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Matrix XA=B, AY=B

  1. Mar 20, 2008 #1
    1. The problem statement, all variables and given/known data

    I have two 3x3 matrices A and B

    and my problem is to find XA=B and AY=B

    Isn't XA=B and AY=B

    2. Relevant equations



    3. The attempt at a solution

    [tex]X=A^-^1 B[/tex]

    [tex]Y=A^-^1 B[/tex]

    or I am wrong?
     
  2. jcsd
  3. Mar 20, 2008 #2

    malawi_glenn

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    remember that matrix multiplication is non-commutative

    so yes you are wrong on the [tex]X=A^-^1 B[/tex], if XA=B
     
  4. Mar 20, 2008 #3
    Can you tell me please, what is the correct formula?
     
  5. Mar 20, 2008 #4

    D H

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    You can pre- or post-multiply a matrix equality by a matrix to yield another matrix equality: If A=B then CA = CB and AD = BD for all matrices C, D of the correct dimensionality. Use this plus the fact that matrix multiplication is associative to find a form that eliminates the matrix A from XA=B.
     
  6. Mar 20, 2008 #5
    Sorry, but I don't understand what you want to say... Can you please write what equals X, and what equals Y?

    [tex]X= ??
    Y=??[/tex]
     
  7. Mar 20, 2008 #6
    Why don't you write out XA = AY, compare the entries and figure out what X and Y could be.
     
  8. Mar 20, 2008 #7

    malawi_glenn

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    We dont give answers here for free!

    You must multiply matrices from the correct order!

    Take this for example:

    XA=B

    How would you do to eliminate X ? By multiplying inv(A) from the left?:
    Then you'll get:

    inv(A)XA = inv(A)B

    And that is NOT you want right? So how would you do it?
     
  9. Mar 20, 2008 #8
    You can multiply from both sides in matrix:
    If you have
    A=B
    Then you can have
    CA = CB
    OR
    AC = BC

    The C can come in on the right or left. But you have to keep track.
     
  10. Mar 20, 2008 #9
    So X=inv(a)B
    and inv(a)AY=inv(a)B , so Y=inv(a)B

    hm....
     
  11. Mar 20, 2008 #10

    malawi_glenn

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    Well the Y is correct, but not the X, try again

    you have:

    XA=B
     
  12. Mar 20, 2008 #11
    XA=B

    inv(x)XA=inv(x)B

    A=inv(x)B ??
     
  13. Mar 20, 2008 #12

    malawi_glenn

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    well yes, but you want to solve for X right?

    Read post #8, by K.J.Healey (T, 19:38)
     
  14. Mar 20, 2008 #13

    D H

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    You are supposed to be solving for X, not A. Why are you insisting on multiplying on the left? XA has two sides (left and right), and so does B.
     
  15. Mar 20, 2008 #14
    XA=B

    XA(inv (a))=B(inv (a))

    X=B(inv(a))

    like this?
     
  16. Mar 20, 2008 #15

    malawi_glenn

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    yes! Good job :-)
     
  17. Mar 20, 2008 #16
    Thank you very much, guys. I love youuuu.... :smile:
     
  18. Mar 24, 2011 #17
    Hi all,

    Pardon me for digging up this old thread, but it is related to what I am asking.
    Is there some matrix manipulation that transforms equation of type xA = B into A'x' = B'? in other words, transform Unknown.KNOWN = KNOWN into KNOWN.Unknown = KNOWN.
    Thanks a bunch!
     
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