Question

Using elementary transformations , find the inverse of each of the matrices,
$\left[\begin{array}{cc}3& -1\\ -4& 2\end{array}\right]$

Answer 1

$A=IA$ [ row transformation ]
$\left[\begin{array}{cc}3& 10\\ 2& 4=7\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A$
$R_1\leftrightarrow 1/3R_1$
$\left[\begin{array}{cc}1& 10/3\\ 2& 7\end{array}\right]=\left[\begin{array}{cc}1/3& 0\\ 0& 1\end{array}\right]A$
$R_2\to R_2-2R_1$
$\left[\begin{array}{cc}1& 10/3\\ 0& 1/3\end{array}\right]=\left[\begin{array}{cc}1/3& 0\\ -2/3& 1\end{array}\right]A$
$R_2\to 3R_2$
$\left[\begin{array}{cc}1& 10/3\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1/3& 0\\ -2& 3\end{array}\right]A$
$R_1\to R_1-10/3R_2$
=$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}7& -10\\ -2& 3\end{array}\right]A$
$\therefore A^{-1}=\left[\begin{array}{cc}7& -10\\ -2& 3\end{array}\right]$