left| begin{array}{ccc} 2 sin frac{pi}{3} & 1 | vedclass

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(D) First,evaluate the trigonometric values: $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$. Substituting the

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$2$

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(D) First,evaluate the trigonometric values: $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$. Substituting these into the determinant: $D = \left| \begin{array}{ccc} 2(\frac{\sqrt{3}}{2}) & 1 & 0 \ 1 & 2(\frac{\sqrt{3}}{2}) & 1 \ 0 & 1 & 2(\frac{\sqrt{3}}{2}) \end{array} ight| = \left| \begin{array}{ccc} \sqrt{3} & 1 & 0 \ 1 & \sqrt{3} & 1 \ 0 & 1 & \sqrt{3} \end{array} ight|$. Expanding along the first row: $D = \sqrt{3} ((\sqrt{3})(\sqrt{3}) - (1)(1)) - 1 ((1)(\sqrt{3}) - (1)(0)) + 0 = \sqrt{3} (3 - 1) - 1 (\sqrt{3}) = \sqrt{3} (2) - \sqrt{3} = 2\sqrt{3} - \sqrt{3} = \sqrt{3}$. Since the calculated value is $\sqrt{3}$,and the provided options were incorrect,the correct value is $\sqrt{3}$.

$\left| \begin{array}{ccc} 2 \sin \frac{\pi}{3} & 1 & 0 \\ 1 & 2 \sin \frac{\pi}{3} & 1 \\ 0 & 1 & 2 \cos \frac{\pi}{6} \end{array} \right| = $ . . . . . .

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