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  1. T

    Computers & Determinants

    Looks like if you pick out all the distinct terms (the ones in the summation that aren't equal to zero) you get (N!). On top of that, you would need to do N multiplications each time (so N multiplications N! times). Yikes Yeah didn't work that one out before
  2. T

    Computers & Determinants

    How would this fare computationally? det(A) = \sum_{{i}_1, {i}_2,...,{i}_n = 1}^{N} \epsilon_{{i}_1, {i}_2,...,{i}_n} a_{{i}_1, 1} \cdot a_{{i}_2, 2} \cdot \cdot \cdot a_{{i}_n, N} Cumbersome and inefficient for a computer algorithm?
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