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Linear Algebra help

  1. Feb 21, 2012 #1
    let A be a 3x3 matrix. Suppose that for every row vector y=[y1 y2 y3] there exists a row vector x=[x1 x2 x3] such that xA=y. Show that A is invertable


    i honestly have no idea where to even go with this. any help would be appreciated (:
     
  2. jcsd
  3. Feb 21, 2012 #2

    jbunniii

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    Hint: A is invertible if and only if there is a 3x3 matrix B such that BA = ?
     
  4. Feb 21, 2012 #3
    such that BA = I?

    i read something about that in my text book but i dont understand what I is. Is it just the inverse?
     
  5. Feb 21, 2012 #4
    i know how to find an inverse, i just dont understand the part with row vectors and where it fits into the equation
     
  6. Feb 21, 2012 #5

    jbunniii

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    No, I is the identity matrix:

    [tex]I = \left[ \begin{array}{ccc}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1 \end{array} \right][/tex]
     
  7. Feb 21, 2012 #6

    jbunniii

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    Try to write BA = I, one row at a time.
     
  8. Feb 21, 2012 #7
    oh, that would make sense.

    so could i say that B is x and I is y?

    i feel like thats completely wrong.
     
  9. Feb 21, 2012 #8

    jbunniii

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    No, B and I are 3x3 matrices, whereas x and y are 1x3.

    If you're having trouble seeing what to do, I suggest naming the elements of the matrix B, for example as follows:

    [tex]B = \left[ \begin{array}{ccc}
    a & b & c \\
    d & e & f \\
    g & h & i \end{array}\right] [/tex]

    Now, what's the first row of BA = I? It is of the form xA = y. What are x and y in this case?
     
  10. Feb 21, 2012 #9
    x would be [a b c] and y would be [1 0 0]?
     
  11. Feb 21, 2012 #10

    jbunniii

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    Right. So now, reverse the argument. You are given the fact that for every row vector y, there is a row vector x such that xA = y. So start by choosing y = [1 0 0], and writing the corresponding x = [a b c]. Now repeat for the remaining two rows. What do you end up with?
     
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