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Determinant of 4 X4

  1. Nov 7, 2008 #1
    1. The problem statement, all variables and given/known data
    The determinant
    a+1 a a a
    b b+1 b b
    c c c+1 c
    d d d d+1
    equals =?
    option : a) 1+ a + b+ c+ d
    b)a + b+ c+ d
    c)a.b.c.d
    d)1 + a.b.c.d



    2. Relevant equations



    3. The attempt at a solution
    [1] tried using finding the co factors.....but the process becomes very length
    [2] tried by row-column operations....as below , but nothing workd....
    1 0 0 0 a a a a
    0 1 0 0 + b b b b
    0 0 1 0 c c c c
    0 0 0 1 d d d d
     
  2. jcsd
  3. Nov 7, 2008 #2

    gabbagabbahey

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    Hi archita, welcome to PF! :smile:

    This forum supports LaTeX, which you can use to write equations and matrices in a much nicer format....For example, writing:

    [$tex]\begin{pmatrix} a+1 & a & a & a \\ b & b+1 & b & b \\ c & c & c+1 & c \\ d & d & d &d+1 \end{pmatrix}[/$tex]

    And deleting the $ signs gives:

    [tex]\begin{pmatrix} a+1 & a & a & a \\ b & b+1 & b & b \\ c & c & c+1 & c \\ d & d & d &d+1 \end{pmatrix}[/tex]

    ...Now, for your question....Hint:

    [tex]\begin{pmatrix} a+1 & a & a & a \\ b & b+1 & b & b \\ c & c & c+1 & c \\ d & d & d &d+1 \end{pmatrix}=\begin{pmatrix} a & a & a & a \\ b & b & b & b \\ c & c & c & c \\ d & d & d &d \end{pmatrix}+I=I+\begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} \begin{pmatrix}1 & 1 & 1 &1 \end{pmatrix}[/tex]

    Where [itex]I[/itex] is the Identity matrix....use Sylvester's detrminant theorem :wink:
     
    Last edited: Nov 7, 2008
  4. Nov 12, 2008 #3
    Then the answer would be one but that is none of the options. shouldn't this Sylvester use his theorem one row at a time?
     
  5. Nov 12, 2008 #4

    gabbagabbahey

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    Sylvester's theorem does not give an answer of one....For two column vectors [itex]u[/itex] and [itex]v[/itex] of equal dimension, Sylvester's theorem states that :

    [tex] \text{det}(I+uv^T)=1+v^T u[/tex]

    In this case you would use [tex]u=\begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}[/tex] and [tex]v^T=\begin{pmatrix}1 & 1 & 1 &1 \end{pmatrix}[/tex]....surely you do not get one using this?
     
  6. Nov 14, 2008 #5
    Now I can see that the problem is solved! Clever Sylvester.
     
  7. Nov 14, 2008 #6
    Still had my doubts. Couldn't there be a solution e: 1+a+b+c+d+ab+ac+ad+bc+bd+cd?

    If I take a=1, b=2, c=3, d=4 and calculate the determinant the old way I get 46
     
  8. Nov 14, 2008 #7

    gabbagabbahey

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    Then you're calculating the determinant incorrectly....using: a=1, b=2, c=3, d=4 and calculating the determinant "the old way" gives:

    [tex]\begin{vmatrix} 2 & 1 & 1 & 1 \\ 2 & 3 & 2 & 2 \\ 3 & 3 & 4 & 3 \\ 4 & 4 & 4 & 5 \end{vmatrix}=\begin{array}{l} 2[3(4*5-3*4)-2(3*5-3*4)+2(3*4-4*4)]-1[2(4*5-3*4)-2(3*5-3*4)+2(3*4-4*4)] \\ +1[2(3*5-3*4)-3(3*5-3*4)+2(3*4-3*4)]-1[2(3*4-4*4)-3(3*4-4*4)+2(3*4-3*4)] \end{array}[/tex]

    [tex]=2[24-6-8]-[16-6-8]+[6-9+0]-[-8+12+0]=20-2-3-4=11 \neq 46[/tex]
     
  9. Nov 14, 2008 #8
    You are right my 46 is not just. Excel gave for the four to four matrix indeed 11 too.
    So my solution e is not just and one of the four a..d is in this case.

    I took (a+1)(b+1)(c+1)(d+1) +3abcd -(a+1)bcd -(b+1)acd -(c+1)abd -(d+1)abc,
    a method wich works for three by three matrixes but apparently not for 4 by 4.
     
  10. Nov 21, 2008 #9
    thanks a ton :smile:
     
  11. Nov 21, 2008 #10

    HallsofIvy

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    Since this is a 4 by 4 matrix (even number of rows and columns) its determinant is the same as the determinant of the transpose,
    [tex]\left|\begin{array}{cccc}a+ 1 & b & c & d \\ a & b+ 1 & c & d \\ a & b & c+ 1 & d \\ a & b & c & d+ 1 \end{array}\right\|[/tex]

    Swapping two rows multiplies the determinant by -1 so doing that twice gives the same determinant,
    [tex]\left|\begin{array}{cccc}a & b & c & d+1 \\ a & b & c+1 & d \\ a & b+1 & c & d \\ a+1 & b & c & d\end{array}\right|[/tex]

    Subtract the first row from each of the other rows,
    [tex]\left|\begin{array}{cccc} a & b & c & d+1 \\ 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & -1 \\ 1 & 0 & 0 & -1\end{array}\right|[/tex]

    and, finally, expand that on the first column.
     
  12. Nov 21, 2008 #11
    So that will be Det(1:4,1:4) = a * det(2:4,2:4) - det(2:4,1:3) ?
    I'll check that!...
     
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