Question

Solve the following equations using Matrix Inversion method.
$2x-3y+6=0$ and $6x+y+8=0$

Answer 1

Solution:

Matrix Inversion Method:

$2x-3y+6=0⟹2x-3y=-6$

$6x+y+8=0⟹6x+y=-8$

The corresponding matrix equation is

$\left[\begin{array}{cc}2& -3\\ 6& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-6\\ -8\end{array}\right]$

i.e. $AX=B$

Where, $A=\left[\begin{array}{cc}2& -3\\ 6& 1\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right],B=\left[\begin{array}{c}-6\\ -8\end{array}\right]$

Hence,premultiplying above matrix equation by $A^{-1}$ on both sides, we get

$A^{-1}AX=A^{-1}B$

or, $IX=A^{-1}B$

or, $X=A^{-1}B$

Now, $A=\left[\begin{array}{cc}2& -3\\ 6& 1\end{array}\right]⟹|A|=2+18=20$

Adjoint of $A=$ $\text{adj}A=\left[\begin{array}{cc}1& 3\\ -6& 2\end{array}\right]$

$A^{-1}=\frac{1}{|A|}\left(\text{adj}A\right)$

or, $=\frac{1}{20}\left[\begin{array}{cc}1& 3\\ -6& 2\end{array}\right]$

$X=\frac{1}{20}\left[\begin{array}{cc}1& 3\\ -6& 2\end{array}\right]\left[\begin{array}{c}-6\\ -8\end{array}\right]$

$=\frac{1}{20}\left[\begin{array}{c}-30\\ 20\end{array}\right]$

$=\left[\begin{array}{c}\frac{-30}{20}\\ \frac{20}{20}\end{array}\right]$

$=\left[\begin{array}{c}-\frac{3}{2}\\ 1\end{array}\right]$

So, $x=-\frac{3}{2},y=1$