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જો $A = \left[ {\begin{array}{*{20}{c}}
1&{\sin \,\theta }&1\\
{ - \,\sin \,\theta }&1&{\sin \,\theta }\\
{ - 1}&{ - \,\sin \,\theta }&1
\end{array}} \right];$ તો દરેક $\theta \, \in \,\left( {\frac{{3\pi }}{4},\frac{{5\pi }}{4}} \right)$ માટે $det (A)$ ની કિમંત મેળવો.
$\left( {1,\left. {\frac{5}{2}} \right]} \right.$
$\left[ {\frac{5}{2},\left. 4 \right)} \right.$
$\left( {\left. {0,\frac{3}{2}} \right]} \right.$
$\left( {\frac{3}{2},\left. 3 \right]} \right.$
Solution
$\left| A \right| = \left| {\begin{array}{*{20}{c}}
1&{\sin \theta }&1\\
{ – \sin \theta }&1&{\sin \theta }\\
{ – 1}&{ – \sin \theta }&1
\end{array}} \right|$
$ = 2\left( {1 + {{\sin }^2}\theta } \right)$
$\theta \in \left( {\frac{{3\pi }}{4},\frac{{5\pi }}{4}} \right) \Rightarrow \frac{1}{{\sqrt 2 }} < \sin \theta < \frac{1}{{\sqrt 2 }}$
$ \Rightarrow 0 \le {\sin ^2}\theta < \frac{1}{2}$
$\therefore \left| A \right| \in \left[ {2,3} \right)$
