Similar problem of Hadamard's maximal determinant problem Hadamard's maximal determinant problem asks for the largest determinant of a matrix with elements equal to 1 or −1.
But what would be the case when the elements are equal to $\pm1\text{ or }0$?
I've tried to look for some reference, but all the reference I found were about elements being equal to 0 or -1 at best.
Anyone knows something about the case for elements being equal to $\pm1\text{ or }0$?
Thanks in advance.
 A: Actually Hadamard's determinant inequality is more general and also applies to your question. Thanks to @hardmath for pointing out the reference here.
The determinant of any $n\times n$ complex matrix $A$ with columns $a_i$ obeys
$$
| \det(A) |\leq \prod_{1\leq i\leq n} \Vert a_i \Vert,
$$
where the norm is the Euclidean norm.
It happens that certain $\pm 1$ entry Hadamard matrices achieve this bound. If you introduce a single zero in such a matrix without making it singular, you can compare the resulting determinant to this upperbound.
Since the proof is via the arithmetic-geometric mean inequality, and achieved for the case of equal norm columns, introducing this zero is unlikely to achieve the upperbound.
In your case of entries from $\{0,\pm 1\}$, the squareroot of the Hamming weight of each row would appear in the right hand side of the bound.
A: The maximum (absolute) determinant of an $n\times n$ (real) matrices with entries in $\{0,\pm 1\}$ is the same as the maximum determinant when their entries are restricted to $\{\pm 1\}$.
This is noted without proof in the OEIS A003432 sequence, "largest determinant of a (real) $\{0,1\}$-matrix of order $n$."

Suppose M = (m(i,j)) is an n X n matrix of real numbers. Let
a(n) = max det M subject to m(i,j) = 0 or 1 [this sequence],
g(n) = max det M subject to m(i,j) = -1 or 1 A003433,
h(n) = max det M subject to m(i,j) = -1, 0 or 1 A003433,
F(n) = max det M subject to 0 <= m(i,j) <= 1 [this sequence],
G(n) = max det M subject to -1 <= m(i,j) <= 1 A003433.
Then a(n) = F(n), g(n) = h(n) = G(n), g(n) = 2^(n-1)*a(n-1). Thus all five problems are equivalent.

Apparently the proof is a simple matter of considering the cofactor expansion of the matrix determinant.  An article "The Hadamard Maximum Determinant Problem," by Joel Brenner (American Math Monthly Vol. 79, No. 6 (Jun. - Jul., 1972), pp. 626-630) available online through JSTOR merely says this:

Consider any real $n\times n$ matrix $A = (a_{ij})$ with $|a_{ij}|\le 1$. Expanding the determinant of any such $A$ by minors along successive rows it is apparent that $\det A$ is dominated by the determinant of a $(-1,1)$ matrix; i.e., a matrix all of whose entries are either $-1$ or $1$.  Since there are finitely many such matrices the maximum determinant problem has a solution for each $n$.

We could elaborate on this by asking for the $\{0,\pm 1\}$-matrix, among those attaining the maximum determinant, having fewest zero entries.  If there were a zero entry, then by the cofactor of that entry being either non-negative or non-positive, we could vary the entry from $0$ to either $1$ or $-1$ respectively to attain the maximum determinant with fewer zero entries.
