Question

By using elementary transformation find the inverse of $A=\left[\begin{array}{c}1\\ 2\end{array}\begin{array}{c}2\\ -1\end{array}\right]$.

Answer 1

Solution :- Given $A=\left[\begin{array}{cc}1& 2\\ 2& -1\end{array}\right]$

Now we find $A^{-1}$ by elementary row transformation

so, we can write

$A=I.A$

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

$R_2\to R_2-2R_1$

$\left[\begin{array}{cc}1& 2\\ 0& -5\end{array}\right]=\left[\begin{array}{cc}1& 0\\ -2& 1\end{array}\right].A$

$R_1\to R_1+\frac{2}{5}{R}_2$

$\left[\begin{array}{cc}1& 0\\ 0& -5\end{array}\right]=\left[\begin{array}{cc}\frac{1}{5}& \frac{2}{5}\\ -2& 1\end{array}\right].A$

$R_2\to \frac{R_2}{-5}$

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{5}& \frac{2}{5}\\ \frac{2}{5}& \frac{-1}{5}\end{array}\right].A$

this form is $I=A^{-1}.A$ So, $A^{-1}=\left[\begin{array}{cc}\frac{1}{5}& \frac{2}{5}\\ \frac{2}{5}& \frac{-1}{5}\end{array}\right]$