Question

Using elementary transformation find the inverse of $A=\left(\begin{array}{cc}1& 3\\ 2& 7\end{array}\right)$

Answer 1

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

$\left[A/I\right]=\left[\begin{array}{cc}1& 3\\ 2& 7\end{array}\right|\begin{array}{cc}1& 0\\ 0& 1\end{array}]\stackrel{R_2-2R}{\to }$,

$=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right|\begin{array}{cc}1& 0\\ -2& 1\end{array}]\stackrel{R_1-3R_2}{\to }\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|\begin{array}{cc}1& -3\\ 0& 1\end{array}]$

$\therefore A^{-1}=\left[\begin{array}{cc}1& -3\\ 0& 1\end{array}\right]$.