3rd order determinant is formed using 0 & 1 as its elements. How many determinant will give value 0.

Question

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.

Answer 1

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.