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Determinant of an n x n matrix

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the determinant of the matrix given by:

    \begin{array}{ccc}
    1 & 2 & 3 & ... & n \\
    2 & 2 & 3 & ... & n \\
    3 & 3 & 3 & ... & n \\
    . & . & . & & . \\
    . & . & . & & . \\
    . & . & . & & . \\
    n & n & n & ... & n \end{array}



    2. Relevant equations

    We use expansion by minors to find determinants.


    3. The attempt at a solution

    So, I computed the matrix for n = 1, 2, 3 and 4 and so reckon I need to show that the determinant is equal to (-1)^(n-1)*n.

    I tried expanding by minors in the last row, but that just seems to give me a load of slightly smaller matrices of which I must find the determinant...
     
  2. jcsd
  3. Apr 8, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Do you have to use expansion by minors on the original matrix? It is easy to simplify this matrix a lot with row (or column) operations, and determine the determinant afterwards in a very simple expansion.
     
  4. Apr 8, 2013 #3
    In what way can you transform the determinant without changing its final value?
     
  5. Apr 8, 2013 #4
    Row operations hadn't crossed my mind! (Feel a bit foolish seeing as the chapter is on determinants and row operations!)

    That's a really good point though. Will have a go using EROs now.

    EDIT: Done! Thanks a lot. Once you use EROs, the answer pretty much drops out. For anyone who's interested, you take the 1st column away from all the other columns and then expand by minors in the bottom row as it has an n as its first entry and then the rest are zeros. So, you get (-1)^(n-1)*n multiplied by the determinant of an upper triangular matrix with ones as the entries for all of its diagonals - so its determinant is one. Thanks again.
     
    Last edited: Apr 8, 2013
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