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Properties of determinants

  1. May 20, 2012 #1

    sharks

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    1. The problem statement, all variables and given/known data
    Use the properties of the determinant of a matrix to show that[tex]\begin{vmatrix}1+x^2 & x & 1 \\ 1+y^2 & y & 1 \\ 1+z^2 & z & 1\end{vmatrix}=(x-y)(x-z)(y-z)[/tex]

    2. Relevant equations
    Properties of determinants. There's 10 of them, according to my notes.


    3. The attempt at a solution
    I used the property where the scalar multiple of -1 of the third column added to the first column gives:[tex]\begin{vmatrix}x^2 & x & 1 \\ y^2 & y & 1 \\ z^2 & z & 1\end{vmatrix}[/tex]
    And then i'm stuck.
     
  2. jcsd
  3. May 20, 2012 #2

    SammyS

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    What do you get when you expand [itex](x-y)(x-z)(y-z)\ ?[/itex]
     
  4. May 20, 2012 #3

    sharks

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    Hi SammyS
    The expansion gives: [tex]x^2y-xy^2-2xyz+y^2z-x^2z-xz^2-yz^2[/tex]But i'm not sure how to relate this to the determinant.

    I grouped the squared terms, as it seemed to me that they formed the cofactor expansion by the first column of the determinant:
    [tex]x^2(y-z) -y^2(x-z) -z^2 (x+y) -2xyz[/tex]But it's different, as the actual cofactor expansion is: [tex]x^2(y-z) -y^2(x-z) +z^2 (x-y)[/tex]
     
    Last edited: May 20, 2012
  5. May 20, 2012 #4

    SammyS

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    Not quite right.

    There's a xyz and a -xyz which cancel .
     
  6. May 20, 2012 #5

    sharks

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    You are correct. :redface: So, to write the solution, meaning how i got to the product of the 3 factors, i just trace back my steps from the factors' expansion.

    Thanks, SammyS.
     
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