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Determinant of large matrix

  1. Mar 3, 2007 #1
    1. The problem statement, all variables and given/known data

    Given the following 101x101 matrix: Akl where k and l are the row and column. Now Akl = (akl)=k^2+1 if k=l and (akl)=2*k*l else. I have to calculate the determinant of this matrix but have no clue how to start working, well I've some idea but it doesn't really help.

    2. Relevant equations

    Formula of determinant, ... (perhaps induction).

    3. The attempt at a solution

    I calculated the determinant of this matrix in case it's a 1x1 matrix, 2x2, 3x3, 4x4 matrix and tried to find some law for the nxn case (then proof with induction), but I did not find a formula.
     
  2. jcsd
  3. Mar 3, 2007 #2

    HallsofIvy

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    If I am not mistaken, the first column consists of the even numbers, in order.
    What happens if you multiply the first row by n and subtract that from the nth row?
     
  4. Mar 4, 2007 #3
    Ah, you mean, I can transform the matrix into a triangle matrix?
     
  5. Mar 4, 2007 #4

    radou

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    The point is to transform it into a triangular matrix, whose determinant is the product of the diagonal elements.
     
  6. Mar 4, 2007 #5

    HallsofIvy

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    And, in fact, that can be done in "one swell foop"!
     
  7. Mar 4, 2007 #6
    Ok, I solved it succesfully by transforming it into a triangular matrix. (I multiplied every row with factor (k+1) or something and then substracted the row with the next row). So I got almost a triangular matrix (except the last row). The last row I treated seperately.

    Thanks for your help
     
  8. Mar 4, 2007 #7

    HallsofIvy

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    I am puzzled as to why you cannot reduce the last row? Doesn't subtracting "k times the first row" from the kth row reduce every row and leave you with a diagonal matrix?
     
  9. Mar 5, 2007 #8
    No, if I do that, then all elements of the last row are zero exzept the first one and the last one (because the calculation of the 1x1 element is another one than the one of the 1x2, 1x3, ... element according to the rule the matrix is built). So I end with almost a diagonal matrix (the first has also nonzero elements)
     
  10. Mar 5, 2007 #9

    radou

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    Did you write your matrix down correct (at least some elements, so you can see what it looks like)? What HallsofIvy suggested should work just fine.
     
  11. Mar 5, 2007 #10

    HallsofIvy

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    In an n by n matrix, constructed according to the rule you give:
    "Akl = (akl)=k^2+1 if k=l and (akl)=2*k*l else"
    The first row is "2 4 6 8 ... 2n". The first row of the first column is 12+ 1= 2 which also happens to be 2*1. The first column is the same (for the same reasons). In particular, the element in the first column of the nth row is 2n.
    In fact the final row is "2n 4n 6n 8n ... n2+ 1". Subtracting, as I said before, "n times the first row" from the final row gives "0 0 0 0 ... n^2+ 1- 2n2" or "0 0 0 0 ... 1- n2".

    The resulting matrix is a diagonal matrix (except for the top row) with 2 in the first row, first column and 1- k2 in the kth row, kth column. The determinant is the product of those diagonal numbers.
     
  12. Mar 5, 2007 #11
    Mhh, you're right, uff, in my calculation 1^2+1 equaled 3. Uff, well, then I haven't to do the last row seperately. What had my assistent thought if he had seen this misscalculation :).

    Thanks
     
  13. Nov 8, 2009 #12
    first off i apologise if it is frowned upon to drag old threads up from out of the blue but this is the most relevant information i can find for what im doing.

    Im trying to find the determinant of an NxN matrix using the triangle matrix method, but i have never been taught this method and the text book i tried to decipher it from ( K.A.STROUD advanced engineering mathematics) was frankly rather vague and useless!

    so i see the technique used for reducing the nth row to zero, but how do i then reduce the previous (n-1 th) row? from the example above, would it be to subtract n-1 times the first row from the n-1 th row?
     
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