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Cofactor expansion

  1. Mar 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the determinant of
    4...3...-5
    5...2...-3
    0..-1....2



    3. The attempt at a solution
    I've tried to get the answer using each of the 3 rows and each time I get a different answer. For the first row I get 14, the second row I get -31, and the third row I get -1. However, I thought that the determinant would be the same value for whichever row (or column) you choose to expand.
     
  2. jcsd
  3. Mar 1, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, it certainly should be!

    Expanding by the first column (since it has only 2 non-zero entries):
    [tex]\left|\begin{array}{ccc}4 & 3 & -5 \\ 6 & 2 & -3 \\ 0 & -1 & 2 \end{array}\right|= 4\left|\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\left|- 5\right|\begin{array}{cc} 3 & -5 \\ -1 & 2\end{array}\right|[/tex]
    [tex]4(4- 3)- 5(6- 5)= 4(1)- 5(1)= -1[/tex]
    The determinant is -1. I notice that is what you got using the third row which also has only two non-zero entries. Perhaps it is that third entry that is confusing you.
     
  4. Mar 2, 2008 #3
    But if you try to work out the determinant using the other rows that do not contain the 0 entry, it does not come out to -1!! Why?
     
  5. Mar 2, 2008 #4
    Ah, I just realized I wasn't multiplying one of the terms with the cofactor expansion!! I feel silly....thanks for your help though!!
     
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