Solution:

(i) Given that $(2 / x)-(3 / y)+(3 / z)=10$
$\begin{array}{l}
(1 / x)+(1 / y)+(1 / z)=10 \\
(3 / x)-(1 / y)+(2 / z)=13
\end{array}$
We can write the given system in matrix form as:
$\left[\begin{array}{lll}
5 & 3 & 1 \\
2 & 1 & 3 \\
1 & 2 & 4
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{l}
16 \\
19 \\
25
\end{array}\right] \operatorname{Or} \mathrm{A} \mathrm{X}=\mathrm{B}$
$\Delta-\left[\begin{array}{lll}5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{array}\right] \mathbf{x}-\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right], \mathrm{and} \mathrm{R}-\left[\begin{array}{c}16 \\ 19 \\ 25\end{array}\right]$
Now, $|A|=5\left|\begin{array}{ll}1 & 3 \\ 2 & 4\end{array}\right|-3\left|\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right|+1\left|\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right|$
$=5(4-6)-3(8-3)+1(4-2)$
$10=15+3$
$-22$
Therefore, the above system of equations has a unique solution, given by
$x=A^{-1} B$
Cofactors of $A$ are $C_{11}=(-1)^{1+1}(4-6)=-2$ $C_{21}=(-1)^{2+1}(12-2)=-10$ $C_{31}=(-1)^{3+1}(9-1)=8$ $C_{12}=(-1)^{1+2}(8-3)=-5$ $C_{22}=(-1)^{2+1} 20-1=19$ $C_{32}=(-1)^{3+1} 15-2=-13$ $C_{13}=(-1)^{1+2}(4-2)=2$ $C_{23}=(-1)^{2+1} 10-3=-7$ $C_{33}=(-1)^{3+1} 5-6=-1$
$\begin{array}{l}
\operatorname{Adj} A=\left[\begin{array}{ccc}
-2 & -5 & -3 \\
-10 & 19 & -7 \\
8 & -13 & -1
\end{array}\right]^{\mathrm{T}} \\
=\left[\begin{array}{ccc}
-2 & -10 & 8 \\
-5 & 19 & 13 \\
3 & -7 & -1
\end{array}\right] \\
\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}
\end{array}$
Now, $X=A^{-1} B=\frac{1}{-22}\left[\begin{array}{ccc}-2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1\end{array}\right]\left[\begin{array}{c}16 \\ 19 \\ 25\end{array}\right]$
$\begin{array}{l}
X=-\frac{1}{-22}\left[\begin{array}{c}
-32-190+200 \\
-80+361-325 \\
48-133-25
\end{array}\right] \\
X=\frac{1}{-22}\left[\begin{array}{c}
-22 \\
-44 \\
-110
\end{array}\right]
\end{array}$
$X=\left[\begin{array}{l}
1 \\
2 \\
5
\end{array}\right]$
As a result, $X=1, Y=2$ and $Z=5$

(ii) Given that $5 x+3 y+z=16$
$\begin{array}{l}
2 x+y+3 z=19 \\
x+2 y+4 z=25
\end{array}$
We can write the given system in matrix form as:
$\begin{array}{l}{\left[\begin{array}{ccc}3 & 4 & 2 \\ 0 & 2 & -3 \\ 1 & -2 & 6\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]=\left[\begin{array}{c}8 \\ 3 \\ -2\end{array}\right] \text { Or } \mathrm{A} \mathrm{X}=\mathrm{B}} \\ \mathrm{A}=\left[\begin{array}{ccc}3 & 4 & 2 \\ 0 & 2 & -3 \\ 1 & -2 & 6\end{array}\right], \mathrm{X}=\left[\begin{array}{c}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{c}8 \\ 3 \\ -2\end{array}\right] \\ \text { Now, }|\mathrm{A}|=3\left|\begin{array}{cc}2 & -3 \\ -2 & 6\end{array}\right|-4\left|\begin{array}{cc}0 & -3 \\ 1 & 6\end{array}\right|+2\left|\begin{array}{cc}0 & 2 \\ 1 & -2\end{array}\right| \\ =3(12-6)-4(0+3)+2(0-2) \\ = & 18-12-4 \\ =2\end{array}$
Therefore, the above system of equations has a unique solution, given by
$X=A^{-1} B$
Cofactors of A are
$\begin{array}{l}
C_{11}=(-1)^{1+1}(12-6)=6 \\
C_{21}=(-1)^{2+1}(24+4)=-28 \\
C_{31}=(-1)^{3+1}(-12-4)=-16 \\
C_{12}=(-1)^{1+2}(0+3)=-3 \\
C_{22}=(-1)^{2+1} 18-2=16 \\
C_{32}=(-1)^{3+1}-9-0=9 \\
C_{13}=(-1)^{1+2}(0-2)=-2 \\
C_{23}=(-1)^{2+1}(-6-4)=10 \\
C_{33}=(-1)^{3+1} 6-0=6
\end{array}$
$\text { Adj } A=\left[\begin{array}{ccc}
6 & -3 & 2 \\
-28 & 16 & 10 \\
-16 & -9 & 6
\end{array}\right]^{\mathrm{T}}$
$A^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}$
Now, $X=A^{-1} B=\stackrel{1}{2}\left[\begin{array}{ccc}6 & -28 & -16 \\ -3 & 16 & -9 \\ 2 & 10 & 6\end{array}\right]\left[\begin{array}{c}8 \\ 3 \\ -2\end{array}\right]$
$X=\frac{1}{2}\left[\begin{array}{c}
48-84+32 \\
-24+48-18 \\
-16+30-12
\end{array}\right]$
$X={ }^{\frac{1}{2}}\left[\begin{array}{c}
-4 \\
6 \\
2
\end{array}\right]$
$\mathrm{X}=\left[\begin{array}{c}
-2 \\
3 \\
1
\end{array}\right]$
As a result, $X=-2, Y=3$ and $Z=1$