Given: \(A^{-1} = \frac{1}{7}\begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix}\)

To find matrix A, we need to find the inverse of \(A^{-1}\).

Let \(B = A^{-1} = \frac{1}{7}\begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix}\)

First, find the determinant of the matrix inside the scalar multiplication:

\(\text{det}\begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix} = (2 \times 2) - (1 \times -3) = 4 + 3 = 7\)

Now, find the inverse of the matrix inside the scalar multiplication:

\(\begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix}^{-1} = \frac{1}{7}\begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix}\)

Since \(B = \frac{1}{7}\begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix}\), then \(B^{-1} = \left(\frac{1}{7}\begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix}\right)^{-1} = 7 \begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix}^{-1} = 7 \cdot \frac{1}{7} \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix}\)

Therefore, \(A = \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix}\)