Question

Using elementary transformations, find the inverse of matrix $\left[\begin{array}{cc}2& -3\\ -1& 2\end{array}\right]$, if it exists.

Answer 1

Let $A=\left[\begin{array}{cc}2& -3\\ -1& 2\end{array}\right]$
We know that $A=IA$
$\Rightarrow \left[\begin{array}{cc}2& -3\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A$
Applying $R_1\to R_1+R_2$
$\Rightarrow \left[\begin{array}{cc}2+(-1)& -3+2\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}1+0& 0+1\\ 0& 1\end{array}\right]A$
$\Rightarrow \left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]A$
Applying $R_2\to R_2+R_1$
$\Rightarrow \left[\begin{array}{cc}1& -1\\ 1+(-1)& 2+(-1)\end{array}\right]=\left[\begin{array}{cc}1& 1\\ 0+1& 1+1\end{array}\right]A$
$\Rightarrow \left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 1\\ 1& 2\end{array}\right]A$
Applying $R_1\to R_1+R_2$
$\Rightarrow \left[\begin{array}{cc}1+0& -1+1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1+1& 1+2\\ 1& 2\end{array}\right]A$
$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]A$
$\Rightarrow I=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]A$
This is similar to $I=A^{-1}A$
Thus, $A^{-1}=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]$