Question

Using elementary tansormations, find the inverse of each of the matrices, if it exists in
$\left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right]$

Answer 1

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

$|A|=3+2=5\ne 0$

Hence, $A^{-1}$ exists.

Now, $A=IA$

$\left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A$

$\left[\begin{array}{cc}1& -1\\ 0& 5\end{array}\right]=\left[\begin{array}{cc}1& 0\\ -2& 1\end{array}\right]A$ $(R_2\to R_2-2R_1)$

$\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ -\frac{2}{5}& \frac{1}{5}\end{array}\right]A$ $(R_2\to \frac{1}{5}{R}_2)$

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}\frac{3}{5}& \frac{1}{5}\\ -\frac{2}{5}& \frac{1}{5}\end{array}\right]A$

Hence, $A^{-1}=\left[\begin{array}{cc}\frac{3}{5}& \frac{1}{5}\\ -\frac{2}{5}& \frac{1}{5}\end{array}\right]$