5 votes

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.

- $3$
- $9$
- $18$
- $27$

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

0 votes

$A = \begin{pmatrix} \sin \frac{\pi}{18} & -\sin \frac{4\pi}{9} \\ \sin \frac{4\pi}{9}& \sin \frac{\pi}{18} \end{pmatrix}$ $= \begin{pmatrix} \sin 10^{\circ} & -\sin 80^{\circ} \\ \sin 80^{\circ} & \sin 10^{\circ} \end{pmatrix}$ $= \begin{pmatrix} \sin (90^{\circ}-80^{\circ}) & -\sin 80^{\circ} \\ \sin 80^{\circ} & \sin (90^{\circ}-80^{\circ}) \end{pmatrix}$

$A= \begin{pmatrix} \cos 80^{\circ} & -\sin 80^{\circ} \\ \sin 80^{\circ} & \cos 80^{\circ} \end{pmatrix}$

Here, note that $A$ is a rotation matrix with $\theta = 80^{\circ}$.

If we have $2$ rotation matrices $A(\theta)$ and $A(\phi)$ then multiplication of these $2$ matrices will be $A(\theta + \phi)$ means corresponding angles will be added in the multiplication of these 2 matrices.

So, if we have a rotation matrix $A(\theta)= \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$ then $A^n(\theta)= \begin{pmatrix} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \end{pmatrix}$

It can be proved using induction on $n.$

So, in this question,

$A^n= \begin{pmatrix} \cos n*80^{\circ} & -\sin n*80^{\circ} \\ \sin n*80^{\circ} & \cos n*80^{\circ} \end{pmatrix} = I$

Assuming, set of natural numbers starts from $1$ and for smallest integer value of $n \in \mathbb{N}$

$n*80^{\circ} = 4*180^{\circ} \Rightarrow n= 9$