Question

Using elementary transformations, find the inverse of matrix $\left[\begin{array}{cc}2& 1\\ 4& 2\end{array}\right]$, if it exists.

Answer 1

Let $A=\left[\begin{array}{cc}2& 1\\ 4& 2\end{array}\right]$
We know that $A=IA$
$\Rightarrow \left[\begin{array}{cc}2& 1\\ 4& 2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A$
Applying $R_1\to R_1-\frac{1}{2}{R}_2$
$\Rightarrow \left[\begin{array}{cc}2-\frac{1}{2}\left(4\right)& 1-\frac{1}{2}\left(2\right)\\ 4& 2\end{array}\right]=\left[\begin{array}{cc}1-\frac{1}{2}\left(0\right)& 0-\frac{1}{2}\left(1\right)\\ 0& 1\end{array}\right]A$
$\Rightarrow \left[\begin{array}{cc}2-2& 1-1\\ 4& 2\end{array}\right]=\left[\begin{array}{cc}1& -\frac{1}{2}\\ 0& 1\end{array}\right]A$
$\Rightarrow \left[\begin{array}{cc}0& 0\\ 4& 2\end{array}\right]=\left[\begin{array}{cc}1& -\frac{1}{2}\\ 0& 1\end{array}\right]A$
Since we have all zeros in the first row of the left hand side matrix of the above equation.
Therefore, $A^{-1}$ does not exist.