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HOw can this determinant be 0 ?

  1. Sep 9, 2009 #1
    HOw can this determinant be 0 ??

    let be a 3x3 matrix with elements

    [tex] a_{1,j}= j [/tex]

    [tex] a_{2,j}= j+3 [/tex]

    [tex] a_{3,j}= 6+j [/tex]

    with i=1,2,3 and j=1,2,3 ...

    but DetA=0 apparently there is no zeros or other condition that points that determinant should be 0, is there any explanation ??
     
  2. jcsd
  3. Sep 9, 2009 #2
    Re: HOw can this determinant be 0 ??

    Because the rows aren't linearly independent:

    2*[4 5 6] - 1*[1 2 3] = [7 8 9]
     
  4. Oct 7, 2009 #3
    Re: HOw can this determinant be 0 ??

    The determinant of a 3x3 matrix

    a b c
    d e f
    g h i

    is: aei+bfg+cdh-afh-bdi-ceg


    In your case

    a b c
    d e f =
    g h i

    1 2 3
    4 5 6
    7 8 9



    Therefore the determinant is:
    1x5x9 + 2x6x7 + 3x4x8 - 1x6x8 - 2x4x9 - 3x5x7
    = 45 + 84 + 96 - 48 -72 - 105
    = 225 - 225
    = 0
     
  5. Oct 7, 2009 #4
    Re: HOw can this determinant be 0 ??

    If you look at the 3x3 matrix as an array of 3 vectors (row or column), all of them are parallel.
     
  6. Oct 8, 2009 #5
    Re: HOw can this determinant be 0 ??

    ?? [1,2,3] and [4,5,6] are parallel?
     
  7. Oct 8, 2009 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: HOw can this determinant be 0 ??

    No, they aren't. But they all lie in the same plane. Another way of looking at this is that that the determinant of a matrix having three vectors, u, v, w, say, as rows (or columns) is a "triple product": [itex]u\cdot(v\times w)[/itex]. And that can be interpreted as the volume of the parallelopiped having the three vectors as edges. That volume is 0 if and only if the three vectors are all in the same plane so that the parallelopiped has no "height".
     
    Last edited: Oct 9, 2009
  8. Oct 8, 2009 #7
    Re: HOw can this determinant be 0 ??

    Sorry about that...mixed up the words...they lie on the same plane as HallsofIvy stated.
     
    Last edited: Oct 8, 2009
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