Question

Using elementary tansormations, find the inverse of each of the matrices, if it exists in
$\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]$

Answer 1

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

$|A|=14-15=-1\ne 0$

Hence, $A^{-1}$ exists.

Now, $A=IA$

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

$\left[\begin{array}{cc}1& \frac{3}{2}\\ 5& 7\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& 1\end{array}\right]A$ $(R_1\to \frac{1}{2}{R}_1)$

$\left[\begin{array}{cc}1& \frac{3}{2}\\ 0& -\frac{1}{2}\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& 0\\ -\frac{5}{2}& 1\end{array}\right]A$ $(R_2\to R_2-5R_1)$

$\left[\begin{array}{cc}1& \frac{3}{2}\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& 0\\ 5& -2\end{array}\right]A$ $(R_2\to -2R_2)$

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}-7& 3\\ 5& -2\end{array}\right]A$ $(R_1\to R_1-\frac{3}{2}{R}_2)$

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