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  • Confused! kindly explain, - Matrices and Determinants - JEE Main-4

Let A=\begin{pmatrix} \cos \alpha - &\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix},(\alpha \: \epsilon \: \mathbb{R}) such that A^{32}=\begin{pmatrix} 0 &-1 \\ 1 &0 \end{pmatrix}. Then a value of \alpha is:
 

  • Option 1)

    \frac{\pi}{16}

  • Option 2)

    0

  • Option 3)

    \frac{\pi}{32}

     

  • Option 4)

    \frac{\pi}{64}

 

Answers (1)

best_answer

A=\begin{pmatrix} \cos \alpha - &\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix},(\alpha \: \epsilon \: \mathbb{R})

A^{2}={\begin{pmatrix} \cos \alpha &-\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}\begin{pmatrix} \cos \alpha &-\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}}

       =\begin{pmatrix} \cos ^{2}\alpha -\sin ^{2}\alpha & -\cos \alpha \sin\alpha-\sin\alpha\cos \alpha\\ \sin\alpha\cos \alpha+\cos \alpha \sin\alpha &-\sin^{2}\alpha+\cos^{2}\alpha \end{pmatrix}

A^{2}=\begin{pmatrix} \cos 2\alpha &-\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{pmatrix}

Similarly we can observe that

A^{n}=\begin{pmatrix} \cos n\alpha &-\sin n\alpha \\ \sin n\alpha & \cos n\alpha \end{pmatrix}

A^{32}=\begin{pmatrix} \cos 32\alpha &-\sin 32\alpha \\ \sin 32\alpha & \cos 32\alpha \end{pmatrix}=\begin{pmatrix} 0 &-1 \\ 1&0 \end{pmatrix}

\cos 32 \: \alpha=0

32\alpha =\frac{\pi}{2}

\alpha =\frac{\pi}{64}

\sin32\: \alpha =1\Rightarrow 32\: \alpha =\frac{\pi}{2}

\alpha=\frac{\pi}{64}


Option 1)

\frac{\pi}{16}

Option 2)

0

Option 3)

\frac{\pi}{32}

 

Option 4)

\frac{\pi}{64}

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