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Let $\theta=\frac{2 \pi}{67}$. Consider the matrix $$A = \begin{pmatrix} \cos\theta & \sin\theta\\ -\sin \theta& \cos \theta \end{pmatrix} $$ Then the matrix $A^{2010}$ is?

My approach $$A^2 = \begin{pmatrix} \cos2\theta & \sin2\theta\\ -\sin 2\theta& \cos 2\theta \end{pmatrix} $$

$2010$ is a multiple of $67$. So I'm trying to convert in such format by taking $A$ matrix as power of $2$ and multiplying. But $2048$ is the nearest $2$ power term. So help please. I dont know if you can understand what I did!! Sorry for being not explaining properly

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  • $\begingroup$ Is this from TIFR-GS 2015 ? $\endgroup$ – Srinivas K Mar 7 '15 at 12:37
  • $\begingroup$ No ISI Sample... $\endgroup$ – N S Mar 7 '15 at 12:38
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Observe that when you multiply the matrices $$\begin{pmatrix} \cos\theta & \sin\theta\\ -\sin \theta& \cos \theta \end{pmatrix} * \begin{pmatrix} \cos\phi & \sin\phi\\ -\sin \phi& \cos \phi \end{pmatrix} = \begin{pmatrix} \cos(\theta+\phi) & \sin(\theta+\phi)\\ -\sin (\theta+\phi)& \cos (\theta+\phi) \end{pmatrix}.$$

(You can show this with the multiangle formulae for $\sin$ and $\cos$.) So we have

$$A^{67} = \begin{pmatrix} \cos67\theta & \sin67\theta\\ -\sin 67\theta& \cos 67\theta \end{pmatrix} = \begin{pmatrix} \cos2\pi & \sin2\pi\\ -\sin 2\pi& \cos 2\pi \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I.$$ $$\therefore A^{2010} = (A^{67})^{30} = I.$$ (I tried to put the end bit as a 'spoiler' which you have to mouse over to view, but I couldn't work out how to!)

As mentioned by other people, applying the matrix $$M_\theta = \begin{pmatrix} \cos\theta & \sin\theta\\ -\sin \theta& \cos \theta \end{pmatrix}$$ has the action of rotation a vector (anticlockwise) by $\theta$. Thus doing it repeatedly simply means that they sum: $M_\theta * M_\phi = M_{\theta + \phi}$.

Hopefully this helps! :)

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By noticing that $67|2010$ you are actually done. This means that the rotation $A^{2010}$ corresponds to an integer number of full rotations. If $A(\theta)$ is the matrix $A$ as above with any angle, then notice that $$(A(\theta))^k = A(k\theta)$$ So $$(A(\frac{2\pi}{67}))^{2010} = A(\frac{2\pi}{67}\cdot 2010) = A(2\pi \cdot 30)$$ Now notice that $A(\theta + 2\pi) = A(\theta)$ to see that $A(\frac{2\pi}{67})^{2010} = A(0) = I_2$

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The matrix $A$ is rotation by $2\pi/67$. Take the vector $(1,0)$. If you rotate it $2010$ times the total angle is $2010*2\pi/67=60\pi$. So it comes back to $(1,0)$.Similar argument shows that the $(0,1)$ also remains at $(0,1)$. So, this is the identity matrix.

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Building on what was said, you can view $R(\theta)$ as a rotation matrix in the complex plane. Indeed, for a vector written as $v=x+iy$ , then the rotation matrix is simply the operator $e^{i\theta}$. This is shown by Euler's formula. From this, you know that ${(e^{i\theta})}^n=(e^{ni\theta})$ which links back to what was shown.

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