To find matrix \(A\), we need to solve for \(A\) in the equation \(A \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}\).

Let \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\). Then the equation becomes:

\[
\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}
\]

Now, perform the matrix multiplication:

\[
\begin{bmatrix} a(0) + b(2) & a(1) + b(-1) \\ c(0) + d(2) & c(1) + d(-1) \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}
\]

Simplify the matrix equation:

\[
\begin{bmatrix} 2b & a - b \\ 2d & c - d \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}
\]

Now, equate corresponding elements:

1. \(2b = 2\)  \(\Rightarrow b = 1\)
2. \(a - b = -1\) \(\Rightarrow a - 1 = -1\)  \(\Rightarrow a = 0\)
3. \(2d = 1\)  \(\Rightarrow d = \frac{1}{2}\)
4. \(c - d = 0\) \(\Rightarrow c - \frac{1}{2} = 0\)  \(\Rightarrow c = \frac{1}{2}\)

Therefore, the matrix \(A\) is:

\[
A = \begin{bmatrix} 0 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}
\]

The correct option is \(\begin{bmatrix} 0 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}\)