Suppose that the elements of an n- order determinant are either $0$ or $1$, I wonder the maximum and the minimum of the determinant.
Furthermore, I wonder all possible values of it. What if I change the question that the elements are either $1$ or $-1$?
Closed forms for the maximum and minimum are not known, but there are decent bounds available. For a $(0,1)$ $n\times n$ matrix $A$, we have $$|\det A|\leq \frac{(n+1)^{(n+1)/2}}{2^n}$$ and for a $(-1,1)$ matrix the maximum value is $2^{n-1}$ times the maximum $(0,1)$ value. See mathworld for more info.
Note that in both cases the minimum value is the additive inverse of the maximum value for $n\geq 2$, since we can permute two rows to achieve this value.
The determinant of a matrix $A=a_{i,j}$,$1 \le i,j \le n$ is given by (where $S_n$ is the set of all permutations of $\{1,\ldots,n\}$ and $\operatorname{sgn}(\sigma)$ is the sign of a permutation $\sigma \in S_n$ ($-1$ for odd, $1$ for even)
$$\det(A)=\sum_{\sigma \in S_n}\operatorname{sgn}(\sigma) \prod_{i=1}^n a_{i, \sigma(i)}$$
so an alternating sum of terms, which are each products of $n$ terms. All these product terms are $0$ or $1$, so the greatest value is achieved when all even permutations yield $1$ while odd permutations yield $0$, e.g. So $\frac{n!}{2}$ is best (not sure if it's reachable though).
With $1,-1$ it's more complicated. Maybe check out this question or this reference
number-theory? – José Carlos Santos Dec 29 '19 at 15:13