Answer
$A^{-1}=\left[\begin{array}{rr}
1 & -1\\
-1 & 2
\end{array}\right]$
Work Step by Step
$1.\quad $Start with $[A|I]$,
$2.\quad $ Row reduce,
$3.\quad $ If the reduced form is $[I|B], $then $B=A^{-1}.$
If not, then A is singular (has no inverse)
----
$\left[\begin{array}{lllll}
2 & 1 & | & 1 & 0\\
1 & 1 & | & 0 & 1
\end{array}\right]\left\{\begin{array}{l}
R_{1}\leftrightarrow R_{2}.\\
.
\end{array}\right.$
$\rightarrow\left[\begin{array}{lllll}
1 & 1 & | & 0 & 1\\
2 & 1 & | & 1 & 0
\end{array}\right]\left\{\begin{array}{l}
.\\
-2R_{1}.
\end{array}\right.$
$\rightarrow\left[\begin{array}{lllll}
1 & 1 & | & 0 & 1\\
0 & -1 & | & 1 & -2
\end{array}\right]\left\{\begin{array}{l}
+R_{1}.\\
\times(-1).
\end{array}\right.$
$\rightarrow\left[\begin{array}{lllll}
1 & 0 & | & 1 & -1\\
0 & 1 & | & -1 & 2
\end{array}\right]$
$A^{-1}=\left[\begin{array}{rr}
1 & -1\\
-1 & 2
\end{array}\right]$
Check: $\left[\begin{array}{ll}
2 & 1\\
1 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & -1\\
-1 & 2
\end{array}\right]=\left[\begin{array}{ll}
2-1 & -2+2\\
1-1 & -1+2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]=I_{2}$
