Question

Using elementary transformation find inverse of each of the following$A=\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]$

Answer 1

$A=\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]$

In order to find inverse by using row elementary transformation we write as $A=IA$.

$\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A$

Apply $R_1\to \frac{1}{2}\times R_1$

$\left[\begin{array}{cc}1& \frac{3}{2}\\ 5& 7\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& 1\end{array}\right]A$

Apply $R_2\to R_2-5R_1$

$\left[\begin{array}{cc}1& \frac{3}{2}\\ 0& \frac{-1}{2}\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& 0\\ \frac{-5}{2}& 1\end{array}\right]A$

Apply $R_2\to 2\times R_2$

$\left[\begin{array}{cc}1& \frac{3}{2}\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& 0\\ 5& -2\end{array}\right]A$

Apply $R_1\to R_1-\frac{3}{2}{R}_2$

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}-7& 3\\ 5& -2\end{array}\right]A$

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