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Solving 3D matrix equation

  1. Jul 21, 2011 #1

    I'm pretty rusty with solving linear equations, my equations are:

    ai = bijkcjk

    i = 1,... n
    j, k = 1,... m

    Would like to know c, given a and b. Need somewhere to start, without having to cover 3 years of notes that I took 5 years ago, thanks if anyone can point me in the right direction!

    Comments: I know I could compact j and k into a single dimension and reduce it to a "simple" matrix equation but this process discards valuable information that I need later on. Or at least makes it difficult to retrieve this information, so I'd like to explore a way of solving the equations outright. Thanks.
  2. jcsd
  3. Jul 21, 2011 #2


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    I'm not sure that you can solve for the individual values of c. It looks to me like you will have more uknown values than equations. If the indices run from 1 to 3, c will have 9 values but because of the contraction on both i and k, you will have only three equations.
  4. Jul 21, 2011 #3
    Good point, I forgot to mention that. Let's say n = m^2, and the equations are consistent.
  5. Jul 21, 2011 #4
    Yep - I think HallsofIvey is correct. The term we could use here is an underdetermined system.
  6. Jul 21, 2011 #5
    Yeah for a system of equations (unless those are complex nums, which I thought on first glance) you need the num of differing equations= the num of variables.
  7. Jul 22, 2011 #6
    Ok, that condition is satisfied by setting n=m^2, now i,j = 1,... m, k = 1,... m^2 so we have m^2 equations in m^2 unknowns.
  8. Jul 22, 2011 #7

    Ben Niehoff

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    If n = m^2, then this is simple. First you need to "unwrap" c_jk into a single column vector c_l, where l now runs from 1 to m^2. b_ijk must be similarly unwrapped into b_il. In both cases, this is easily accomplished by defining

    l = 3(j-1) + (k-1) + 1 = 3j + k - 3

    l now runs from 1 to 9 as j and k run from 1 to 3, with each value of j and k being mapped to a unique l. Notice all I've done here is expand l in base 3. For general m^2, you would have

    l = m(j - 1) + (k - 1) + 1 = mj + k - m

    Once you have re-written your equation as

    a_i = b_il c_l

    it is easy to solve by standard linear algebra.

    Finally, since each (j,k) pair maps to a unique l, it is also easy to get back c_jk the way you want it.
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