# Rotation of matrices

I am doing rotation of matrices at the moment, I know that if I want to rotate a point, let's say (2,1) 90 degrees clockwise, I have to multiply the matrix [ 2 1 ] * [0 1, -1 0] , but how do I find these points? if Iam asked for instance to rotate it 54 degrees anticlock wise, what matrix would I have to multiply it for? is there any formula for that?

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You would use the following rotation matrix, $A_\theta = \left[ \begin{matrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{matrix} \right]$, where $\theta$ represents an anticlockwise rotation. $(2,1)$ can be represented as a vector $[\begin{matrix} 2 \\ 1\end{matrix}]$ and multiplied on the right by the rotation matrix to get your answer.

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cheeers ;) , also..I have to do it through linear transformation,it says my example that to find a clockwise rotation of 90 degrees I have to find the image of the unit vectors under the tranformation. It follows by saying: [ 1 0] to [0 -1] and that maps the vector [0 1] to [1 0] and thus the matrix is [0 1 , -1 0]. And that's it, it is all what my book says..why [ 1 0] to start with? and not [ 1 1] , then where does the [ 0 -1] coming from? and the others? sorry, but my coursebook is a bit lane and when I go online to try to find answers I just come up with kilometric formulas .. –  Maximilian1988 Sep 24 '13 at 4:14
I'll help you if you checkmark my answer, as it is clearly more thorough than Tpofofn's answer! I need to increase my reputation here ;) –  Anonymous Sep 24 '13 at 4:15
If you multiply the column vector [1 0] by the rotation matrix with theta as 270 degrees, because counterclockwise, then you will get the resultant column vector of dimension 1 x 2: [0 -1]. The same happens for [0 1] to [1 0]. The matrix is [0 1, -1 0] because cos (270) = 0, -sin (270)= -1, sin(270) = 1, and cos (270) = 0. –  Anonymous Sep 24 '13 at 4:23
$a_{11}=0$, $a_{12}=1$, $a_{21}=-1$, $a_{22}=0$. –  Anonymous Sep 24 '13 at 4:34
It's coming from the rotation matrix in my answer, with $\theta=270$ degrees. $a_{11}=\cos(270)$, $a_{12}=-\sin(270)$, $a_{21}=\sin(270)$, $a_{22}=\cos(270)$. Look on bouma's wikipedia link for more details. –  Anonymous Sep 24 '13 at 4:39

Use http://en.wikipedia.org/wiki/Rotation_matrix and plug in the appropriate angle (in radians).

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Yes, you would use the matrix

$$\left[ \begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array} \right]$$

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