Question

Solve the following by the method of inversion

$2x+3y=5$ and $3x+y=3$

Answer 1

$2x+3y=5$

$3x+y=3$

$\left(\begin{array}{cc}2& 3\\ 3& 1\end{array}\right)\left(\begin{array}{cc}x& y\end{array}\right)=\left(\begin{array}{c}5\\ 3\end{array}\right)$

Determinant

$\left(\begin{array}{cc}2& 3\\ 3& 1\end{array}\right)=-7$

Inverse

$\left(\begin{array}{cc}2& 3\\ 3& 1\end{array}\right)=\left(\begin{array}{cc}\frac{-1}{7}& \frac{3}{7}\\ \frac{3}{7}& \frac{-2}{7}\end{array}\right)$

Multiply inverse on both sides

$\left(\begin{array}{cc}2& 3\\ 3& 1\end{array}\right)\left(\begin{array}{cc}\frac{-1}{7}& \frac{3}{7}\\ \frac{3}{7}& \frac{-2}{7}\end{array}\right)$ $\left(\begin{array}{c}x\\ y\end{array}\right)$ $=\left(\begin{array}{cc}\frac{-1}{7}& \frac{3}{7}\\ \frac{3}{7}& \frac{-2}{7}\end{array}\right)$ $\left(\begin{array}{c}5\\ 3\end{array}\right)$

$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}\frac{4}{7}\\ \frac{9}{7}\end{array}\right)$

$\therefore x=\frac{4}{7},y=\frac{9}{7}$