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Matrices(column operations)

  1. Jan 3, 2008 #1

    rock.freak667

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    If I have a matrix

    [tex]
    A= \left(
    \begin{array}{ccc}
    1 & 2 & 3\\
    0 & -1 & 4\\
    1 & 1 & 6
    \end{array}
    \right)
    [/tex]
    and I need to find [itex]A^{-1}[/itex] I would just augment with the identity matrix and then do row operations. But if I want to use column operations instead does it work in the same manner? because I think if use the column operations, the matrix A would be reduced to RRE form but nothing will happen to the identity matrix.
    (Not too sure if I was clear about my problem.)
     
  2. jcsd
  3. Jan 4, 2008 #2
    what are column operations?
     
  4. Jan 4, 2008 #3

    Hurkyl

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    You need to adjoin an identity matrix below the original, not to its right.
     
  5. Jan 4, 2008 #4

    Hurkyl

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    You mean reduced column echelon form.
     
  6. Jan 4, 2008 #5

    rock.freak667

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    Yeah sorry about that.That is what I meant.
    Below it? So there is no way to both row and column operations at the same time to find the inverse or whatever?Also how would I solve a system of equations using column operations only?
     
  7. Jan 4, 2008 #6
    wouldn't these column operations be the same thing as row reducing the transpose?
     
  8. Jan 4, 2008 #7

    rock.freak667

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    It is the same thing basically except that column operations uses the columns of the matrix,
     
  9. Jan 4, 2008 #8

    Hurkyl

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    Not with this particular bookkeeping method. Of course you can use both kinds of operations to find an inverse: algebraically, if R is the matrix denoting your row operations and C is the one denoting your column operations, then if you reduce your matrix to the identity, that says
    RAC = I​
    which you can easily solve for A.
     
  10. Jan 4, 2008 #9

    rock.freak667

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    oh okay then thanks
     
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