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The number of binary matrices of order  $N*N$  whose determinant is exactly zero.
asked in Linear Algebra by Active (5k points) | 31 views
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@Mk Utkarsh 

yeah i checked that but i didn't find any method or may be i overlooked it.

+1

let X(n) be number of singular matrices of Number of singular $N \times N$ rational (0,1) matrices.

let Y(n) be number of binary matrices of order  $N \times N$  whose determinant is exactly zero.

$Y(n) =$$\large  2^{(n^2)} - n! * \binom{2^n -1}{n} + n! * X(n)$ 

I don't think it is required to to remember all these things for GATE. 

 

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@Mk Utkarsh 

actually i thought it also have some trivial method as odd/even #determinant matrices has but someone on SO mentioned it as somehow tough job so yeah may be it wont be asked.

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