This matrix question for 2x2 Mx, how to find A^2010?

[karimkhan]

[h=3]If y is 2pi/67 and 2x2 matrix : A= ( cos(y) sin(y), -sin(y) cos(y) ). Then A^2010?[/h]
I also have uploaded the image here..

 

[karimkhan]

I tried and got ans 1 by (A^2)^1005=1,

but in option A, B, and D there's same answer 1!!!

 

[daon2]

I tried and got ans 1 by (A^2)^1005=1,

but in option A, B, and D there's same answer 1!!!

This is a rotation matrix of order 67. Since 67 divides 2010, it is true that A^2010 = (A^67)^30 = I^30 = I.

 

[karimkhan]

This is a rotation matrix of order 67. Since 67 divides 2010, it is true that A^2010 = (A^67)^30 = I^30 = I.

Thanks Daon,

But i didnt get completely your solution, can you please elaborate it???
how can we say it rotation matrix of order 67?

(A^2)^1005 also giving I ... so

please explain ....

 

[karimkhan]

here using theta we can rotate the matrix, thats why its rotational i think, if im not wrong...

 

[karimkhan]

ok i got it

A^k =
[cos(ky) sin(ky)]
[-sin(ky) cos(ky)].

So, A^2010 equals
[cos(2010π/67) sin(2010π/67)]
[-sin(2010π/67) cos(2010π/67)].

Since sine and cosine have period 2π, and 2010/67 = 30, we can rewrite this as
[cos(30π) sin(30π)]
[-sin(30π) cos(30π)] =

[1 0]
[0 1].

 

[HallsofIvy]

Yes, the matrix \(\displaystyle A= \begin{bmatrix}cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta)\end{bmatrix}\) is a rotation matrix (rotation around the z-axis through angle \(\displaystyle -\theta\)) and so \(\displaystyle A^n\) is rotation through angle \(\displaystyle -n\theta\).
That would be, of course, \(\displaystyle A^{2010}= \begin{bmatrix}cos(2010\theta) & sin(2010\theta) \\ -sin(2010\theta) & cos(2010\theta)\end{bmatrix}\).