Let [.],{.} and operatorname{sgn}(.) denote t | vedclass

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(B) First,evaluate the individual terms in the determinant:$1$. $[\pi] = 3$ (Greatest integer less than or equal to $\pi$).$2$. $\operatorname{amp}(1

Vedclass · Accepted Answer

$\frac{5\pi}{3} - \frac{\pi^2}{3} - 5$

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(B) First,evaluate the individual terms in the determinant:$1$. $[\pi] = 3$ (Greatest integer less than or equal to $\pi$).$2$. $\operatorname{amp}(1 + i\sqrt{3}) = 	an^{-1}(\frac{\sqrt{3}}{1}) = \frac{\pi}{3}$.$3$. $\operatorname{sgn}(\cot^{-1}x) = 1$ (Since $\cot^{-1}x > 0$ for all real $x$).$4$. $\{\pi\} = \pi - [\pi] = \pi - 3$.Substituting these values into the determinant:$D = \left| \begin{array}{ccc} 3 & \pi/3 & 1 \ 1 & 0 & 2 \ 1 & 1 & \pi-3 \end{array} ight|$Expanding along the second row:$D = -1 \cdot \left| \begin{array}{cc} \pi/3 & 1 \ 1 & \pi-3 \end{array} ight| + 0 \cdot \dots - 2 \cdot \left| \begin{array}{cc} 3 & \pi/3 \ 1 & 1 \end{array} ight|$$D = -1 \cdot (\frac{\pi}{3}(\pi-3) - 1) - 2 \cdot (3 - \frac{\pi}{3})$$D = -(\frac{\pi^2}{3} - \pi - 1) - (6 - \frac{2\pi}{3})$$D = -\frac{\pi^2}{3} + \pi + 1 - 6 + \frac{2\pi}{3}$$D = -\frac{\pi^2}{3} + \frac{5\pi}{3} - 5$.

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