Question

Convert $\left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right]$ into an identity matrix by suitable row transformations.

Answer 1

convert $\left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right]$ into identify matrix by suitable row motion

$\therefore A{A}^{-1}=I$

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

Applying Elementary Row operation.

$R_1\to R_1+\frac{1}{3}{R}_2$

$A6-1\left[\begin{array}{cc}\frac{5}{3}& 0\\ 2& 3\end{array}\right]=\left[\begin{array}{cc}1& \frac{1}{3}\\ 0& 1\end{array}\right]$

$R_1\to \frac{3}{5}{R}_1$

$A^{-1}\left[\begin{array}{cc}1& 0\\ 2& 3\end{array}\right]=\left[\begin{array}{cc}\frac{3}{5}& \frac{1}{5}\\ 0& 1\end{array}\right]$

$A_1\left[\begin{array}{cc}1& 0\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}\frac{3}{5}& \frac{1}{5}\\ \frac{-6}{5}& \frac{3}{5}\end{array}\right]$

$R_2\to \frac{1}{3}{R}_2$

$A^{-1}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}\frac{3}{5}& \frac{1}{5}\\ \frac{-2}{5}& \frac{1}{5}\end{array}\right]$

So, $A^{-1}\left[\begin{array}{cc}\frac{3}{5}& \frac{1}{5}\\ \frac{-2}{5}& \frac{1}{5}\end{array}\right]$