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Matrix fun

  1. Apr 13, 2009 #1
    Any help solving this determinant:

    0 1 1 1
    1 0 1 1
    1 1 0 1
    1 1 1 0

    My calc says the answer is -3 but there is supposed to be a quicker way than doing all the individual calculations, I did all the calculations and got -3 but there is supposed to be a quicker way. Anyone?
     
  2. jcsd
  3. Apr 13, 2009 #2

    dx

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    What way did you use?
     
  4. Apr 13, 2009 #3
    a11a22a33a44
    - a11a22a34a43
    + a11a23a34a42
    - a11a23a32a44
    + a11a24a32a43
    - a11a24a33a42
    - a12a23a34a41
    + a12a23a31a44
    - a12a24a31a43
    + a12a24a33a41
    - a12a21a33a44
    + a12a21a34a43
    + a13a24a31a42
    - a13a24a32a41
    + a13a21a32a44
    - a13a21a34a42
    + a13a22a34a41
    - a13a22a31a44
    - a14a21a32a43
    + a14a21a33a42
    - a14a22a33a41
    + a14a22a31a43
    - a14a23a31a42
    + a14a23a32a41
     
  5. Apr 13, 2009 #4

    dx

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    You can use row and column operations to simplify the determinant. For example,

    R2 --> R2 - R3
    R3 --> R3 - R4

    This makes the first column 0 0 0 1.
     
  6. Apr 13, 2009 #5
    I get:

    0 1 1 1
    0 -1 1 0
    0 0 -1 1
    1 1 1 0

    which then a co-factor expansion would give:

    0+0+0+0+1*determinant of

    1 1 1
    -1 1 0
    0 -1 1

    wheres the -1 come from cause i get an answer of 3? is it supposed to be 0+0+0+0-1?
     
  7. Apr 13, 2009 #6

    dx

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    Yes.
     
  8. Apr 13, 2009 #7
    Don't forget that the equation for the co-factor expansion includes the term [tex](-1)^{i+j}[/tex] where i is the row and j is the column.

    In this case, [tex]i = 4[/tex] and [tex]j = 1[/tex], so this term is [tex](-1)^{4+1} = (-1)^5 = -1[/tex].
     
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