# Math Help - Find image of u under matrix transformation

1. ## 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.

2. 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.

3. ## My calculation of second coordinate

. If

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

4. Originally Posted by jennifer1004
. If

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.