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Differents answers to the same determinant

  1. Mar 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Lets find the determinant of

    1 1 1 1 ... 1
    1 2 2 2 ... 2
    1 2 3 3 ... 3
    1 2 3 4 ... 4
    ................
    1 2 3 4 ... n

    3. The attempt at a solution

    In class, we solved it by subtracting the previous line of every single line, ending up with

    1 1 1 1 ... 1
    1 2 2 2 ... 2 II-I
    1 2 3 3 ... 3 III-II
    1 2 3 4 ... 4 IV-III =
    ................ ...
    1 2 3 4 ... n n-(n-1)

    1 1 1 1 ... 1
    0 1 1 1 ... 1
    0 0 1 1 ... 1
    0 0 0 1 ... 1
    ................
    0 0 0 0 ... 1

    And from the multiplication of the main diagonal elements we get that its equal to 1.
    Makes sense.

    Yet when I started looking at the problem at home, I tried a solution like this:

    1 1 1 1 ... 1
    1 2 2 2 ... 2 II-n
    1 2 3 3 ... 3 III-n
    1 2 3 4 ... 4 IV-n =
    ................ ...
    1 2 3 4 ... (n-2) (n-2)-n
    1 2 3 4 ... (n-1) (n-1)-(n-2)
    1 2 3 4 ... n n-(n-1)

    1 1 1 1 ... 1
    0 0 -1 -2 ... (-n+2)
    0 0 0 -1 ... (-n+3)
    0 0 0 0 ... (-n+4)
    ..........................
    0 0 0 0 ... -2
    0 0 0 0 ... -1
    0 0 0 0 ... 1

    This way we get zeros below the main diagonal, and we can calculate the determinant by multiplying the elements of the main diagonal. Since it contains zeros, we get it equal to 0.

    I end up with the same result when subtracting the first line times the line number from every other line, for example

    II - 2*I
    III - 3*I
    IV - 4*I
    ...
    n - n*I

    because this way I get a triangle of zeros above the main diagonal, which is basically the same thing as below. The main diagonal also contains zeros, giving the answer 0.

    As Im new to matrixes, Im probably missing something fundamental here. Would like to know what it is. Thanks.

    EDIT: Didnt know how to insert the determinants here properly, so changed the color of the first side of the equation to red, to make it a tad better to read. Sorry about that.
     
    Last edited: Mar 6, 2013
  2. jcsd
  3. Mar 6, 2013 #2

    Ray Vickson

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    In the second way you have newRow2 = Row2-Row_n, newRow3 = Row3 - Row_n, .... But this still leaves Row_n unchanged, so you do not have a diagonal matrix.
     
  4. Mar 6, 2013 #3
    No. I changed the colors of what I did to the rows to blue. Look at my post again.
     
  5. Mar 6, 2013 #4

    Ray Vickson

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    I cannot tell how you managed to get a final row [0,0,0,...,1]. For example, for n = 4 your initial matrix is
    [tex]A = \pmatrix{1&1&1&1\\1&2&2&2\\1&2&3&3\\1&2&3&4}[/tex]
    After subtracting row 4 from rows 2 and 3 we have
    [tex]A_1 = \pmatrix{1&1&1&1\\0&0&-1&-2\\0&0&0&-1\\1&2&3&4}[/tex]
    At this point I don't see any row operations that can "zero-out" the first three components of row 4. Please explain further.
     
  6. Mar 7, 2013 #5
    [tex]A = \pmatrix{1&1&1&1\\1&2&2&2\\1&2&3&3\\1&2&3&4}[/tex]

    During the same step we
    subtract Row4 from Row2
    subtract Row2 from Row3
    subtract Row3 from Row4

    ending up with

    [tex]A_1 = \pmatrix{1&1&1&1\\0&0&-1&-2\\0&0&1&1\\0&0&0&1}[/tex]

    And the answer to be 0.
     
  7. Mar 7, 2013 #6

    Ray Vickson

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    I am assuming you meant that you applied those steps to A. When I follow the steps exactly as you have written them I get
    [tex] \pmatrix{1&1&1&1\\0&0&-1&2\\1&2&4&5\\0&0&-1&-1}[/tex]
    with determinant equal to 1.
     
  8. Mar 7, 2013 #7
    So what you're saying is that Im supposed to use the "new"/changed Row2 when doing further calculations to the rows?
    That wasnt followed in the following example (in class with the teacher):

    [tex] \pmatrix{0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0}[/tex] = [tex] - \pmatrix{1&0&1&1\\0&1&1&1\\1&1&0&1\\1&1&1&0}[/tex] = [tex] - \pmatrix{1&0&1&1\\0&1&1&1\\0&1&-1&0\\0&0&1&-1}[/tex] (III-I ; IV - III) = [tex] - \pmatrix{1&0&1&1\\0&1&1&1\\0&0&-2&-1\\0&0&1&-1}[/tex] (III-II) = [tex] - \pmatrix{1&0&1&1\\0&1&1&1\\0&0&-2&-1\\0&0&0&-1,5}[/tex] (IV+1/2*III) = -3
     
  9. Mar 7, 2013 #8
    Right, row operations are performed one at a time and you can only use the updated rows at each step. In the original teacher's example, this is possible if you do the operations from the bottom up.
     
  10. Mar 7, 2013 #9
    Oh, so IV - III is calculated before III-I in the previous example. Makes perfect sense to me now. Thats why my version of the original problem was wrong.

    Thanks guys.
     
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