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Let $A$ be the $2 \times 2$ matrix $\begin{pmatrix} \sin\frac{\pi}{18}&-\sin \frac{4\pi}{9} \\ \sin \frac{4\pi}{9}&\sin \frac {\pi}{18} \end{pmatrix}$. Then the smallest number $n \in \mathbb{N}$ such that $A^{n}=1$ is.

1. $3$
2. $9$
3. $18$
4. $27$
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+1 vote

By solving Characteristic Equation

$\lambda = sin(\pi/18) +\mathbf{i} sin(4\pi/9 )$

$\lambda = sin(\pi/18) -\mathbf{i} sin(4\pi/9 )$

By cayley hamilton theorem

$\lambda ^n=1$

and solving for n(minimum) get n= 9

so Option  B is correct

by (223 points)

+1 vote