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Tips and tricks for finding determinant of 3x3 matrix?

  1. Jul 6, 2010 #1
    I have a midterm tomorrow and I find I'm quite slow at finding the determinant of a 3x3 matrix. Basically I'll only need to find the determinant to find the characteristic polynomial (at least for this class) and my prof on the board does it so fast, I'm wondering if there's some trick I missed out on. I feel like it takes me too much time to get the determinant of 3x3 matrices. Also my first time posting, long time lurker lol. Thanks.
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  3. Jul 6, 2010 #2


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    Exercise. In general, if the matrix contains numerical values (specially values of not so high precision, since it takes time to plug in, for example, 1.03427*10^-28), you should be able to calculate the determinant fairly quickly. Of course, review all the properties of the determinant which can help you get over with your calculations more quickly and elegant.
  4. Jul 6, 2010 #3
    Yea there will be no decimal values as there is no calculator on the midterm. Also, once I find the 3x3 matrix I end up having to solve a cubic (in most cases) which also takes time. Is there no easier way to get the eigenvalues of a 3x3 (with whole numbers eigenvalues) rather then just solving the Characteristic polynomial?
  5. Jul 6, 2010 #4


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    I bet your teacher won't give you a cubic which you can't arrange in a way which is fairly easy to solve. :wink:
  6. Jul 6, 2010 #5


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    You know how to write it as the sum of the determinants of three 2x2 matrices, right? If there are any zeros in your 3x3 matrix, you can expand along that row or column to eliminate one or two of those 2x2 matrices.

    If all the numbers below or above the diagonal are zeros, the determinant is just the product of the diagonal elements.

    Also, if one of the rows (or columns) is a linear combination of the other two, the determinant is zero.
  7. Jul 6, 2010 #6
    Thanks frederick, that definitely helps a lot. Let me make sure I understand:

    If A = [2 0 0; 1 5 4; 1 2 4] (arranged by column) then my determinant is simply 2 * 5 * 4?
  8. Jul 6, 2010 #7
    Nevermind I understood you wrong lol. I'm guessing if A is something like: [a 0 0; 0 b 0; 0 0 c].
  9. Jul 6, 2010 #8
    It doesn't have to be as restrictive as diagonal. Upper or lower triangular form is sufficient.
    So for example if [tex]A=(a_{ij}) : a_{ij}=0, i>j [/tex] This is upper triangle form. And Similarly for [tex]i<j[/tex]
  10. Jul 6, 2010 #9


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    No, that's incorrect. Expanding on the first row the determinant is
    [tex]\left|\begin{array}{ccc}2 & 0 & 0\\1 & 5 & 4\\ 1 & 2 & 4\end{array}\right|= 2\left|\begin{array}{cc}5 & 4 \\ 2 & 4\end{array}\right|= 2(20- 8)= 24[/tex]
    because all of the other multipliers, the other numbers in the first row, are 0.

    If a matrix is "upper triangular" (all numbers below the main diagonal are 0), you can expand on the first column so you have only the number in "first row first column" times a smaller determinant, which can be expanded on the first column, etc. giving just the product of the numbers on the diagonal:
    [tex]\left|\begin{array}{ccc}a & b & c \\ 0 & d & e \\ 0 & 0 & f\end{array}\right|= a\left|\begin{array}{cc}d & e \\ 0 & f\end{array}\right|= a(d f)= adf[/tex]

    If a matrix is "lower diagonal" (all numbers above the main diagonal are 0), then expanding on the first row repeatedly gives the same thing.

    Finally, you can always "row reduce" a matrix until it is "upper triangular" remembering that:
    1) Swapping two rows multiplies the determinant by -1
    2) Multiplying an entire row by a number multiplies the determinant by that same number (so you have to divide the determinant resulting triangular matrix by that number to recover the determinant of the original matrix).
    3) Adding or subtracting a multiple of one row from another does not change the determinant.
  11. Jul 6, 2010 #10
    Prop: Gaussian Elimination does not change the determinant
    Prop: product of the diagonal entries of an upper triangular matrix = determinant
    Prop: If n row exchanges are performed when finding the triangular form, then determinant = (product of diagonals)(-1)^n
    Prop: If a scalar c multiplies a row, then multiply the determinant by 1/c, c!=0
    Last edited: Jul 6, 2010
  12. Jul 6, 2010 #11
    You can do what I do, which is just to expand by minors. It's a very simple method of breaking down the determinant of a larger matrix to that of a smaller matrix and consequently works wondrously in the 3x3 case. Probably best illustrated by example: http://ceee.rice.edu/Books/LA/det/det2.html"
    Last edited by a moderator: Apr 25, 2017
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