Solution:

(i) Given that $3 x+4 y-5=0$
$x-y+3=0$
We can write the above system of equations as
$\left[\begin{array}{cc}
3 & 4 \\
1 & -1
\end{array}\right]\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{Y}
\end{array}\right]=\left[\begin{array}{c}
5 \\
-3
\end{array}\right] \text { Or } \mathrm{AX}=\mathrm{B}$
Where $A=\left[\begin{array}{cc}3 & 4 \\ 1 & -1\end{array}\right]_{B}=\left[\begin{array}{c}5 \\ -3\end{array}\right]$ and $X=\left[\begin{array}{l}X \\ Y\end{array}\right]$ $|A|=-3-4=-7$
Therefore, the above system of equations has a unique solution, given by
$X=A^{-1} B$
Suppose $\mathrm{C}_{i j}$ be the cofactor of $a_{i j}$ in $A$, then
$\begin{array}{l}
C_{11}=(-1)^{1+1}-1=-1 \\
C_{12}=(-1)^{1+2} 1=-1 \\
C_{21}=(-1)^{2+1} 4=-4 \\
C_{22}=(-1)^{2+2} 3=3
\end{array}$
Also, $\operatorname{adj} A=\left[\begin{array}{cc}-1 & -1 \\ -4 & 3\end{array}\right]^{\mathrm{T}}$
$\begin{array}{l}
=\left[\begin{array}{cc}
-1 & -4 \\
-1 & 3
\end{array}\right] \\
A^{-1}=\frac{1}{|A|} \operatorname{adj} A \\
A^{-1}=\frac{1}{7}\left[\begin{array}{cc}
-1 & -4 \\
-1 & 3
\end{array}\right]
\end{array}$
Now, $X=A^{-1} B$
$\begin{array}{l}
{\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{Y}
\end{array}\right]=\frac{1}{7}\left[\begin{array}{cc}
-1 & -4 \\
-1 & 3
\end{array}\right]\left[\begin{array}{c}
5 \\
-3
\end{array}\right]} \\
{\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{Y}
\end{array}\right]=\frac{1}{7}\left[\begin{array}{c}
-5+12 \\
-5-9
\end{array}\right]} \\
{\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{Y}
\end{array}\right]=\frac{1}{7}\left[\begin{array}{c}
7 \\
-14
\end{array}\right]}
\end{array}$
$\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{Y}
\end{array}\right]=\left[\begin{array}{c}
1 \\
-2
\end{array}\right]$
As a result, $X=1 Y=-2$,

(ii) Given that $3 x+y=19$
$3 x-y=23$
We can write the above system of equations as
$\left[\begin{array}{cc}
3 & 1 \\
3 & -1
\end{array}\right]\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{Y}
\end{array}\right]=\left[\begin{array}{l}
19 \\
23
\end{array}\right] \text { Or } \mathrm{AX}=\mathrm{B}$
Where $A=\left[\begin{array}{cc}3 & 1 \\ 3 & -1\end{array}\right]_{B}=\left[\begin{array}{l}19 \\ 23\end{array}\right]$ and $X=\left[\begin{array}{l}\mathrm{X} \\ \mathrm{Y}\end{array}\right]$
$|A|=-3-3=-6$
Therefore, the above system of equations has a unique solution, given by
$\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}$
Suppose $C_{i j}$ be the cofactor of $a_{i j}$ in $A$, then
$\begin{array}{l}
C_{11}=(-1)^{1+1}-1=-1 \\
C_{12}=(-1)^{1+2} 3=-3 \\
C_{21}=(-1)^{2+1} 1=-1 \\
C_{22}=(-1)^{2+2} 3=3
\end{array}$
Also, $\operatorname{adj} \mathrm{A}=\left[\begin{array}{cc}-1 & -3 \\ -1 & 3\end{array}\right]^{\mathrm{T}}$
$\begin{array}{l}
=\left[\begin{array}{cc}
-1 & -1 \\
-3 & 3
\end{array}\right] \\
\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \text { adj } \mathrm{A} \\
\mathrm{A}^{-1}=\frac{1}{-6}\left[\begin{array}{cc}
-1 & -1 \\
-3 & 3
\end{array}\right]
\end{array}$
Now, $X=A^{-1} B$
$\begin{array}{l}
{\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{Y}
\end{array}\right]=\frac{1}{-6}\left[\begin{array}{cc}
-1 & -1 \\
-3 & 3
\end{array}\right]\left[\begin{array}{l}
19 \\
23
\end{array}\right]} \\
{\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{Y}
\end{array}\right]=\frac{1}{-6}\left[\begin{array}{c}
-19-23 \\
-57+69
\end{array}\right]}
\end{array}$
$\begin{array}{l}
{\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{Y}
\end{array}\right]=\frac{1}{-6}\left[\begin{array}{c}
-42 \\
14
\end{array}\right]} \\
{\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{Y}
\end{array}\right]=\left[\begin{array}{c}
7 \\
-2
\end{array}\right]}
\end{array}$
As a result, $x=7$ and $y=-2$