Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

up vote 2 down vote accepted

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.

share|improve this answer
    
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
1  
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
1  
$a_{11}=0$, $a_{12}=1$, $a_{21}=-1$, $a_{22}=0$. –  Anonymous Sep 24 '13 at 4:34
1  
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).

share|improve this answer

Yes, you would use the matrix

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.