Results 1 to 4 of 4

Math Help - Find image of u under matrix transformation

  1. #1
    Banned
    Joined
    Jan 2009
    Posts
    58

    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

    f(-2)= -1/2 ......-rad3/2....... -2 = (-1/2)(-2)+(-rad3/2)(-3)
    (-3).... rad3/2.... -1/2... .*.. -3... (rad3/2)+(-1/2)(-3)

    = (2+3rad3)/2 = 3.59808
    = (3-2rad3)/2 = -.23205

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2008
    Posts
    36
    Thanks
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Jan 2009
    Posts
    58

    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2008
    Posts
    36
    Thanks
    1
    Quote Originally Posted by jennifer1004 View Post
    . 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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 30th 2011, 04:36 PM
  2. Complex analysis, find the image of a transformation
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: July 14th 2010, 11:54 AM
  3. Image of a linear transformation matrix...
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: January 18th 2010, 11:34 AM
  4. Replies: 1
    Last Post: October 27th 2009, 01:09 AM
  5. Image of a transformation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 25th 2009, 12:14 AM

Search Tags


/mathhelpforum @mathhelpforum