Question

If $A=\left[{}_3^1{}_4^2\right]$ find $A^{-1}$by elementary transformation.?

Answer 1

$A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$

We know that $A=AI$

$\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]=A\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$C_2\to \frac{C_2}{2}$

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

$C_2\to C_2-C_1$

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

$C_2\to C_2\times (-1)$

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

$C_1\to C_1-3C_2$

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

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