Question

If $A=\left[\begin{array}{cc}1& -2\\ 2& -6\end{array}\right]$, find $A^{-1}$ using elementary row operations only.

Answer 1

$\begin{array}{c}A=\left[\begin{array}{cc}1& -2\\ 2& -6\end{array}\right]\\ \left[\begin{array}{cc}1& -2\\ 2& -6\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right].A\\ \Rightarrow \left[\begin{array}{cc}1& -2\\ 0& -2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ -2& 1\end{array}\right].A\left[R_2\to R_2-2R_1\right]\\ \Rightarrow \left[\begin{array}{cc}1& -2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 1& \frac{-1}{2}\end{array}\right].A\left[R_2\to \left(\frac{-1}{2}\right)R_2\right]\\ \Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}3& -1\\ 1& -\frac{1}{2}\end{array}\right].A\left[R_1\to R_1+2R_2\right]\\ Hence,\\ A^{-1}=\left[\begin{array}{cc}3& -1\\ 1& -\frac{1}{2}\end{array}\right]\end{array}$