Consider an n*n Matrix called A, where the elements within A are either 1 or 0. As n approaches Infinity, What percent of the possible matrices have det(A)=0?
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As a start, it is easier to consider the percentage of matrices that have an even determinant; of course if this percent is zero then you have your answer. – Ben Grossmann Aug 31 '17 at 04:04
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1In terms of $n$, that answer happens to be $$ \prod_{k=1}^n(1 - 2^{-k}) \approx 0.28879 $$ which, notably, is not zero. – Ben Grossmann Aug 31 '17 at 04:08
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It's because the probability that 2 row's are the same decrease exponentially as more rows/columns are added. – mtheorylord Aug 31 '17 at 04:10
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But the number of rows/columns increase linearly. – mtheorylord Aug 31 '17 at 04:11
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2Actually, my answer above should have been $1 - 0.28879$, which is perhaps surprisingly high. – Ben Grossmann Aug 31 '17 at 04:15
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Related (https://mathoverflow.net/questions/18547/number-of-unique-determinants-for-an-nxn-0-1-matrix) (http://mathworld.wolfram.com/01-Matrix.html). – Jean Marie Aug 31 '17 at 04:29