Let's find the correct eigenvectors of the matrix $A = \begin{bmatrix} \frac{3}{2} & -1 \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix}$ by determining the eigenvalues and then using them to find corresponding eigenvectors.

Step 1: Find the Eigenvalues

The characteristic equation is:

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

where $\lambda$ represents the eigenvalues. This equation can be expressed as:

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

Calculating the determinant:

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

Simplifying this:

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

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

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

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

Solving this 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 and match with the correct option from:

A: $v_1 = [-1 - \sqrt{5}, 1], v_2 = [-1 + \sqrt{5}, 1]$.

B: $v_1 = [1 - \sqrt{3}, 1], v_2 = [1 + \sqrt{3}, 1]$.

C: $v_1 = [1 - \sqrt{5}, 1], v_2 = [1 + \sqrt{5}, 1]$.

D: $v_1 = [-1 - \sqrt{3}, 1], v_2 = [-1 + \sqrt{3}, 1]$.