Question

Using elementary transformations, find the inverse of the followng matrix.

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

Answer 1

Let $\left[\begin{array}{cc}1& 3\\ 2& 7\end{array}\right]$. We know that A=IA
$\left[\begin{array}{cc}1& 3\\ 2& 7\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A\Rightarrow \left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ -2& 1\end{array}\right]A$ (Using $R_2\to R_2-2R_1)$
$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}7& -3\\ -2& 1\end{array}\right]A$ (Using $R_1\to R_1-3R_2)$
$A^{-1}=\left[\begin{array}{cc}7& -3\\ -2& 1\end{array}\right]$