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any help or light on this question would be hugely appreciated!!
thank you
That shows, I think, that the minimum number of 1s in such an invertible n by n matrix is n. But the question was about the maximum number. We do know that if a matrix has two columns the same, it is not invertible. I think we can avoid that by having a 0, at a different row, in each column. That implies that the maximum number of 1s in such an invertible n by n matrix is $\displaystyle n^2- n$.