# Find image of u under matrix transformation

• Feb 2nd 2009, 10:12 AM
jennifer1004
Find image of u under matrix transformation
Find the image of u under the matrix formation f:

f: $R^2-->R^2$ is a counterclockwise rotation through $2/3$ pi raidans; u = the vector
-2
-3

I started by writing f(u)= cos(2/3 pi) -sin(2/3 pi)
sin(2/3 pi) cos(2/3 pi) * u

My book does not explain how to do this at all. I was actally using the answer to another problem in a solutions manual as a guide. Can anyone confirm if I did this problem right? Thanks!

P.S. I used all the periods to try and space the matrix and vector out since I'm not sure how to type it out correctly.
• Feb 2nd 2009, 12:19 PM
mylestone
You got the first coordinate of the image of $u$ correct, but I think there is an error with the second. I got (roughly) -3.23.

In general, the matrix of transformation for a counterclockwise rotation of $\theta$ (about the origin) in $\mathbb{R} ^2$ is $\left[ \begin{array}{cc} cos \theta & -sin \theta \\ sin \theta & cos \theta \end{array} \right]$. If $u = \left[ \begin{array}{c} -2 \\ -3 \end{array} \right]$, and you want to find the image of $u$ under this transformation all you have to do is multiply matrix*vector. Looks like you know this already, but I'm not sure where your matrix came from and thought that a clean presentation would be enough to clear-up whatever arithmetic errors may have occurred.
• Feb 2nd 2009, 01:23 PM
jennifer1004
My calculation of second coordinate
http://www.mathhelpforum.com/math-he...377d09f3-1.gif. If http://www.mathhelpforum.com/math-he...63711069-1.gif

I replaced the thetas with (2/3 pi) radians and to get the second coordinate, I calcuated sin(2/3 pi)(-2)+cos(2/3 pi)(-3)=(radical3)/2)*(-2)+(-1/2)(-3)=[3-2(radical3)]/2= -.23205
• Feb 2nd 2009, 01:41 PM
mylestone
Quote:

Originally Posted by jennifer1004
http://www.mathhelpforum.com/math-he...377d09f3-1.gif. If http://www.mathhelpforum.com/math-he...63711069-1.gif

I replaced the thetas with (2/3 pi) radians and to get the second coordinate, I calcuated sin(2/3 pi)(-2)+cos(2/3 pi)(-3)=(radical3)/2)*(-2)+(-1/2)(-3)=[3-2(radical3)]/2= -.23205

My slip--you're right. And now I get the notation for your matrix as well. Nice work.