Question

Solve system of linear equations, using matrix method.
$5x+2y=3$
$3x+2y=5$

Answer 1

Simplification of given data
Given: The system of equations is
$5x+2y=3$
$3x+2y=5$
Writing above equation as $AX=B$
$\left[\begin{array}{cc}5& 2\\ 3& 2\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}3\\ 5\end{array}\right]$
Hence, $A=\left[\begin{array}{cc}5& 2\\ 3& 2\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right]$ and $B=\left[\begin{array}{c}3\\ 5\end{array}\right]$
Calculating $|A|$
$|A|=\left|\begin{array}{cc}5& 2\\ 3& 2\end{array}\right|$
$=5\left(2\right)-3\left(2\right)=10-6=4$
Since $|A|\ne 0$
The system of equations is consistent and has a unique solution.
Calculate $A^{-1}$
$A=\left[\begin{array}{cc}5& 2\\ 3& 2\end{array}\right]$
$adjA=\left[\begin{array}{cc}2& -2\\ -3& 5\end{array}\right]$
Now,
$A^{-1}=\frac{1}{|A|}adjA$
$A^{-1}=\frac{1}{4}\left[\begin{array}{cc}2& -2\\ -3& 5\end{array}\right]$
Solve for the values of $x,y$
$AX=B$
$\Rightarrow X=A^{-1}B$
$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{4}\left[\begin{array}{cc}2& -2\\ -3& 5\end{array}\right]\left[\begin{array}{c}3\\ 5\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}2\left(3\right)+(-2)5\\ -3\left(3\right)+5\left(5\right)\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}6-10\\ -9+25\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}-4\\ 16\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-1\\ 4\end{array}\right]$
$\therefore x=-1$ and $y=4$