Question
If \[A = \left[ \begin{array}{l}
\cos \theta \,\,\,\,\,\,\,\, - \sin \theta \\
\sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\cos \theta
\end{array} \right],\] then the matrix $${A^{ - 50}}$$ when $$\theta = \frac{\pi }{{12}},$$ is equal to:
A.
\[\left[ \begin{array}{l}
\,\frac{1}{2}\,\,\,\,\,\,\,\,\,\, - \frac{{\sqrt 3 }}{2}\\
\frac{{\sqrt 3 }}{2}\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{2}
\end{array} \right]\]
B.
\[\left[ \begin{array}{l}
\frac{{\sqrt 3 }}{2}\,\,\,\,\,\,\,\,\,\, - \frac{1}{2}\\
\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{\sqrt 3 }}{2}
\end{array} \right]\]
C.
\[\left[ \begin{array}{l}
\,\,\,\frac{{\sqrt 3 }}{2}\,\,\,\,\,\,\,\,\,\,\,\frac{1}{2}\\
- \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\frac{{\sqrt 3 }}{2}
\end{array} \right]\]
D.
\[\left[ \begin{array}{l}
\,\,\,\,\,\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\frac{{\sqrt 3 }}{2}\\
- \frac{{\sqrt 3 }}{2}\,\,\,\,\,\,\,\,\,\,\,\frac{1}{2}
\end{array} \right]\]
Answer :
\[\left[ \begin{array}{l}
\,\,\,\frac{{\sqrt 3 }}{2}\,\,\,\,\,\,\,\,\,\,\,\frac{1}{2}\\
- \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\frac{{\sqrt 3 }}{2}
\end{array} \right]\]
Solution :
\[A = \left[ \begin{array}{l}
\cos \theta \,\,\,\,\,\,\, - \sin \theta \\
\sin \theta \,\,\,\,\,\,\,\,\,\,\cos \theta
\end{array} \right]\]
$$ \Rightarrow \,\,\left| A \right| = 1$$
\[adj\left( A \right) = {\left[ {\begin{array}{*{20}{c}}
{ + \cos \theta }&{ - \sin \theta }\\
{ + \sin \theta }&{ + \cos \theta }
\end{array}} \right]^T}\]
\[\,\,\,\,\, = \left[ \begin{array}{l}
\,\,\cos \theta \,\,\,\,\,\,\,\,\,\sin \theta \\
- \sin \theta \,\,\,\,\,\,\,\,\cos \theta
\end{array} \right]\]
\[ \Rightarrow \,\,{A^{ - 1}} = \left[ \begin{array}{l}
\,\cos \theta \,\,\,\,\,\,\,\,\,\,\sin \theta \\
- \sin \theta \,\,\,\,\,\,\,\,\cos \theta
\end{array} \right] = B\]
\[{B^2} = \left[ \begin{array}{l}
\,\,\cos \theta \,\,\,\,\,\,\,\,\,\sin \theta \\
- \sin \theta \,\,\,\,\,\,\,\,\cos \theta
\end{array} \right]\left[ \begin{array}{l}
\,\,\cos \theta \,\,\,\,\,\,\,\,\,\sin \theta \\
- \sin \theta \,\,\,\,\,\,\,\,\cos \theta
\end{array} \right]\]
\[ = \left[ \begin{array}{l}
\,\,\cos 2\theta \,\,\,\,\,\,\,\,\,\sin 2\theta \\
- \sin 2\theta \,\,\,\,\,\,\,\,\cos 2\theta
\end{array} \right]\]
\[ \Rightarrow \,\,{B^3} = \left[ \begin{array}{l}
\,\,\cos 3\theta \,\,\,\,\,\,\,\,\,\sin 3\theta \\
- \sin 3\theta \,\,\,\,\,\,\,\,\cos 3\theta
\end{array} \right]\]
\[ \Rightarrow \,\,{A^{ - 50}} = {B^{50}} = \left[ \begin{array}{l}
\,\,\,\,\cos \left( {50\theta } \right)\,\,\,\,\,\,\,\sin \left( {50\theta } \right)\\
\, - \sin \left( {50\theta } \right) \,\,\,\,\,\,\,\,\,\cos \left( {50\theta } \right)
\end{array} \right]\]
\[{\left( {{A^{ - 50}}} \right)_{\theta = \frac{\pi }{12}}} = \left[ \begin{array}{l}
\,\frac{{\sqrt 3 }}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{2}\\
- \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\frac{{\sqrt 3 }}{2}
\end{array} \right]\]
$$\left[ {\because \,\,\cos \left( {\frac{{50\pi }}{{12}}} \right) = \cos \left( {4\pi + \frac{\pi }{6}} \right) = \cos \frac{\pi }{6} = \frac{{\sqrt 3 }}{2}} \right]$$