Question

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

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

Answer 1

Let $A=\left[\begin{array}{cc}2& -3\\ -1& 2\end{array}\right]$. We know that A=IA
$\therefore \left[\begin{array}{cc}2& -3\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A\Rightarrow \left[\begin{array}{cc}1& -\frac{3}{2}\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& 1\end{array}\right]A$ (Using $R_1\to \frac{1}{2}{R}_1)$
$\Rightarrow \left[\begin{array}{cc}1& -\frac{3}{2}\\ 0& \frac{1}{2}\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& 0\\ \frac{1}{2}& 1\end{array}\right]A$ (Using $R_2\to R_2+R_1)$
$\Rightarrow \left[\begin{array}{cc}1& -\frac{3}{2}\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& 0\\ 1& 2\end{array}\right]A$ (Using $R_2\to 2R_2)$
$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]A$ (Using $R_1\to R_1+\frac{3}{2}{R}_2)$
$\therefore A^{-1}=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right](\because A{A}^{-1}=I)$