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Using permutations explain how for a triangular matrix only one term can be non zero.

Please do not include any proofs using the cofactor method.

Edit (OP's attempt as written in the comment section):
My [attempted] proof states that for a lower triangular matrix we start with $n$ choices but $n-1$ of them are zero. We must choose the upper left hand corner because the other choices will lead to a zero term and therefore a zero permutation. Moving on to the next row we are faced with $n-1$ choices, but $n-2$ of them are zero. Therefore we again must make the choice along the diagonal. The proof continues like this until we arrive at the lower right hand corner. Then I declare that $\det(A)= \det(A')$ and therefore it is proven for any triangular matrix.

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  • $\begingroup$ Your question title says "diagonal matrix", but your question details say "triangular matrix". $\endgroup$
    – M. Vinay
    Jul 8 '16 at 5:47
  • $\begingroup$ What are your thoughts on the problem? What have you tried? $\endgroup$ Jul 8 '16 at 15:12
  • $\begingroup$ Hint: Use the Liebniz expansion. Any permutation besides the identity permutation will satisfy $\sigma(i) > i$ and $\sigma(j) < j$ for some $i$ and $j$. $\endgroup$ Jul 8 '16 at 15:15
  • $\begingroup$ @Omnomnomnom My proof states that for a lower triangular matrix we start with n choices but n-1 of them are zero. We must choose the upper left hand corner because the other choices will lead to a zero term and therefore a zero permutation. Moving on to the next row we are faced with n-1 choices, but n-2 of them are zero. Therefore we again must make the choice along the diagonal. The proof continues like this until we arrive at the lower right hand corner. Then I declare that Det(A)=Det(A') and therefore it is proven for any triangular matrix. $\endgroup$
    – Mike K
    Jul 9 '16 at 6:47
  • $\begingroup$ @M.Vinay It should say triangular matrix in both cases. I don't know how to edit it. $\endgroup$
    – Mike K
    Jul 9 '16 at 6:48

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