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Finding determinant through Gaussian elimination

  1. Oct 18, 2014 #1
    If I switch 2 rows, do I have to multiple by -1 each time?

    For example, I have

    0%20%26%20-1%5C%5C%203%20%26%201%20%26%201%5C%5C%200%20%26%20-1%20%26%20-1%20%5Cend%7Bvmatrix%7D.gif

    If I switch row 2 and 3, will it become this:
    0%20%26%20-1%5C%5C%200%20%26%20-1%20%26%20-1%5C%5C%203%20%26%201%20%26%201%20%5Cend%7Bvmatrix%7D.gif
    Or this?
    0%20%26%20-1%5C%5C%200%20%26%20-1%20%26%20-1%5C%5C%203%20%26%201%20%26%201%20%5Cend%7Bvmatrix%7D.gif

    Each time I make a switch, do I have to also put a negative sign?

    Edit: Not really related to Gaussian elimination, but this is from a Gaussian elimination problem; I just wanted to know whether I have to put a negative each time I make the switch.
     
  2. jcsd
  3. Oct 18, 2014 #2

    SteamKing

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    It's not clear what switching rows does for you in terms of finding the determinant of this matrix.

    Using the original matrix, you would want to eliminate element a21 = 3 first. After that, you can work on eliminating a32, leaving you with an upper triangular matrix. The determinant of an upper triangular matrix is easy to calculate.

    FYI, the elementary row operations affect the determinant as discussed in the following article:

    http://en.wikipedia.org/wiki/Gaussian_elimination

    See the section 'Computing determinants' under Applications.
     
  4. Oct 20, 2014 #3

    HallsofIvy

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    Actually, it is "clear what switching rows does for you in terms of finding the determinant of this matrix."

    Anytime you swap two rows in a determinant, you multiply the determinant by -1.

    In this case, the original determinant is [tex]\left|\begin{array}{ccc}2 & 0 & -1 \\ 3 & 1 & 1 \\ 0 & -1 & -1 \end{array}\right|[/tex]. If you "expand by minors" on the first row, you get [tex]2\left|\begin{array}{cc}1 & 1 \\ -1 & -1\end{array}\right|- 1\left|\begin{array}{cc} 3 & 1 \\ 0 & -1\end{array}\right|= 2(-1+ 1)- (-3)= 3[/tex]

    Switching the first two rows you get [tex]\left|\begin{array}{ccc}3 & 1 & 1 \\ 2 & 0 & -1 \\ 0 & -1 & -1 \end{array}\right|[/tex]. If you "expand by minors" on the second row, you get exactly the same thing except that, because your leading coefficients are from the second row, their sign is changed: [tex]-2\left|\begin{array}{cc}1 & 1 \\ -1 & -1\end{array}\right|+ 1\left|\begin{array}{cc} 3 & 1 \\ 0 & -1\end{array}\right|= -2(-1+ 1)+ (-3)= -3[/tex].

    Switching the first and third rows and expanding on the third row will give you almost exactly the same thing. This time the leading coefficients will be the same as the first time but the two rows in the sub-determinants are reversed, reversing the sign there:
    [tex]\left|\begin{array}{ccc}0 & -1 & -1\\ 3 & 1 & 1 \\ 2 & 0 & -1 \end{array}\right|[/tex]. If you "expand by minors" on the third row, you get [tex]2\left|\begin{array}{cc}-1 & -1 \\ 1 & 1\end{array}\right|- 1\left|\begin{array}{cc} 0 & -1 \\ 3 & 1\end{array}\right|= 2(-1+ 1)- (3)= -3[/tex]
     
  5. Oct 20, 2014 #4

    SteamKing

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    That's fine, but the OP wanted to find the determinant of the matrix by using elimination, not by expanding minors. Switching rows as he did originally did not make his transformed matrix closer to upper triangular form; it complicated the elimination by adding extra steps.

    To find the determinant of a 3x3 matrix is a trivial exercise anyway; one can always calculate the determinant using the formula of Sarrus.
     
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