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How to find unique solution of X

  1. Dec 31, 2013 #1
    Hi All,

    I have a simple problem

    consider two matrix equations

    A1 and A2 are 3X3 matrix. Moreover, A2 is a positive, definite, symmetric matrix. B1 and B2 are two 3X1 column vectors. The A1, A2, B1, B2 are known. I need to find unique solution of X vector from these two equations. Is there any suitable way to do that?

    Can I do it by using Lagrange multiplier? It is not a homework problem. I am struggling this by two days but didn't get success.

    Thanks in advance

  2. jcsd
  3. Dec 31, 2013 #2


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    Your system is over-defined, Each of your two matrix equations represents three equations, so you have a total of 6 equations and only three unknowns. Let's assume A1 and A2 are non-singular. Unless:
    [tex]A2^{-1}B2 = A1^{-1} B1[/tex]
    there is no solution. If this is satisfied, then
    [tex] X = A2^{-1}B2 = A1^{-1} B1[/tex]
  4. Dec 31, 2013 #3
    Thank you very much for your reply.

    Actually, I want to use first equation is a constraint for second one. If so how do I combine the first equation with the second equation by some constrained minimization method like Lagrange multiplier method.

    in fact, my two matrices are
    A1=[[-1 -1 0] [1 0 -1] [0 1 1]]
    A2=[[a11 a12 a13] [a12 a22 a23] [a13 a23 a33]]

    can you give some clue, how do I proceed?
  5. Dec 31, 2013 #4


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    Do you know B1 and B2? If so, what are they?
  6. Dec 31, 2013 #5


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    Homework Helper

    it A2 is positive definite, it s nonsingular, you have a unique solution of A2X = B2.

    If that solution does not satisfy A1X = B1, you need to reconsider what you are really trying to do here. You can't "do something impossible" by trying to use A1X = B1 as constraint equations, or whatever.
  7. Dec 31, 2013 #6
    I know the values of B1 and B2 but they are different
    B1=[-12.212 0.56 0.65]

    B2=[-68.49 -68.49 0.0]
    Also the A2 matrix is
    A2=[[151.8 2.4 -149.4][2.4 151.8 149.4] [-149.4 149.4 298.8]]

    These values are related with my system
    Last edited: Dec 31, 2013
  8. Dec 31, 2013 #7


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    It's easy to see that no solutions to your problem exist. Expand your equation A1 X = B1, to get:
    -x1 - x2 = b1
    x1 - x3 = b2
    x2 + x3 = b3

    Now add all three equations together, giving 0 = b1 + b2 + b3. Your B1 does not satisfy this equation. Therefore no solutions exist.
  9. Jan 1, 2014 #8
    Yes, phyzguy, you are right. I know that. Moreover, A1 is rank deficient and determinant of A2 is 0. therefore, it will not be giving a solution in a simple way. but i need that solution. please help me out in this regard. please give me some clue if anyone aware about this kind of problem.
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