Answer
$A^{-1}=\left[\begin{array}{rr}
-1 & 1\\
1 & 0
\end{array}\right]$
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)
----
$\left[\begin{array}{lllll}
0 & 1 & | & 1 & 0\\
1 & 1 & | & 0 & 1
\end{array}\right]\left\{\begin{array}{l}
r_{1}\leftrightarrow r_{2}.\\
.
\end{array}\right.$
$\rightarrow\left[\begin{array}{lllll}
1 & 1 & | & 0 & 1\\
0 & 1 & | & 1 & 0
\end{array}\right]\left\{\begin{array}{l}
-r_{2}.\\
.
\end{array}\right.$
$\rightarrow\left[\begin{array}{lllll}
1 & 0 & | & -1 & 1\\
0 & 1 & | & 1 & 0
\end{array}\right]$
$A^{-1}=\left[\begin{array}{ll}
-1 & 1\\
1 & 0
\end{array}\right]$
Check: $\left[\begin{array}{ll}
0 & 1\\
1 & 1
\end{array}\right]\left[\begin{array}{ll}
-1 & 1\\
1 & 0
\end{array}\right]$=$\left[\begin{array}{ll}
0+1 & 0+0\\
-1+1 & 1+0
\end{array}\right]=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]$