Question

Solve system of linear equations, using matrix method.
$2x-y=-2$
$3x+4y=3$

Answer 1

Simplification of given data
Given: The system of equations is
$2x-y=-2$
$3x+4y=3$
Write above equation as $AX=B$
$\left[\begin{array}{cc}2& -1\\ 3& 4\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-2\\ 3\end{array}\right]$
Hence, $A=\left[\begin{array}{cc}2& -1\\ 3& 4\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right]$ and $B=\left[\begin{array}{c}-2\\ 3\end{array}\right]$
$A=\left[\begin{array}{cc}2& -1\\ 3& 4\end{array}\right]=2\left(4\right)-(-1)\left(3\right)=8+3=11$
$\therefore |A|\ne 0$
Thus, system of equations is consistent and has a unique solution
Calculate $A^{-1}$
$A=\left[\begin{array}{cc}2& -1\\ 3& 4\end{array}\right]$
$adjA=\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]$
Now,
$A^{-1}=\frac{1}{|A|}adjA$
$A^{-1}=\frac{1}{11}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]$
Solve for the values of $x,y$ and $z$
$AX=B$
$X=A^{-1}B$
$\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]\left[\begin{array}{c}-2\\ 3\end{array}\right]$
$\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}4(-2)+1\left(3\right)\\ -3(-2)+2\left(3\right)\end{array}\right]$
$\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}-8+3\\ 6+6\end{array}\right]$
$\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}-5\\ 12\end{array}\right]$
$\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{-5}{11}\\ \frac{12}{11}\end{array}\right]$
$\therefore x=\frac{-5}{11}$ and $y=\frac{12}{11}$