MCQ
Let $A = \left( {\begin{array}{*{20}{c}}
{\cos \,\alpha }&{ - \sin \,\alpha }\\
{\sin \,\alpha }&{\cos \,\alpha }
\end{array}} \right)$, $\left( {\alpha  \in R} \right)$ such that ${A^{32}} = \left( {\begin{array}{*{20}{c}}
0&{ - 1}\\
1&0
\end{array}} \right)$. Then a value of $\alpha $ is
  • A
    $0$
  • B
    $\frac{\pi }{{16}}$
  • C
    $\frac{\pi }{{32}}$
  • $\frac{\pi }{{64}}$

Answer

Correct option: D.
$\frac{\pi }{{64}}$
d
$A = \left[ {\begin{array}{*{20}{c}}
{\cos \alpha }&{ - \sin \alpha }\\
{\sin \alpha }&{\cos \alpha }
\end{array}} \right]$

${A^2} = \left[ {\begin{array}{*{20}{c}}
{\cos \alpha }&{ - \sin \alpha }\\
{\sin \alpha }&{\cos \alpha }
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{\cos \alpha }&{ - \sin \alpha }\\
{\sin \alpha }&{\cos \alpha }
\end{array}} \right]$

$ = \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 $n = 0$

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