Question

Solve system of linear equations, using matrix method.
$4x-3y=3$
$3x-5y=7$

Answer 1

Simplification of given data
Given: The system of equations is
$4x-3y=3$
$3x-5y=7$
Writing equation as $AX=B$
$\left[\begin{array}{cc}4& -3\\ 3& -5\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}3\\ 7\end{array}\right]$
Hence $A=\left[\begin{array}{cc}4& -3\\ 3& -5\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right]$ and $B=\left[\begin{array}{c}3\\ 7\end{array}\right]$
Calculating $|A|$
$|A|=\left|\begin{array}{cc}4& -3\\ 3& -5\end{array}\right|=4(-5)-3(-3)$
$=-20+9=-11$
Since $|A|\ne 0$
$\therefore$ System of equations is consistent & has a unique solution .
Calculate $A^{-1}$
$A=\left[\begin{array}{cc}4& -3\\ 3& -5\end{array}\right]$
$adjA=\left[\begin{array}{cc}-5& 3\\ -3& 4\end{array}\right]$
Now,
$A^{-1}=\frac{1}{|A|}adjA$
$A^{-1}=\frac{1}{-11}\left[\begin{array}{cc}-5& 3\\ -3& 4\end{array}\right]$
$=\frac{1}{11}\left[\begin{array}{cc}5& -3\\ 3& -4\end{array}\right]$
Solve for the values of $x,y$
$AX=B$
$X=A^{-1}B$
$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{cc}5& -3\\ 3& -4\end{array}\right]\left[\begin{array}{c}3\\ 7\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}5\left(3\right)+(-3)7\\ 3\left(3\right)+(-4)7\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}15-21\\ 9-28\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{(-6)}{11}\\ \frac{-19}{11}\end{array}\right]$
$\therefore x=\frac{-6}{11}$ and $y=\frac{-19}{11}$