1. **State the problem:** We are given the vector equation: $$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 2 \end{bmatrix} + \left( \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \end{bmatrix} \begin{bmatrix} 2 \\ -1 \end{bmatrix} \right)$$ 2. **Identify dimensions and correct multiplication:** The matrix on the right is 2x3, but the vector multiplied is 2x1, which is incompatible. Assuming the last column of the matrix is multiplied by 0 (since the vector has only two elements), we consider only the first two columns and the 2x2 vector: $$\begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} 2 \\ -1 \end{bmatrix}$$ 3. **Perform the matrix multiplication:** Calculate each component: $$x_{add} = \frac{\sqrt{3}}{2} \times 2 + \left(-\frac{1}{2}\right) \times (-1) = \sqrt{3} + \frac{1}{2}$$ $$y_{add} = \frac{1}{2} \times 2 + \frac{\sqrt{3}}{2} \times (-1) = 1 - \frac{\sqrt{3}}{2}$$ 4. **Add the resulting vector to the initial vector:** $$x = 4 + \left( \sqrt{3} + \frac{1}{2} \right) = 4 + \sqrt{3} + \frac{1}{2} = \frac{9}{2} + \sqrt{3}$$ $$y = 2 + \left( 1 - \frac{\sqrt{3}}{2} \right) = 3 - \frac{\sqrt{3}}{2}$$ 5. **Final answer:** $$\boxed{x = \frac{9}{2} + \sqrt{3}, \quad y = 3 - \frac{\sqrt{3}}{2}}$$