If A = left[ {begin{array}{*{20}{c}}
1&{ | vedclass

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Question

$\left| A ight| = \left| {\begin{array}{*{20}{c}}
1&{\sin 	heta }&1\
{ - \sin 	heta }&1&{\sin 	heta }\
{ - 1}&{ - \sin \

Vedclass · Accepted Answer

$\left( {\frac{3}{2},\left. 3 \right]} \right.$

Vedclass · Answer

$\left| A ight| = \left| {\begin{array}{*{20}{c}}
1&{\sin 	heta }&1\
{ - \sin 	heta }&1&{\sin 	heta }\
{ - 1}&{ - \sin 	heta }&1
\end{array}} ight|$

$ = 2\left( {1 + {{\sin }^2}	heta } ight)$

$	heta  \in \left( {\frac{{3\pi }}{4},\frac{{5\pi }}{4}} ight) \Rightarrow \frac{1}{{\sqrt 2 }} < \sin 	heta  < \frac{1}{{\sqrt 2 }}$

$ \Rightarrow 0 \le {\sin ^2}	heta  < \frac{1}{2}$

$	herefore \left| A ight| \in \left[ {2,3} ight)$

If $A = \left[ {\begin{array}{*{20}{c}}
1&{\sin \,\theta }&1\\
{ - \,\sin \,\theta }&1&{\sin \,\theta }\\
{ - 1}&{ - \,\sin \,\theta }&1
\end{array}} \right];$ then for all $\theta \, \in \,\left( {\frac{{3\pi }}{4},\frac{{5\pi }}{4}} \right),$ det $(A)$ lies in the interval

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