Solution:

(i) Given that $x+y-z=3$
$\begin{array}{l}
2 x+3 y+z=10 \\
3 x-y-7 z=1
\end{array}$
We can write the given system in matrix form as:
$\begin{array}{c}
{\left[\begin{array}{ccc}
1 & 1 & -1 \\
2 & 3 & 1 \\
3 & -1 & -7
\end{array}\right]\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{y} \\
\mathrm{Z}
\end{array}\right]=\left[\begin{array}{c}
3 \\
10 \\
1
\end{array}\right]_{\text {Or }} \mathrm{A} \mathrm{X}=\mathrm{B}} \\
\mathrm{A}=\left[\begin{array}{ccc}
1 & 1 & -1 \\
2 & 3 & 1 \\
3 & -1 & -7
\end{array}\right], \mathrm{X}=\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{c}
3 \\
10 \\
1
\end{array}\right]
\end{array}$
Now, $|\mathrm{A}|=1\left|\begin{array}{cc}3 & 1 \\ -1 & -7\end{array}\right|-1\left|\begin{array}{cc}2 & 1 \\ 3 & -7\end{array}\right|-1\left|\begin{array}{cc}2 & 3 \\ 3 & -1\end{array}\right|$
$\begin{array}{l}
=(-20)-1(-17)-1(11) \\
=-20+17+11=8
\end{array}$

Therefore, the above system of equations has a unique solution, given by
$X=A^{-1} B$
Cofactors of $\mathrm{A}$ are
$\begin{array}{l}
C_{11}=(-1)^{1+1}-21+1=-20 \\
C_{21}=(-1)^{2+1}-7-1=8 \\
C_{31}=(-1)^{3+1} 1+3=4 \\
C_{12}=(-1)^{1+2}-14-3=17 \\
C_{22}=(-1)^{2+1}-7+3=-4 \\
C_{32}=(-1)^{3+1} 1+2=-3 \\
C_{13}=(-1)^{1+2}-2-9=-11 \\
C_{23}=(-1)^{2+1}-1-3=4 \\
C_{33}=(-1)^{3+1} 3-2=1 \\
\qquad \operatorname{dj} A=\left[\begin{array}{ccc}
-20 & 17 & -11 \\
8 & -4 & 4 \\
4 & -3 & 1
\end{array}\right]^{T}
\end{array}$
$=\left[\begin{array}{ccc}
-20 & 8 & 4 \\
17 & -4 & -3 \\
-11 & 4 & 1
\end{array}\right]$
Now, $X=A^{-1} B=\frac{1}{8}\left[\begin{array}{ccc}-20 & 8 & 4 \\ 17 & -4 & -3 \\ -11 & 4 & 1\end{array}\right]\left[\begin{array}{c}3 \\ 10 \\ 1\end{array}\right]$
$\begin{array}{r}
x=\left[\begin{array}{c}
-60+80+4 \\
51-40-3 \\
-33+40+1
\end{array}\right] \\
X=\left[\begin{array}{c}
24 \\
8 \\
8
\end{array}\right]=\left[\begin{array}{l}
3 \\
1 \\
1
\end{array}\right]
\end{array}$
As a result, $X=3, Y=1$ and $Z=1$

(ii) Given that $x+y+z=3$
$\begin{array}{l}
2 x-y+z=-1 \\
2 x+y-3 z=-9
\end{array}$
We can write the given system in matrix form as:
$\left[\begin{array}{ccc}
1 & 1 & 1 \\
2 & -1 & 1 \\
2 & 1 & -3
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{Z}
\end{array}\right]=\left[\begin{array}{c}
3 \\
-1 \\
-9
\end{array}\right] \text { Or } \mathrm{A} \mathrm{X}=\mathrm{B}$
$A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & -1 & 1 \\ 2 & 1 & -3\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}3 \\ -1 \\ -9\end{array}\right]$
Now, $|A|=1\left|\begin{array}{cc}-1 & 1 \\ 1 & -3\end{array}\right|-1\left|\begin{array}{cc}2 & 1 \\ 2 & -3\end{array}\right|+1\left|\begin{array}{cc}2 & 1 \\ 2 & 1\end{array}\right|$
$\begin{array}{l}
=(3-1)-1(-6-2)+1(2+2) \\
=2+8+4 \\
=14
\end{array}$
Therefore, the above system of equations has a unique solution, given by
$X=A^{-1} B$
Cofactors of $A$ are
$\begin{array}{l}
C_{11}=(-1)^{1+1} 3-1=2 \\
C_{21}=(-1)^{2+1}-3-1=4 \\
C_{31}=(-1)^{3+1} 1+1=2 \\
C_{12}=(-1)^{1+2}-6-2=8
\end{array}$
$\begin{array}{l}
C_{22}=(-1)^{2+1}-3-2=-5 \\
C_{32}=(-1)^{3+1} 1-2=1 \\
C_{13}=(-1)^{1+2} 2+2=4 \\
C_{23}=(-1)^{2+1} 1-2=1 \\
C_{33}=(-1)^{3+1}-1-2=-3
\end{array}$
$\operatorname{Adj} A=\left[\begin{array}{ccc}2 & 8 & 4 \\ 4 & -5 & 1 \\ 2 & 1 & -3\end{array}\right]^{\mathrm{T}}$
$=\left[\begin{array}{ccc}
2 & 4 & 2 \\
8 & -5 & 1 \\
4 & 1 & -3
\end{array}\right]$
Now, $X=A^{-1} B={ }^{\frac{1}{14}}\left[\begin{array}{ccc}2 & 4 & 2 \\ 8 & -5 & 1 \\ 4 & 1 & -3\end{array}\right]\left[\begin{array}{c}3 \\ -1 \\ -9\end{array}\right]$
$X=\frac{1}{14}\left[\begin{array}{c}
-16 \\
20 \\
38
\end{array}\right]$
$X=\left[\frac{1}{7}\left[\begin{array}{c}
\frac{-8}{7} \\
\frac{10}{7} \\
\frac{19}{7}
\end{array}\right]\right.$
As a result, $X=\frac{-8}{7}, Y=\frac{10}{7}$ and $Z=\frac{19}{7}$