To find the correct eigenvectors of the matrix $A = \begin{bmatrix} \frac{3}{2} & -1 \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix}$, we first need to compute the eigenvalues of the matrix. Then, we find the corresponding eigenvectors.

Step 1: Find the Eigenvalues

The eigenvalues $\lambda$ of matrix $A$ are found by solving the characteristic equation:

$\text{det}(A - \lambda I) = 0,$

where $I$ is the identity matrix. The characteristic equation becomes:

$\begin{vmatrix} \frac{3}{2} - \lambda & -1 \\ -\frac{1}{2} & \frac{1}{2} - \lambda \end{vmatrix} = 0.$

Calculating the determinant:

$(\frac{3}{2} - \lambda)(\frac{1}{2} - \lambda) - (-1)(-\frac{1}{2}) = 0.$

Simplifying this:

$(\frac{3}{2} - \lambda)(\frac{1}{2} - \lambda) - \frac{1}{2} = 0.$

$\left(\frac{3}{2} \cdot \frac{1}{2}\right) - \left(\frac{3}{2} \lambda\right) - \left(\frac{1}{2} \lambda\right) + \lambda^2 - \frac{1}{2} = 0.$

$\frac{3}{4} - 2\lambda + \lambda^2 - \frac{1}{2} = 0.$

Simplify to form a standard quadratic equation:

$\lambda^2 - 2\lambda + \frac{1}{4} = 0.$

Solve this equation using the quadratic formula:

$\lambda = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(\frac{1}{4})}}{2(1)}.$

After solving, plug back to find corresponding eigenvectors then compare to answer choices.