Question

Using elementary transformations, find the inverse of matrix $\left[\begin{array}{cc}2& 5\\ 1& 3\end{array}\right]$, if it exists.

Answer 1

Let $A=\left[\begin{array}{cc}2& 5\\ 1& 3\end{array}\right]$
We know that $A=IA$
$\Rightarrow \left[\begin{array}{cc}2& 5\\ 1& 3\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A$
Applying $R_1\to R_1-R_2$
$\Rightarrow \left[\begin{array}{cc}2-1& 5-3\\ 1& 3\end{array}\right]=\left[\begin{array}{cc}1-0& 0-1\\ 0& 1\end{array}\right]A$
$\Rightarrow \left[\begin{array}{cc}1& 2\\ 1& 3\end{array}\right]=\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right]A$
Applying $R_2\to R_2-R_1$
$\Rightarrow \left[\begin{array}{cc}1& 2\\ 1-1& 3-2\end{array}\right]=\left[\begin{array}{cc}1& -1\\ 0-1& 1-(-1)\end{array}\right]A$
$\Rightarrow \left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]A$
Applying $R_1\to R_1-2R_2$
$\Rightarrow \left[\begin{array}{cc}1-2\left(0\right)& 2-2\left(1\right)\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1-2(-1)& -1-2\left(2\right)\\ -1& 2\end{array}\right]A$
$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}3& -5\\ -1& 2\end{array}\right]A$
$\Rightarrow I=\left[\begin{array}{cc}3& -5\\ -1& 2\end{array}\right]A$
This is similar to $I=A^{-1}A$
Thus, $A^{-1}=\left[\begin{array}{cc}3& -5\\ -1& 2\end{array}\right]$