Let R = begin{bmatrix} x & 0 & 0 0 & y & 0 | vedclass

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(C) Given $x \sin 	heta = y \sin \left(	heta + \frac{2 \pi}{3}ight) = z \sin \left(	heta + \frac{4 \pi}{3}ight) = \lambda 
eq 0$.Since $\sin \

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Only $(II)$ is true

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(C) Given $x \sin 	heta = y \sin \left(	heta + \frac{2 \pi}{3}ight) = z \sin \left(	heta + \frac{4 \pi}{3}ight) = \lambda 
eq 0$.Since $\sin 	heta + \sin \left(	heta + \frac{2 \pi}{3}ight) + \sin \left(	heta + \frac{4 \pi}{3}ight) = 0$,we have $x = \frac{\lambda}{\sin 	heta}$,$y = \frac{\lambda}{\sin(	heta + 2\pi/3)}$,$z = \frac{\lambda}{\sin(	heta + 4\pi/3)}$.$	ext{Trace}(R) = x + y + z = \lambda \left( \frac{1}{\sin 	heta} + \frac{1}{\sin(	heta + 2\pi/3)} + \frac{1}{\sin(	heta + 4\pi/3)} ight)$.Using the identity $\frac{1}{\sin A} + \frac{1}{\sin(A+2\pi/3)} + \frac{1}{\sin(A+4\pi/3)} = \frac{-4 \sin(3A)}{\sin(3A)} = -4$ is incorrect here; rather,the sum is $\frac{-3 \sin(3	heta)}{\sin(3	heta)} = -3$ is not applicable. Actually,$x+y+z = \lambda \frac{\sum \sin(	heta+2\pi/3)\sin(	heta+4\pi/3)}{\prod \sin(	heta+2\pi/3)} = \lambda \frac{-3/4}{-1/4 \sin(3	heta)} = \frac{3\lambda}{\sin(3	heta)} 
eq 0$.Thus,Statement $(I)$ is False.For $(II)$,$	ext{adj}(	ext{adj}(R)) = |R| R = (xyz) R = \begin{bmatrix} x^2yz & 0 & 0 \ 0 & xy^2z & 0 \ 0 & 0 & xyz^2 \end{bmatrix}$.$	ext{Trace}(	ext{adj}(	ext{adj}(R))) = xyz(x+y+z)$. Since $x, y, z 
eq 0$ and $x+y+z 
eq 0$,the trace is non-zero.$A$ conditional statement "If $P$,then $Q$" is true if $P$ is false. Since the hypothesis "trace$(	ext{adj}(	ext{adj}(R))) = 0$" is false,the statement $(II)$ is vacuously true.

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