Answer
$A^{-1}=\left[\begin{array}{cc}
1 / 2 & 0 \\
-1 / 2 & 1
\end{array}\right]
$
Work Step by Step
since the Matrix $A=\left[\begin{array}{ll}2 & 0 \\ 1 & 1\end{array}\right]$
To find: Inverse of $A,$ i.e; $A^{-1}$ using elementary matrices.
Applying $R_{1} \rightarrow \frac{R_{1}}{2},$ we reduce $A$ to $B=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$ and corresponding elementary matrix is $E_{1}=\left[\begin{array}{cc}1 / 2 & 0 \\ 0 & 1\end{array}\right]$
Now, by applying $R_{2} \rightarrow R_{2}-R_{1},$ we further reduce $B$ to $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ and the corresponding elementary matrix is $E_{2}=\left[\begin{array}{cc}1 & 0 \\ -1 & 1\end{array}\right]$
Therefore, we have $I=E_{2} E_{1} A \Longrightarrow A^{-1}=E_{2} E_{1} I$. Hence
\[
A^{-1}=E_{2} E_{1}=\left[\begin{array}{cc}
1 & 0 \\
-1 & 1
\end{array}\right] \times\left[\begin{array}{cc}
1 / 2 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 / 2 & 0 \\
-1 / 2 & 1
\end{array}\right]
\]
