Results 1 to 7 of 7
Like Tree1Thanks
  • 1 Post By emakarov

Math Help - Finding the Determinant

  1. #1
    Member
    Joined
    Oct 2013
    From
    Washington state
    Posts
    77
    Thanks
    1

    Finding the Determinant

    Prove that the determinant of the matrix

     \begin{bmatrix} 1&n&n&...&n\\ n&2&n&...&n\\ ...&...&...&...&...\\ n&n&n&...&n \end{bmatrix}

    is equal to (-1)^{n-1}n!.


    Do I approach this with induction? where do I start?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,559
    Thanks
    785

    Re: Finding the Determinant

    Subtract the last row from every other row.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,232
    Thanks
    1795

    Re: Finding the Determinant

    emakarov is suggesting a row operation. Do you know how row operations on a matrix change the determinant?

    If you swap two rows, you multiply the determinant by -1.

    If you multiply a row by a constant, you multiply the determinant by that constant.

    If you add a multiple of a row to another row (such as adding 1 times the last two to each other row) does not change the determinant.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2013
    From
    Washington state
    Posts
    77
    Thanks
    1

    Re: Finding the Determinant

    Quote Originally Posted by emakarov View Post
    Subtract the last row from every other row.
    THANK YOU! I would not have caught that trick. So here is what I did:

     \begin{bmatrix} 1&n&n&...&n&n\\ n&2&n&...&n&n\\ n&n&3&...&n&n\\ ...&...&...&...&...&...\\ n&n&n&...&n&n \end{bmatrix}

     = \begin{bmatrix} 1-n&0&0&...&0&0\\ 0&2-n&0&...&0&0\\ 0&0&3-n&...&0&0\\ ...&...&...&...&...&...\\ 0&0&0&...&(n-1)-n&0 \\ 0&0&0&...&0&0 \end{bmatrix}

     = \begin{bmatrix} -1(n-1)&0&0&...&0&0\\ 0&-1(n-2)&0&...&0&0\\ 0&0&-1(n-3)&...&0&0\\ ...&...&...&...&...&...\\ 0&0&0&...&-1&0 \\ 0&0&0&...&0&0 \end{bmatrix}

    Which of course is a diagonal matrix so the determinant is the product of the diagonal terms, however the last one is 0, which throws this off. What did I do wrong?

    I can see that the first n-1 terms give you (-1)^{n-1}(n-1)! but where does the last n come from?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,559
    Thanks
    785

    Re: Finding the Determinant

    Quote Originally Posted by emakarov View Post
    Subtract the last row from every other row.
    The hint was to subtract the last row from every row except the last one. Subtracting a row v from a different row w, (as HallsofIvy wrote, formally this is replacing w with the sum w + (-1)v) does not change the determinant. Subtracting a row from itself definitely changes the determinant in general: it turns it to zero!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2013
    From
    Washington state
    Posts
    77
    Thanks
    1

    Re: Finding the Determinant

    Quote Originally Posted by emakarov View Post
    The hint was to subtract the last row from every row except the last one. Subtracting a row v from a different row w, (as HallsofIvy wrote, formally this is replacing w with the sum w + (-1)v) does not change the determinant. Subtracting a row from itself definitely changes the determinant in general: it turns it to zero!
    So if you don't subtract the last row from itself then you would have n|diagonalmatrix| -n|diagonalmatrix| +n|diagonalmatrix| ... Which does not seem to give n!.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,559
    Thanks
    785

    Re: Finding the Determinant

    I'll borrow your wonderful LaTeX code.

    Expanding the determinant along the last column (Laplace's formula), we get

    \begin{vmatrix} -1(n-1)&0&0&\dots&0&0\\ 0&-1(n-2)&0&\dots&0&0\\ 0&0&-1(n-3)&\dots&0&0\\ \dots&\dots&\dots&\dots&\dots&\dots\\ 0&0&0&\dots&-1&0 \\ n&n&n&\dots&n&n \end{vmatrix}={}

    n\begin{vmatrix} -1(n-1)&0&0&\dots&0\\ 0&-1(n-2)&0&\dots&0\\ 0&0&-1(n-3)&\dots&0\\ \dots&\dots&\dots&\dots&\dots\\ 0&0&0&\dots&-1\end{vmatrix}={}

    n(-1)^{n-1}(n-1)!=(-1)^{n-1}n!.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. finding determinant
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 8th 2012, 09:16 PM
  2. Finding Determinant
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 12th 2010, 07:57 PM
  3. finding determinant of inverse
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 22nd 2010, 06:32 AM
  4. Finding the determinant
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 12th 2008, 11:57 AM
  5. Finding The Determinant
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: October 21st 2006, 08:56 AM

Search Tags


/mathhelpforum @mathhelpforum