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4x4 matrix determinant

  1. Nov 13, 2006 #1
    Use row and/or column operations to simplify the determinant of the following matrix A, by reduction to upper triangular form, then evaluate.
    [tex]A = \left(\begin{array}{cccc}
    2 & 3 & 4 & 5\\
    0 & -1 & 2 & 1\\
    0 & 0 & 2 & 4\\
    0 & 3 & -6 & 0
    \end{array}
    \right)[/tex]

    Is there an simpler way to find the determinant so that I don't have to expand cofactors etc? Because it would be: 2|(3x3 matrix)| - 3|(3x3 matrix| + 4|(3x3 matrix)| - 5|(3x3 matrix)| and then I have to find the determinants of each 3x3 matrix...
     
  2. jcsd
  3. Nov 13, 2006 #2

    radou

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    Homework Helper

    Yes, there is a simpler way, and the text of the problem itself suggests which way it is. Investigate row and column operations and try to solve the problem.
     
  4. Nov 13, 2006 #3
    I have reduced to this:


    [tex]A = \left(\begin{array}{cccc}
    1 & 1.5 & 2 & 2.5\\
    0 & 1 & -2 & -1\\
    0 & 0 & 1 & 2\\
    0 & 0 & 0 & 1
    \end{array}
    \right)[/tex]

    How would I determine the determinant using this?
     
  5. Nov 13, 2006 #4

    radou

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    Good work. You can evaluate the determinant by using the fourth row now.
     
    Last edited: Nov 13, 2006
  6. Nov 13, 2006 #5
    The determinant would be 1.

    However, the answer is listed as -12, and the second part is to check my answer by expanding the third row of the determinant. I'm at a loss as to what to do now.
     
  7. Nov 13, 2006 #6

    radou

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    Then something went wrong with your row/column operations. Actually, I reduced it to an upper triangular form in 3 steps with row operations only. Its easy, go through it one more time.
     
  8. Nov 13, 2006 #7

    HallsofIvy

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    You had better have kept track of your divisors while row reducing. Dividing one row by a number divides the determinant by that number.
    Swapping two rows multiplies the determinant by -1. Add a multiple of one row to another does not change the determinant.

    Starting with
    [tex]\left(\begin{array}{cccc}2 & 3 & 4 & 5\\0 & -1 & 2 & 1\\0 & 0 & 2 & 4\\0 & 3 & -6 & 0\end{array}\right)[/tex]
    since the first column is already 0's (except for the first row), start on the second column: add 3 times the second row to the fourth row to get
    [tex]\left(\begin{array}{cccc}2 & 3 & 4 & 5\\0 & -1 & 2 & 1\\0 & 0 & 2 & 4\\0 & 0 & 0& 3\end{array}\right)[/tex]
    Now we have an upper triangular matrix: it's determinant is just the product of the numbers on the main diagonal:2(-1)(2)(3)= -12. Since the only row operation used was "add a multiple of one row to another", that is the determinant of the original matrix.
     
  9. Nov 13, 2006 #8
    I realised that I reduced to upper triangular row echelon form rather than just upper triangular form.

    The next part asks me to check this determinant by expanding the third row. How do I go about expanding it? Would it be using cofactors etc.?
     
  10. Nov 14, 2006 #9

    HallsofIvy

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    Yes, that's what "expanding" a determinant means. Using the third row is particularly easy since it has only 2 non-zero numbers.
     
  11. Jul 24, 2008 #10
    I thought I'd mention
    that the -12 seems to
    be the right answer, because
    I am getting -48 + 48 -12 + many zero terms,
    so -12 seems right.
    I am using a pictoral technique
    from post#7 from this forum:
    http://www.mathisfunforum.com/viewtopic.php?pid=95302#p95302
    (I made that red/black diagram, so I was checking
    my work with this example)
     
  12. Jul 25, 2008 #11

    HallsofIvy

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    Jack, this thread is two years old! And it was said two years ago that -12 is the determinant.
     
  13. Oct 17, 2010 #12
    You should know that you can take the determinant along any column as well as any row, so the obvious choice here is to thake the det. along column 2,0,0,0.

    this way you end up with 2|(3x3 matrix)| - 0|(3x3 matrix| + 0|(3x3 matrix)| - 0|(3x3 matrix)|
    so now the det is 2 times the determinant of the remaining matrix

    -1, 2, 1
    0, 2, 4
    0, 0, 3

    Which is -6, and -6*2= -12.
     
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