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Thread: How do I find the determinant of this matrix?

  1. #1
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    How do I find the determinant of this matrix?

    I have the following matrix:
    A - square matrix of n*n as follows:

    How do I find the determinant of this matrix?-1.jpeg

    I figured out that the determinant is $\displaystyle (-1)^{n-1}(n-1)$, but how do I prove it?

    P.S: no, induction doesn't work...
    Last edited by DMan16; Nov 28th 2016 at 03:46 AM.
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  2. #2
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    Re: How do I find the determinant of this matrix?

    The matrix has 0's along the main diagonal, 1's everywhere else? Then $\displaystyle (-1)^n(n- 1)$ is NOT correct. When n= 2, the matrix is $\displaystyle \begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}$ which has determinant -1, not $\displaystyle (-1)^2(2- 1)= 1$. Perhaps it is $\displaystyle (-1)^{n-1}(n-1)$.
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  3. #3
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    Re: How do I find the determinant of this matrix?

    yes, my bad, forgot the -1 in the power - changed it.
    any thoughts on how to solve it?
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  4. #4
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    Re: How do I find the determinant of this matrix?

    Quote Originally Posted by DMan16 View Post
    yes, my bad, forgot the -1 in the power - changed it.
    any thoughts on how to solve it?
    One thing I notice about it is that given $n$ the eigenvalues are $n-1$, and $n-1$ repeated values of $-1$

    Given this clearly the determinant is just $(-1)^{n-1}(n-1)$

    it's not too hard to see that the eigenvector of all 1's produces the eigenvalue of $n-1$, it's just the sum of the row values of $C$ which is clearly $n-1$

    the eigenvectors corresponding to the eigenvalue of $-1$ are of the form

    $(-1, 0, 0, 0, \dots , 1, 0, \dots, 0)$

    where the $1$ appears in $n-1$ different rows corresponding to $n-1$ different eigenvectors.

    I know this doesn't solve your problem but it might help.

    cheers
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