Question

Find the inverse of the following matrix using elementary row transformation.
$\left[\begin{array}{cc}1& 2\\ 2& -1\end{array}\right]$

Answer 1

$A=IA$

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

$R_1\to R_1+2R_2$

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

$R_1\to R_1/5$

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

$R_2\to R_2-2R_1$

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

$R_2\to R_2/-1$

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

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