}-2+3=-1 \\
C_{22}=(-1)^{2+1}-4-3=-7 \\
C_{32}=(-1)^{3+1}-2-1=3 \\
C_{13}=(-1)^{1+2} 1-9=-8 \\
C_{23}=(-1)^{2+1} 2-3=-1 \\
C_{33}=(-1)^{3+1} 6-1=5
\end{array}$
$\begin{array}{l}
\text { Adj } A=\left[\begin{array}{ccc}
-5 & -1 & -8 \\
3 & -7 & 1 \\
-4 & 3 & 5
\end{array}\right]^{T} \\
=\left[\begin{array}{ccc}
-5 & 3 & -4 \\
-1 & -7 & 3 \\
-8 & 1 & 5
\end{array}\right]
\end{array}$
$\mathrm{A}^{-1}=\stackrel{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}$
Now, $X=A^{-1} B=\frac{1}{-19}\left[\begin{array}{ccc}-5 & 3 & -4 \\ -1 & -7 & 3 \\ -8 & 1 & 5\end{array}\right]\left[\begin{array}{l}2 \\ 5 \\ 6\end{array}\right]$
$\begin{array}{l}
X=^{\frac{1}{-19}}\left[\begin{array}{c}
-10+15-24 \\
-2-35+18 \\
-16+5+30
\end{array}\right] \\
X=^{\frac{1}{-19}}\left[\begin{array}{c}
-19 \\
-19 \\
19
\end{array}\right] \\
X=\left[\begin{array}{c}
1 \\
1 \\
-1
\end{array}\right]
\end{array}$
As a result, $X=1, Y=1$ and $Z=-1$
(ii) Given that $2 x+y+z=2$
$\begin{array}{l}
x+3 y-z=5 \\
3 x+y-2 z=6
\end{array}$
We can write the given system in matrix form as:
$\left[\begin{array}{ccc}
2 & 1 & 1 \\
1 & 3 & -1 \\
3 & 1 & -2
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{l}
2 \\
5 \\
6
\end{array}\right]_{\text {Or } A \mathrm{X}=\mathrm{B}}$
$A=\left[\begin{array}{ccc}2 & 1 & 1 \\ 1 & 3 & -1 \\ 3 & 1 & -2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}2 \\ 5 \\ 6\end{array}\right]$
Now, $|A|=2\left|\begin{array}{ll}3 & -1 \\ 1 & -2\end{array}\right|-1\left|\begin{array}{ll}1 & -1 \\ 3 & -2\end{array}\right|+1\left|\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right|$
$\begin{array}{l}
=2(-6+1)-1(-2+3)+1(1-9) \\
=-10-1-8 \\
=-19
\end{array}$
Therefore, the above system of equations has a unique solution, given by
$X=A^{-1} B$
Cofactors of $A$ are
$\begin{array}{l}
C_{11}=(-1)^{1+1}-6+1=-5 \\
C_{21}=(-1)^{2+1}(24+4)=-28 \\
C_{31}=(-1)^{3+1}-1-3=-4 \\
C_{12}=(-1)^{1+2}-2+3=-1 \\
C_{22}=(-1)^{2+1}-4-3=-7 \\
C_{32}=(-1)^{3+1}-2-1=3 \\
C_{13}=(-1)^{1+2} 1-9=-8 \\
C_{23}=(-1)^{2+1} 2-3=-1 \\
C_{33}=(-1)^{3+1} 6-1=5
\end{array}$
$\begin{array}{l}
\text { Adj } A=\left[\begin{array}{ccc}
-5 & -1 & -8 \\
3 & -7 & 1 \\
-4 & 3 & 5
\end{array}\right]^{T} \\
=\left[\begin{array}{ccc}
-5 & 3 & -4 \\
-1 & -7 & 3 \\
-8 & 1 & 5
\end{array}\right] \\
A^{-1}=\frac{1}{|A|} a d j A
\end{array}$
Now, $X=A^{-1} B=\frac{1}{-19}\left[\begin{array}{ccc}-5 & 3 & -4 \\ -1 & -7 & 3 \\ -8 & 1 & 5\end{array}\right]\left[\begin{array}{l}2 \\ 5 \\ 6\end{array}\right]$
$\begin{array}{l}
X=\frac{1}{-19}\left[\begin{array}{c}
-10+15-24 \\
-2-35+18 \\
-16+5+30
\end{array}\right] \\
X=\frac{1}{-19}\left[\begin{array}{c}
-19 \\
-19 \\
19
\end{array}\right]
\end{array}$
$\mathrm{X}=\left[\begin{array}{c}
1 \\
1 \\
-1
\end{array}\right]$
As a result, $X=1, Y=1$ and $Z=-1$