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Solving for a Matrix

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Skjermbilde_2012_03_08_kl_11_13_40_AM.png

    3. The attempt at a solution

    I think I'm oversimplifying this problem. Why can't I just write:

    AX + B =CA
    X + A'B = C
    X = C - A'B

    ?
     
  2. jcsd
  3. Mar 8, 2012 #2

    lanedance

    User Avatar
    Homework Helper

    does your notation mean [itex] A' = A^{-1}[/itex]?

    ... if so, you can't do what you propse because matrix multiplication does not necesarrily commute. ionce you fix that, you're not far off the answer
     
  4. Mar 8, 2012 #3
    Yes, that's what I intended my notation to convey.

    Do I have to write:

    AX + B = CA
    AX = CA - B
    A'(AX) = A'(CA - B)
    IX = A'CA - A'B
    X = A'CA - A'B

    ?
     
  5. Mar 10, 2012 #4

    lanedance

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    Homework Helper

    that looks better
     
  6. Mar 11, 2012 #5

    Mark44

    Staff: Mentor

    The notation A' is sometimes used to mean the transpose of matrix A. For inverses, I don't think I've ever seen ' used to indicate the inverse.

    It's just as easy to write an exponent of -1 as an exponent of 2 or 3, and the intent is much clearer.

    At the bottom of the input pane, click the Go Advanced button. This opens a menu of icons at the top of the input pane. The X2 button lets you write exponents, which it does by inserting [ sup ] and [ /sup ] tags (without the spaces).

    You can also do this manually, like so: A[noparse]-1[/noparse]. I have inserted some other tags so that you could see the sup tags. Without those other tags, what I wrote renders like this: A-1.
     
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