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Matrix (2 actually)

  1. Sep 23, 2008 #1
    1. The problem statement, all variables and given/known data
    #1. Solve the system:
    3cos(a) + cos(b) +5sin(2a)=2
    cos(a)=cos(b)+sin(2a)
    7cos(a) + 13sin(2a)=-7cos(b)

    3. The attempt at a solution
    #1. I decided to substitute variables for sines and cosines. So I made cos(a) = x, cos(b)=y and sin(2a)=z, as a result of which, my equations now look like:
    3x+y+5z=2
    x-y-z=0
    7z+7y+13z=0
    after which I created the matrix, here is a picture of how I solved it. So I got -3/8 for "x" and entered it (it only asks for cos(a)), but it tells me I'm wrong. Am I being dumb and making some obvious error? Because I entered all equations with variable values into my calculator and they work out...
    [​IMG]
    Thanks...
     
    Last edited by a moderator: Sep 24, 2008
  2. jcsd
  3. Sep 24, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    "cos" has no value! Do you mean cos(a) or cos(b)?


    You can't do that. cos(a) and sin(2a) are not independent. sin(2a)= 2sin(a)cos(a)
    Are you sure the system has a solution? There are 3 equations in only two variables, a and b. In order to have a solution one equation must be dependent on the other two. (Not necessarily linearly dependent- perhaps through a trig identity.)
    I think I would be inclined to solve the second equation for cos(b): cos(b)= cos(a)- sin(2a) and then substitute that into the first and third equations. Now you have two equations for the single variable, a. The third equation looks like it should become especially simple. Again, that is possible only if the two equations are dependent.

     
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