Question

Using elementary transformations, find the inverse of the matrix $\left[\begin{array}{cc}3& 10\\ 2& 7\end{array}\right]$

Answer 1

$A=\left[\begin{array}{cc}3& 10\\ 2& 7\end{array}\right]$

We know that $A=IA$ (using row operations)

$\left[\begin{array}{cc}3& 10\\ 2& 7\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A$

Using operation $R_1\Rightarrow R_1-R_2$

$\left[\begin{array}{cc}1& 3\\ 2& 7\end{array}\right]=\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right]A$

Using operation $R_2\Rightarrow R_2-2R_1$

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

Using operation $R_1\Rightarrow R_1-3R_2$

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}7& -10\\ -2& 3\end{array}\right]A$

Therefore,

$A^{-1}=\left[\begin{array}{cc}7& -10\\ -2& 3\end{array}\right]$