Solve the following system of equations by matrix method: (i) 5x + 7y + 2 = 0 4x + 6y + 3 = 0 (ii) 5x + 2y = 3 3x + 2y = 5

Solve the following system of equations by matrix method:
(i) 5x + 7y + 2 = 0
4x + 6y + 3 = 0
(ii) 5x + 2y = 3
3x + 2y = 5

Solve the following system of equations by matrix method:
(i) 5x + 7y + 2 = 0
4x + 6y + 3 = 0
(ii) 5x + 2y = 3
3x + 2y = 5

Solution:

(i) Given that $5 x+7 y+2=0$ and $4 x+6 y+3=0$
We can write the above system of equations as
$\left[\begin{array}{ll} 5 & 7 \\ 4 & 6 \end{array}\right]\left[\begin{array}{l} \mathrm{X} \\ \mathrm{Y} \end{array}\right]=\left[\begin{array}{l} -2 \\ -3 \end{array}\right] \text { Or } \mathrm{AX}=\mathrm{B}$
Where $A=\left[\begin{array}{ll}5 & 7 \\ 4 & 6\end{array}\right]_{B}=\left[\begin{array}{l}-2 \\ -3\end{array}\right]$ and $X=\left[\begin{array}{l}\mathrm{X} \\ \mathrm{Y}\end{array}\right]$ $|A|=30-28=2$
Therefore, the above system of equations has a unique solution, given by $X=A^{-1} B$
Suppose $C_{i j}$ be the cofactor of $a_{i j}$ in A, then
$\begin{array}{l} C_{11}=(-1)^{1+1} 6=6 \\ C_{12}=(-1)^{1+2} 4=-4 \\ C_{21}=(-1)^{2+1} 7=-7 \\ C_{22}=(-1)^{2+2} 5=5 \end{array}$
Also, adj $A=\left[\begin{array}{cc}6 & -4 \\ -7 & 5\end{array}\right]^{\mathrm{T}}$
$\begin{array}{l} =\left[\begin{array}{cc} 6 & -7 \\ -4 & 5 \end{array}\right] \\ A^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A} \\ A^{-1}=\frac{1}{2}\left[\begin{array}{cc} 6 & -7 \\ -4 & 5 \end{array}\right] \end{array}$
Now, $X=A^{-1} B$
$\begin{array}{l} {\left[\begin{array}{l} \mathrm{X} \\ \mathrm{Y} \end{array}\right]=\frac{1}{2}\left[\begin{array}{cc} 6 & -7 \\ -4 & 5 \end{array}\right]\left[\begin{array}{l} -2 \\ -3 \end{array}\right]} \\ {\left[\begin{array}{l} \mathrm{X} \\ \mathrm{Y} \end{array}\right]=\frac{1}{2}\left[\begin{array}{c} -12+21 \\ 8-15 \end{array}\right]} \\ {\left[\begin{array}{l} \mathrm{X} \\ \mathrm{Y} \end{array}\right]=\left[\begin{array}{c} \frac{9}{2} \\ \frac{-7}{2} \end{array}\right]} \end{array}$
As a result, $x=9 / 2$ and $y=-7 / 2$

(ii) Given that $5 x+2 y=3$
$3 x+2 y=5$
We can write the above system of equations as
$\left[\begin{array}{ll} 5 & 2 \\ 3 & 2 \end{array}\right]\left[\begin{array}{l} \mathrm{X} \\ \mathrm{Y} \end{array}\right]=\left[\begin{array}{l} 3 \\ 5 \end{array}\right]_{\text {Or } \mathrm{AX}=\mathrm{B}} \end{array}$
Where $A=\left[\begin{array}{ll}5 & 2 \\ 3 & 2\end{array}\right]_{B}=\left[\begin{array}{l}3 \\ 5\end{array}\right]$ and $X=\left[\begin{array}{l}\mathrm{X} \\ \mathrm{Y}\end{array}\right]$ $|\mathrm{A}|=10-6=4$
Therefore, the above system of equations has a unique solution, given by
$X=A^{-1} B$
Suppose $C_{i j}$ be the cofactor of $a_{i j}$ in $A$ , then
$\begin{array}{l} C_{11}=(-1)^{1+1} 2=2 \\ C_{12}=(-1)^{1+2} 3=-3 \\ C_{21}=(-1)^{2+1} 2=-2 \\ C_{22}=(-1)^{2+2} 2=5 \end{array}$
Also, adj $A=\left[\begin{array}{cc}2 & -3 \\ -2 & 5\end{array}\right]^{\mathrm{T}}$
$\begin{array}{l} =\left[\begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array}\right] \\ A^{-1}=\frac{1}{|A|} \operatorname{adj} A \\ A^{-1}=\frac{1}{4}\left[\begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array}\right] \end{array}$
Now, $X=A^{-1} B$
$\begin{array}{l} {\left[\begin{array}{l} \mathrm{X} \\ \mathrm{Y} \end{array}\right]=\frac{1}{4}\left[\begin{array}{cc} 2 & -2 \\ -3 & 5 \end{array}\right]\left[\begin{array}{l} 3 \\ 5 \end{array}\right]} \\ {\left[\begin{array}{l} \mathrm{X} \\ \mathrm{Y} \end{array}\right]=\frac{1}{4}\left[\begin{array}{c} 6-10 \\ -9+25 \end{array}\right]} \\ {\left[\begin{array}{l} \mathrm{X} \\ \mathrm{Y} \end{array}\right]=\frac{1}{4}\left[\begin{array}{l} -4 \\ 16 \end{array}\right]} \\ {\left[\begin{array}{l} \mathrm{X} \\ \mathrm{Y} \end{array}\right]=\left[\begin{array}{c} -1 \\ 4 \end{array}\right]} \end{array}$
As a result, $x=-1$ and $y=4$