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જો $A = \left( {\begin{array}{*{20}{c}}
{\cos \,\alpha }&{ - \sin \,\alpha }\\
{\sin \,\alpha }&{\cos \,\alpha }
\end{array}} \right)$, $\left( {\alpha \in R} \right)$ આપલે છે કે જેથી ${A^{32}} = \left( {\begin{array}{*{20}{c}}
0&{ - 1}\\
1&0
\end{array}} \right)$ તો $\alpha $ ની કિમંત મેળવો.
$0$
$\frac{\pi }{{16}}$
$\frac{\pi }{{32}}$
$\frac{\pi }{{64}}$
Solution
$A = \left[ {\begin{array}{*{20}{c}}
{\cos \alpha }&{ – \sin \alpha }\\
{\sin \alpha }&{\cos \alpha }
\end{array}} \right]$
${A^2} = \left[ {\begin{array}{*{20}{c}}
{\cos \alpha }&{ – \sin \alpha }\\
{\sin \alpha }&{\cos \alpha }
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{\cos \alpha }&{ – \sin \alpha }\\
{\sin \alpha }&{\cos \alpha }
\end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}}
{\cos 2\alpha }&{ – \sin 2\alpha }\\
{\sin 2\alpha }&{\cos 2\alpha }
\end{array}} \right]$
${A^3} = \left[ {\begin{array}{*{20}{c}}
{\cos 2\alpha }&{ – \sin 2\alpha }\\
{\sin 2\alpha }&{\cos 2\alpha }
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{\cos \alpha }&{ – \sin \alpha }\\
{\sin \alpha }&{\cos \alpha }
\end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}}
{\cos 3\alpha }&{ – \sin 3\alpha }\\
{\sin 3\alpha }&{\cos 3\alpha }
\end{array}} \right]$
Simiarly ${A^{32}} = \left[ {\begin{array}{*{20}{c}}
{\cos 32\alpha }&{ – \sin 32\alpha }\\
{\sin 32\alpha }&{\cos 32\alpha }
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&{ – 1}\\
1&0
\end{array}} \right]$
$ \Rightarrow \cos 32\alpha = 0$ and $\sin 32\alpha = 0$
$ \Rightarrow 32\alpha = \left( {4n + 1} \right)\frac{\pi }{2},n \in 1$
$\alpha = \left( {4n + 1} \right)\frac{\pi }{{64}},n \in 1$
$\alpha = \frac{\pi }{{64}}$ for $n = 0$
