Results 1 to 4 of 4

Math Help - Find Determinant by Upper Trianglular matrix

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    65

    Find Determinant by Upper Trianglular matrix

    So the matrix is this:
    [ 1 1 1 1 ]
    [ a b c d ]
    [ a^2 b^2 c^2 d^2 ]
    [ a^3 b^3 c^3 d^3 ]

    The method I was using was to multiple row 2, 3, 4 by 1/a, 1/a^2, 1/a^3, then subtract from row 1. Then multiply row 3 and 4 such that I can subtract from row 2, etc.

    The problem with this method is that I'm getting a lot of computation and i'm starting to wonder if i'm do this the right way. Is there a better way of doing this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by hashshashin715 View Post
    So the matrix is this:
    [ 1 1 1 1 ]
    [ a b c d ]
    [ a^2 b^2 c^2 d^2 ]
    [ a^3 b^3 c^3 d^3 ]

    The method I was using was to multiple row 2, 3, 4 by 1/a, 1/a^2, 1/a^3, then subtract from row 1. Then multiply row 3 and 4 such that I can subtract from row 2, etc.

    The problem with this method is that I'm getting a lot of computation and i'm starting to wonder if i'm do this the right way. Is there a better way of doing this?
    When I was in school 55 years ago they taught us to do it this way. You can see that the determinant must be a polynomial of degree 6 in the variables a,b,c,d. If a=b then the first and second columns of the matrix are equal and so the determinant is 0. Therefore (by the factor theorem) a-b is a factor of the determinant. For the same reason, a-c, a-d, b-c, b-d and c-d are also factors. Thus the determinant is a multiple of (a-b)(a-c)(a-d)(b-c)(b-d)(c-d). But since that is already a polynomial of degree 6, there cannot be any further factors apart from a numerical constant.

    Thus \begin{vmatrix}1&1&1&1\\ a&b&c&d\\ a^2&b^2&c^2&d^2\\ a^3&b^3&c^3&d^3\end{vmatrix} = k(a-b)(a-c)(a-d)(b-c)(b-d)(c-d). To find the numerical constant k, compare the coefficients of  bc^2d^3 on both sides of that equation, and you find that k=-1. So \begin{vmatrix}1&1&1&1\\ a&b&c&d\\ a^2&b^2&c^2&d^2\\ a^3&b^3&c^3&d^3\end{vmatrix} = -(a-b)(a-c)(a-d)(b-c)(b-d)(c-d).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2008
    Posts
    65
    Thanks for this method, but the problem states that I have to find the determinant by transforming the matrix to an upper triangular matrix. Sorry for not stating that in OP.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,791
    Thanks
    1531
    First reduce the first row by subtracting a times the first row from the second, a^2 times the first row from the third row, and a^3 times the first row from the fourth to get
    \begin{bmatrix}1 & 1 & 1 & 1 \\0 & b-a & c-a & d-a\\ 0 & b^2- a^2 & c^2- a^2 & d^2- a^2 \\ 0 & b^3- a^3 & c^3- a^3 & d^3- a^3\end{bmatrix}

    Now [tex]b^2- a^2= (b- a)(b+ a) so we need to multiply b+ a times the second row and subract that from the third row.
    [tex](c^2- a^2)- (c- a)(b+ a)= (c- a)(c+ a)- (c- a)(b+ a)= (c- a)(c- b)[tex] and (d^2- a^2)- (d- a)(b+ a)= (d-a)(d- b)

    [tex]b^3- a^3= (b- a)(b^2+ ab+ a) so need to multiply b^2+ ab+ a^2 times the second row and subtract that from the fourth row.
    (c^3- a^3)- (c- a)(b^2+ ab+ a^2)= (c-a)(c^2+ ac+ a^2)- (c-a)(b^2+ ab+ a^2)= (c- a)(c^2- b^2+ ac-ab)= (c-a)(c-b)(a+ b+ c)

    Similarly, (d^3- a^3)- (d-a)(b^2+ ab+ a^2)= (d-a)(d-b)(a+ b+ d)
    Clearing the second column, we have
    \begin{bmatrix}1 & 1 & 1 &1 \\ 0 & b-a & c-a & d-a \\ 0 & 0 & (c-a)(c-b) & (d-a)(d-b) \\ 0 & 0 & (c-a)(c-b)(a+ b+ c) & (d-a)(d-b)(a+ b+ d)\end{bmatrix}.

    Now you try the third column. (Multiply the third row by (a+ b+ c) and subtract from the fourth row.)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 27th 2011, 09:27 PM
  2. Replies: 5
    Last Post: June 16th 2011, 07:32 AM
  3. Replies: 2
    Last Post: January 18th 2011, 08:24 PM
  4. unitary and upper triangular matrix => diagonal matrix
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 10th 2009, 06:52 PM
  5. Replies: 3
    Last Post: June 15th 2009, 02:16 PM

Search Tags


/mathhelpforum @mathhelpforum