Question

Using elementary transformations, find the inverse of matrix $\left[\begin{array}{cc}6& -3\\ -2& 1\end{array}\right]$, if it exists.

Answer 1

Let $A=\left[\begin{array}{cc}6& -3\\ -2& 1\end{array}\right]$
We know that $A=IA$
$\Rightarrow \left[\begin{array}{cc}6& -3\\ -2& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A$
Applying $R_1\to \frac{1}{6}{R}_1$
$\Rightarrow \left[\begin{array}{cc}\frac{6}{6}& \frac{-3}{6}\\ -2& 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{6}& \frac{0}{6}\\ 0& 1\end{array}\right]A$
$\Rightarrow \left[\begin{array}{cc}1& \frac{-1}{2}\\ -2& 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{6}& 0\\ 0& 1\end{array}\right]A$
Applying $R_2\to R_2+2R_1$
$\Rightarrow \left[\begin{array}{cc}1& -\frac{1}{2}\\ -2+2\left(1\right)& 1+2\left(\frac{-1}{2}\right)\end{array}\right]=\left[\begin{array}{cc}\frac{1}{6}& 0\\ 0+2\left(\frac{1}{6}\right)& 1+2\left(0\right)\end{array}\right]A$
$\Rightarrow \left[\begin{array}{cc}1& -\frac{1}{2}\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}\frac{1}{6}& 0\\ \frac{1}{3}& 1\end{array}\right]A$
$\because$ Left hand side matrix involves all zeroes in the second row.
$\therefore A^{-1}$ does not exist.