Question

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

$\left[\begin{array}{cc}6& -3\\ -2& 1\end{array}\right]$

Answer 1

Let $A=\left[\begin{array}{cc}6& -3\\ -2& 1\end{array}\right]$. We know that A =IA
$\therefore \left[\begin{array}{cc}6& -3\\ -2& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A\Rightarrow \left[\begin{array}{cc}1& -\frac{1}{2}\\ -2& 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{6}& 0\\ 0& 1\end{array}\right]A$ (Using $R_1\to \frac{1}{6}{R}_1)$
$\Rightarrow \left[\begin{array}{cc}1& -\frac{1}{2}\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}\frac{1}{6}& 0\\ \frac{1}{3}& 1\end{array}\right]A(Using{R}_2\to R_2+2R_1)$

Now, in the above equation, we can see all the elements are zero in the second row of the matrix on the LHS. Therefore, $A^{-1}$ does not exist.

Note Suppose A=IA, ofter applying the elementary transformation, if any row or column of a matrix on LHS is zero, then $A^{-1}$ does not exist.