Question

Solve the following equations by the inversion method:
$2x+3y=-5$ and $3x+y=3$

Answer 1

We know if A and X and B are $3$ matrix

such that $AX=B$ then $X=A^{-1}B$

$2x+3y=-5$

$3x+y=3$

$\therefore A=\left[\begin{array}{cc}2& 3\\ 3& 1\end{array}\right]X=\left[\begin{array}{c}x\\ y\end{array}\right]B=\left[\begin{array}{c}-5\\ 3\end{array}\right]$

$\therefore \left[\begin{array}{c}x\\ y\end{array}\right]=A^{-1}\left[\begin{array}{c}-5\\ 3\end{array}\right]$

$A=\left[\begin{array}{cc}2& 3\\ 3& 1\end{array}\right]$

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

For $2\times 2$ matrix, we exchange the principal diagonal elements and multiply the other two by $-1$

$\therefore A^{-1}=\frac{1}{2\times 1-3\times 3}\left[\begin{array}{cc}1& -3\\ -3& 2\end{array}\right]$

$\therefore A^{-1}=\frac{1}{-7}\left[\begin{array}{cc}1& -3\\ -3& 2\end{array}\right]$

$x=A^{-1}B=\frac{1}{-7}\left[\begin{array}{cc}1& -3\\ -3& 2\end{array}\right]\left[\begin{array}{c}-5\\ 3\end{array}\right]$

$\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{-7}\left[\begin{array}{c}-5-9\\ 15+6\end{array}\right]=\left[\begin{array}{c}\frac{-14}{-7}\\ \frac{21}{-7}\end{array}\right]=\left[\begin{array}{c}2\\ -3\end{array}\right]$

$\therefore x=2,y=-3$