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Inverse of a 3x3 matrix of trigometric functions

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data
    I need to find the inverse of the following matrix:
    [(cos*sin) (-cos) (sin2)]
    [(cos2) (sin) (-cos*sin)]
    [(sin) 0 (cos)]

    2. Relevant equations
    gauss-jordan elimination

    3. The attempt at a solution
    I know that in general, gauss-jordan elimination can be used to solve the inverse of a 3x3, but given this matrix, I don't see how that method could be particularly useful or efficient. I tried to get started using gauss-jordan but expressions quickly became to large to deal with. Are there any slightly less involved ways to find the inverse here?
  2. jcsd
  3. Apr 7, 2010 #2


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    Homework Helper

    It might ease the pain to make substitutions a = sin, b = cos, then substitute back when you are done. I'm guessing (w/o actually solving it) that some of the terms will simplify in the end.
  4. Apr 8, 2010 #3
    ok, crisis averted. it probably would have been helpful to note that the first part of the problem was to prove that the matrix is orthogonal, and then find the inverse. What I didn't realize was that for an orthogonal matrix A-1=AT which makes the problem much easier

    thanks for you help anyway
  5. Apr 8, 2010 #4

    D H

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    Staff Emeritus
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    You sure about that? The matrix presented in the original post is *not* an orthogonal matrix. This is:

    \cos x \sin x & -\cos x & \sin^2x \\ \cos^2x &\sin x & \cos x \sin x \\ \sin x & 0 & -\cos x

    Did you mean that instead (or something like it)?
  6. Apr 8, 2010 #5
    yes, as a matter of fact I did mean something like that. The upper right hand term in the original matrix should be -sin2(x), not just sin2(x). Thanks for catching that.
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