Let $P_0 = (3,1)$ and define $P_{n+1} = (x_n, y_n)$ for $n \ge 0$ by $$x_{n+1} = - \frac{3x_n - y_n}{2}, y_{n+1} = - \frac{x_n + y_n}{2}$$Find the area of the quadrilateral formed by the points $P_{96}, P_{97}, P_{98}, P_{99}$.
1. **Define the sequence and transformation matrix:**
   Let \( P_0 = (3,1) \) and define \( P_{n+1} = (x_n, y_n) \) for \( n \ge 0 \) by:
   \[
   x_{n+1} = - \frac{3x_n - y_n}{2}, \quad y_{n+1} = - \frac{x_n + y_n}{2}
   \]
   We can represent this transformation using a matrix \( M \):
   \[
   \vec{v}_n = \begin{pmatrix} x_n \\ y_n \end{pmatrix}, \quad M = \begin{pmatrix} -\frac{3}{2} & \frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} \end{pmatrix}
   \]
   The recurrence relation can be written as:
   \[
   \vec{v}_{n+1} = M \vec{v}_n
   \]

2. **Calculate the determinant of the transformation matrix:**
   The determinant of \( M \) is:
   \[
   \text{det}(M) = \left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right) - \left(-\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{3}{4} - \left(-\frac{1}{4}\right) = \frac{3}{4} + \frac{1}{4} = 1
   \]
   Since the determinant is 1, the area of any shape transformed by \( M \) remains unchanged.

3. **Initial points and area calculation:**
   Calculate the first few points:
   \[
   P_0 = (3, 1)
   \]
   \[
   P_1 = M \vec{v}_0 = \begin{pmatrix} -\frac{3}{2} & \frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -\frac{9}{2} + \frac{1}{2} \\ -\frac{3}{2} - \frac{1}{2} \end{pmatrix} = \begin{pmatrix} -4 \\ -2 \end{pmatrix}
   \]
   \[
   P_2 = M \vec{v}_1 = \begin{pmatrix} -\frac{3}{2} & \frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} -4 \\ -2 \end{pmatrix} = \begin{pmatrix} 6 - 1 \\ 2 + 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}
   \]
   \[
   P_3 = M \vec{v}_2 = \begin{pmatrix} -\frac{3}{2} & \frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} 5 \\ 3 \end{pmatrix} = \begin{pmatrix} -\frac{15}{2} + \frac{3}{2} \\ -\frac{5}{2} - \frac{3}{2} \end{pmatrix} = \begin{pmatrix} -6 \\ -4 \end{pmatrix}
   \]

4. **Area calculation using the Shoelace formula:**
   The area of the quadrilateral formed by \( P_0, P_1, P_2, P_3 \) can be calculated using the Shoelace formula:
   \[
   \text{Area} = \frac{1}{2} \left| x_0y_1 + x_1y_2 + x_2y_3 + x_3y_0 - (y_0x_1 + y_1x_2 + y_2x_3 + y_3x_0) \right|
   \]
   Substituting the points:
   \[
   \text{Area} = \frac{1}{2} \left| 3(-2) + (-4)3 + 5(-4) + (-6)1 - (1(-4) + (-2)5 + 3(-6) + (-4)3) \right|
   \]
   \[
   = \frac{1}{2} \left| -6 - 12 - 20 - 6 - (-4 - 10 - 18 - 12) \right|
   \]
   \[
   = \frac{1}{2} \left| -44 - (-44) \right| = \frac{1}{2} \left| -44 + 44 \right| = \frac{1}{2} \left| 0 \right| = 0
   \]
   This calculation seems incorrect. Let's re-evaluate the points and the area.

5. **Re-evaluate the area:**
   The points \( P_0, P_1, P_2, P_3 \) form a parallelogram. The area of a parallelogram can also be calculated using the determinant of the matrix formed by the vectors:
   \[
   \text{Area} = \left| \text{det} \begin{pmatrix} x_1 - x_0 & x_2 - x_0 \\ y_1 - y_0 & y_2 - y_0 \end{pmatrix} \right|
   \]
   Using \( P_0 = (3,1), P_1 = (-4,-2), P_2 = (5,3) \):
   \[
   \text{Area} = \left| \text{det} \begin{pmatrix} -4 - 3 & 5 - 3 \\ -2 - 1 & 3 - 1 \end{pmatrix} \right| = \left| \text{det} \begin{pmatrix} -7 & 2 \\ -3 & 2 \end{pmatrix} \right|
   \]
   \[
   = \left| (-7)(2) - (2)(-3) \right| = \left| -14 + 6 \right| = \left| -8 \right| = 8
   \]

6. **Conclusion:**
   Since the determinant of \( M \) is 1, the area of the quadrilateral remains unchanged for any \( i \). Therefore, the area of the quadrilateral formed by \( P_{96}, P_{97}, P_{98}, P_{99} \) is the same as that formed by \( P_0, P_1, P_2, P_3 \).

The final answer is \(\boxed{8}\)