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Linear Algebra question: n x n determinants

  1. Mar 12, 2006 #1
    I know what a determinant is and how to solve for it, but in the case of this n x n matrix on my homework, I'm at a complete loss as to how to even begin starting this. I don't expect anyone to work it out for me, just to give me a solid method for how to solve it and prove my method for the n by n case.

    The problem states to find the determinant and prove our result for the n by n case for the square n by n matrix defined as:


    This is the square (n x n) tridiagonal matrix with 'b' on the first and third diagonal and 'a' on the primary diagonal. Thanks everyone for your help in advance :)
    Last edited: Mar 13, 2006
  2. jcsd
  3. Mar 12, 2006 #2
    Call the upper non-zero submatrix for U and the lower for L. Then the determinant is the product of their determinants, i.e det(U)det(L)
    Last edited: Mar 12, 2006
  4. Mar 12, 2006 #3
    Can anyone else verify if this method works and if so why? Thanks for your time, incredible.
  5. Mar 12, 2006 #4
    This method will work.
    say you have a matrix
    [A 0] say A is n x n and B is m x m then you can write
    [0 B]

    [A 0] = [A 0] [I 0] so you're matrix is in the form of the product of these
    [0 B] [0 I] [0 B]

    two matrices. then you can use the product formula for determinants
    then and you know that
    det[A 0]=detA and the same for B.
    [0 I]
    Last edited: Mar 12, 2006
  6. Mar 13, 2006 #5
    I miscopied the problem it would seem, it is actually a tridiagonal matrix rather than the one that previously appeared here. The picture above has been updated to reflect this.

    I again have no idea how to do this problem, any and all help is appreciated as before :)
  7. Mar 13, 2006 #6


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    You can find a recursive formula by cofactor expansion. The expansion will be in terms of a, b, the determinant of the same type of matrix of size n-1, and the determinant of a matrix of size n-1 that is almost the same type except it has a b in the top left instead of an a.

    But you know, I did the first few determinants on my calculator and I don't see a simple pattern. You might want to use a calculator to solve the recurrence.
    Last edited: Mar 13, 2006
  8. Mar 13, 2006 #7
    try changing it into an upper diagonal matrix (a matrix where all entries below the main diagonal are zero) using these properties
    1) interchange two rows which gives a sign change in the determinant
    2) multiply all elements in a row by a nonzero scalar which multiplies the determinant by the nonzero scalar
    3)adding to one row a scalar multiple of the other which leaves the determinant unchanged

    then the determinant is the product of the diagonal entries. Don't forget about what operations you used on the matrix to turn it into an upper diagonal because you'll still need to change the determinant accordingly.
    i.e. property 1 change the sign of the determinant everytime you interchange two rows etc.
  9. Mar 13, 2006 #8
    How can I use row operations for the n x n matrix when I don't know how many rows there will be? I don't understand how that yields a formula for the n x n case.
  10. Mar 13, 2006 #9
    sorry, i wasn't thinking for the n x n case. try using cofactor expansion and i think you'll start to recognize a pattern developing which should be able to help you.
  11. Mar 14, 2006 #10


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    Expand by minors using the first column (or, equivalently, the first row). You should see a recursive method for that.
  12. Mar 14, 2006 #11
    Thanks everybody
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