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The value of determinant can either be -2,-1,0,1,2 where number of determinant with value 2 will be equal to that of determinant with value -2. Same goes with 1 and -1.

Tushar chaturvedi
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  • Exhaustive search : you have "only" $2^9$ determinants to compute... Is there a "clever" method ? – Jean Marie Jun 14 '19 at 20:28
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    @JeanMarie See my answer below-- don't know how "clever" it is, at least better than a full computer search. [IF it's right...] – coffeemath Jun 14 '19 at 22:40

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The set of $0,1$ matrices dimension $3 \times 3$ which have nonzero determinant mod $2$ (when matrix entries viewed as in the group of integers mod $2$) form a group of order $168,$ which is isomorphic to the group of collineations of the Fano plane (a projective plane of seven points).

Now there are six $3 \times 3$ matrices whose actual determinant is not zero but is zero mod $2.$ For $\det=+2,$ one must choose two of the three downward diagonals to make all $1$'s, and the remaining downward diagonal must be all $0$'s. So 3 of this type, and another 3 like these but using upward for previous downward to get $\det=-2,$ for the total six mentioned.

So to count the number with determinant actually zero we compute $2^9-168=344$ giving the number having determinant even, and subtract the above mentioned 6 having even but nonzero determinant. So the result is $344-6=338,$ if I counted right.

coffeemath
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    You have counted right. Besides, there are $84$ matrices with determinant 1 (and $84$ as well for det -1) and only $3$ with det=2 (and $3$ with det -2). Total : $2^9=512$. Results obtained by brute force (using a computer, not by hand). – Jean Marie Jun 14 '19 at 23:12