Question

Find the inverse of the matrices , if it exists.
$\left[\begin{array}{cc}2& 3\\ 5& 7\end{array}\right]$

Answer 1

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

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

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

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

$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& -\frac{1}{2}\end{array}\right]=\left[\begin{array}{cc}-7& 3\\ -\frac{5}{2}& 1\end{array}\right]A(R_1\to R_1-3R_2)$

$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& \frac{1}{2}\end{array}\right]=\left[\begin{array}{cc}-7& 3\\ \frac{5}{2}& -1\end{array}\right]A(R_2\to -R_2)$

$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}-7& 3\\ 5& -2\end{array}\right]A(R_2\to 2R_2)$

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