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Question about matracies

  1. Sep 27, 2009 #1
    If AB = AC, it's possible that B /= C

    conversely:
    if BA = CA, is it possible that C /= B?

    /= means doesn't equal
     
  2. jcsd
  3. Sep 27, 2009 #2

    nicksauce

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    Certainly. In both cases let A = 0, B = Anything, C = Anything else.
     
  4. Sep 27, 2009 #3

    HallsofIvy

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    More generally, if A is a matrix that does not have an inverse, then we can have A(B- C)= 0 without B- C being equal to 0. And, similarly, (B- C)A= 0 without B- C= 0.
     
  5. Sep 27, 2009 #4
    hallsofivy

    why does A have to be singular?
     
  6. Sep 28, 2009 #5

    lurflurf

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    say
    AB=AC
    suppose A has an inverse D such that AD=DA=
    DAB=DAC
    B=C
    thus if B/=C A must be singular

    Also the matrices might not be square
    [[1,1,1]]*[[1,1,1]]=[[1,1,1]]*[[3,0,0]]=[[3]]
    where * is the adjoint
    [[1,1,1]]/=[[3,0,0]]
     
  7. Sep 28, 2009 #6
    A basic example for you...

    [itex] \begin{pmatrix}0 & 0 & 0\\0 &1 & 0\\ 0 & 0 &0\end{pmatrix} = \begin{pmatrix}1 & 0 & 1\\0 &1 & 0\\ 1 & 0 &1\end{pmatrix} \begin{pmatrix}1 & 0 & -1\\0 &1 & 0\\ -1 & 0 &1\end{pmatrix} = \begin{pmatrix}1 & 0 & 1\\0 &1 & 0\\ 1 & 0 &1\end{pmatrix} \begin{pmatrix}2 & 0 & -5\\0 &1 & 0\\ -2 & 0 &5\end{pmatrix} =
    \begin{pmatrix}0 & 0 & 0\\0 &1 & 0\\ 0 & 0 &0\end{pmatrix}
    [/itex]

    But

    [itex] \begin{pmatrix}1 & 0 & -1\\0 &1 & 0\\ -1 & 0 &1\end{pmatrix} \neq \begin{pmatrix}2 & 0 & -5\\0 &1 & 0\\ -2 & 0 &5\end{pmatrix}[/itex]

    If you look at the rank of A and the vector that is in the nullspace of A, you can see why this holds with the columns of the B and C are chosen as such.
     
    Last edited: Sep 28, 2009
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