A probability problem by Parth Sankhe

Each number is counted twice, once in its row and once in its column. This means the number of 1s and -1s must be equal for the result to be 0. 25x25 = 625 which can not be halved evenly, so the # of 1s and -1s can never be equal.

For any matrix $M,$ we can denote $S(M)=r_1+r_2+...+r_{25}+c_1+c_2+...+c_{25}$ . Let $N$ represent the $25\times 25$ -matrix consisting only of 1s. Obviously, $S(N)=50.$ Now for any matrix $M$ like the one defined in the problem, there is a finite sequence of matrices $M_1, M_2, M_3, ..., M_n,$ such that $M_1=N,$ $M_n=M,$ and every $M_k$ can be obtained from $M_{k-1}$ by changing the sign of only one of the entries of the latter matrix. It is easy to see that $S(M_k)$ is either equal to $S(M_{k-1}),$ $S(M_{k-1})+4,$ or $S(M_{k-1})-4.$ That is why all the members of the sequence are $(S(M_k))_{1\leq k \leq n}$ are congruent modulus 4. Then $S(M)\equiv S(N) \equiv 50 \equiv 2$ mod 4. Therefore, $S(M)\neq 0$ and the answer to the problem is $\boxed{0}.$

The sum of these 50 terms to be zero suggests that there are 25 $1$ s and 25 $-1$ s. [ $r_i$ and $c_j$ can either be 1 or -1]

This tells us that the product of these 50 terms = $(1)^{25}\cdot (-1)^{25}=-1$ .

But, in the product of these terms each $r_i$ will be multiplied to each $c_j$ , making each term $a_{i,j}$ being multiplied to itself.

Thus, the whole product would become product of each $(a_{i,j})^2$ . This a product of 50 squares, and thus a product of 50 positive terms, which can never be $-1$ .

Therefore, no such matrix exists.