Bhavya,

Write out the matrix for (say) n=5 - this won't take you very long.

The expansion of the determinant has n! terms, one for every permutation p() of {1,2,..n}. The term is equal to

A[1,p(1)]*A[2,p(2)]*...*A[n,p(n)]*(-1)^sign(p)

The only mysterious part here is the sign of the permutation. This is the number of pair swaps needed to achieve it starting from the identity; a slightly deep theorem of combinatorics says that while this is not well-defined, it IS well defined modulo 2, which is all you need in order to raise -1 to that power. [In other words, a position reachable in an odd number of swaps cannot be reached in an even number of swaps. Or, as swaps are self-inverse, no odd sequence of swaps can return to the original position.]

Equivalently there is one term for every subset S of the matrix such that S has exactly one element in each row and in each column. This is visually appealing but doesn't give an easy expression for the sign.

OK now - for your 5x5 matrix how many of these 120 terms are nonzero? This won't take you long either.

You should now be able to find the answer.

Good Hunting!
RD