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Linear Algebra: If AX=B, and X and B are known, find A.

  1. Oct 3, 2008 #1
    I wasn't sure if linear algebra was precal or not... hopefully I posted this in the right place.

    1. The problem statement, all variables and given/known data
    A, X and B are all matrices.

    X and B are known.

    AX=B

    How do you find A?


    2. Relevant equations
    None.


    3. The attempt at a solution
    I know that if you needed to find X, you could construct a matrix where:
    [A|B], transform to RREF, and read off the solutions which would be A, right? But since AX does not equal XA, I guess you can't do that here.

    Maybe not...

    Does it have something to do with matrix inverses?

    I can't find this in my textbook anywhere!! Thanks in advance for any help.
     
  2. jcsd
  3. Oct 3, 2008 #2

    gabbagabbahey

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    If [itex]X[/itex] is invertable, then [itex]AX=B \Rightarrow AXX^{-1}=A=BX^{-1}[/itex] right? ;)
     
  4. Oct 3, 2008 #3
    Ohh yeah... Because the XX^-1 will give you the identity, which will be kind of like one, right?

    But if I were to multiply the inverse of X from on the left of X and B instead of on the right, wouldn't that affect the answer?

    (Because shouldn't BX^-1 be different from (X^-1)B?)

    Thank you so much, I really really appreciate it!!!
     
  5. Oct 3, 2008 #4

    gabbagabbahey

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    Yes, you need to make sure that you multiply both sides from the right, order is important in matrix multiplication. If you were to multiply from the left you would have [itex]X^{-1}AX=X^{-1}B[/itex] but this wouldn't be of any use since [itex]X^{-1}AX \neq A[/itex].
     
  6. Oct 3, 2008 #5
    And you can't multiply so that the inverse of X is between A and X?
     
  7. Oct 3, 2008 #6

    gabbagabbahey

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    No, you have to multiply the whole expression either from the left or from the right, not just part of the expression.
     
  8. Oct 3, 2008 #7
    here it is a method if you are interested to know how to find A in case X is not invertible. Because this would narrow down your choices, since your matrix X would need to be a square one. So, in general here it is what u would need to do:

    just write the matrix A how it should look like, then multiply it with the matrix X, and use the def. of the equality of two matrices. YOu will get some systems of lin. equations, solve those systmes, and the solutions will be the entries of matrix A.

    P.S. Say someone says. let matrix A me an (mxn) matrix, matrix X be an (nxp) matrix, and B be an (mxp) matrix. then if

    AX=B, find A where X, and B are known. YOu cannot apply the method you are deriving above in this case , for the sole reason that X does not have an inverse.
     
  9. Oct 3, 2008 #8
    Okay, thank you both so much!
     
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