Question

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

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

Answer 1

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