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Find 3x3 matrix for XA=B

  1. Sep 10, 2012 #1
    The problem statement, all variables and given/known data

    Find the 3x3 matrix X such that XA=B, where

    Code (Text):

          1 1  1
    A=   -1 0 -2
          1 0 -1

          1 3  0
    B=   -2 1 -2
          1 4 -1
     
    The attempt at a solution

    I understand how to do AX=B just fine, but XA=B is confusing me. I tried doing [x y z] A, but not sure if that's even right or what to do after that. Yes, I know it can be solved using inverses or whatever, but the professor does not want us to solve it that way (haven't gone over inverses yet).

    The solution for X such that AX=B involved finding individual columns, so does that mean XA=B will find the individual rows?

    Any help is appreciated!
     
    Last edited: Sep 10, 2012
  2. jcsd
  3. Sep 10, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, find the individual rows of X. How would you do it?

    ehild
     
  4. Sep 10, 2012 #3
    If I multiply

    \begin{array}{cccc}
    1 & 1 & 1 \\
    -1 & 0 & -2 \\
    1 & 0 & -1 \\
    \end{array}

    by
    \begin{array}{cccc}
    x & y & z
    \end{array}

    I get

    \begin{array}{cccc}
    (x-y+z) & (x) & (x-2y-z) \\
    \end{array}

    That's where I get stuck. I don't know what to do with it. Would I transpose it to this?:


    \begin{array}{cccc}
    (x-y+z) \\
    (x) \\
    (x-2y-z) \\
    \end{array}
     
  5. Sep 10, 2012 #4
    Transpose both sides.
     
  6. Sep 11, 2012 #5
    Okay, I transposed both sides and I see how to get each column of X now. I think I get it :). Here's my augmented matrix:

    \begin{array}{cccc}
    1 & 0 & 0 | & 3 & 1 & 4 \\
    0 & 1 & 3 | & 1 & 0 & 2 \\
    0 & 0 & -4 | & 1 & 3 & 1 \\
    \end{array}

    The last row seems wrong because of the -4, but if I change it to a 1 using row operations, then the answer is wrong. The correct answer is:

    \begin{array}{cccc}
    3 & 1 & 1 \\
    1 & 0 & 3 \\
    4 & 2 & 1\\
    \end{array}
     
    Last edited: Sep 11, 2012
  7. Sep 11, 2012 #6
    Edit: another way to solve this is to multiply both sides by ##A^{-1}## from the right.

    Sorry, it seems that I almost forgot one property of matrix inverse :)
     
    Last edited: Sep 11, 2012
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