1. **State the problem:** We need to find the vector \( \begin{bmatrix} x \\ y \end{bmatrix} \) given by the equation $$ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 2 \end{bmatrix} + \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2} & 0 \end{pmatrix} \begin{bmatrix} 2 \\ -1 \end{bmatrix} $$ 2. **Check dimensions:** The matrix is 2x3 but the vector to multiply is 2x1, which is incompatible for multiplication. The matrix multiplication requires the number of columns in the matrix to equal the number of rows in the vector. 3. **Assumption:** The matrix should be 2x2 to multiply with the 2x1 vector \( \begin{bmatrix} 2 \\ -1 \end{bmatrix} \). We ignore the third column (zeros) for multiplication. 4. **Perform matrix multiplication:** $$ \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \begin{bmatrix} 2 \\ -1 \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} \times 2 + \left(-\frac{1}{2}\right) \times (-1) \\ \frac{1}{2} \times 2 + \frac{\sqrt{3}}{2} \times (-1) \end{bmatrix} = \begin{bmatrix} \sqrt{3} + \frac{1}{2} \\ 1 - \frac{\sqrt{3}}{2} \end{bmatrix} $$ 5. **Add the vector \( \begin{bmatrix} 4 \\ 2 \end{bmatrix} \):** $$ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4 \\ 2 \end{bmatrix} + \begin{bmatrix} \sqrt{3} + \frac{1}{2} \\ 1 - \frac{\sqrt{3}}{2} \end{bmatrix} = \begin{bmatrix} 4 + \sqrt{3} + \frac{1}{2} \\ 2 + 1 - \frac{\sqrt{3}}{2} \end{bmatrix} = \begin{bmatrix} 4.5 + \sqrt{3} \\ 3 - \frac{\sqrt{3}}{2} \end{bmatrix} $$ 6. **Final answer:** $$ \boxed{\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4.5 + \sqrt{3} \\ 3 - \frac{\sqrt{3}}{2} \end{bmatrix}} $$