Answer
$A^{-1} =\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & \frac{1}{2} & -1/2 \\ 0 & 0 & 1\end{array}\right]$
Work Step by Step
Denote the Elementary Matrix $E_{i}$ and the number of row of the matrix is $R_{i}$
Using the Matrix Operation to get
\[
\begin{array}{c}
{\left[\begin{array}{ccc}
1 & 0 & -2 \\
0 & 1 & \frac{1}{2} \\
0 & 0 & 1
\end{array}\right]} & \frac{R_{2}}{2} \rightarrow R_{2} \quad E_{1}=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \frac{1}{2} & 0 \\
0 & 0 & 1
\end{array}\right] \\
{\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & \frac{1}{2} \\
0 & 1 & 1
\end{array}\right]} & R_{1}+2 R_{3} \rightarrow R_{1} \quad E_{2}=\left[\begin{array}{ccc}
1 & 0 & 2 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
{\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]} & R_{2}-\frac{R_{3}}{2} \rightarrow R_{2} \quad E_{3}=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & -\frac{1}{2} \\
0 & 0 & 1
\end{array}\right]
\end{array}
\]
$\begin{aligned} \therefore A^{-1} &=E_{3} E_{2} E_{1} \\ &=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & 1\end{array}\right] \\ &=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 &1\end{array}\right] \end{aligned}
$
then $A^{-1} =\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & \frac{1}{2} & -1/2 \\ 0 & 0 & 1\end{array}\right]$
