The answer is “Yes”.

Performing \(R_2 \leftarrow R_2 -2R_1\) to \([A \mid I]\) gives \(\left[\begin{array}{cc|cc} 1 & 2 & 1 & 0\\ 0 & -1 & -2 & 1\end{array}\right]\).

Then, performing \(R_1 \leftarrow R_1 +2R_1\) gives \(\left[\begin{array}{cc|cc} 1 & 0 & -3 & 2\\ 0 & -1 & -2 & 1\end{array}\right]\).

Finall, performing \(R_2 \leftarrow -R_2\) gives \(\left[\begin{array}{cc|cc} 1 & 0 & -3 & 2\\ 0 & 1 & 2 &-1\end{array}\right]\).

Since the matrix is in RREF and the first two columns give the identity matrix, \(A^{-1}\) exists and is given by the last two columns of the final matrix above.