Let the maximum value of $n \times n$ determinant with $n \in\mathbb{N} >1$ be $D_n$ . If each of the elements in determinant is either $+1$ or $-1$ , Then the successive values of $D_n$ makes a Geometric Progression .
$D_2$
$\begin{vmatrix} 1&-1 \\ 1&1 \\ \end{vmatrix}=2$
$D_3$
$\begin{vmatrix} 1&-1&\pm1 \\ 1&1&-1 \\ 1&1&1 \end{vmatrix}=4$
$D_4$
$\begin{vmatrix} 1&-1&1&-1 \\ 1&-1&-1&\pm1 \\ 1&1&1&-1\\ 1&1&1&1\\ \end{vmatrix}=8$
$D_5$
$\begin{vmatrix} 1&1&-1&-1&1 \\ -1&1&-1&1&1 \\ 1&1&1&1&1\\ -1&1&1&1&-1\\ 1&-1&1&1&-1\\ \end{vmatrix}=16$
$D_6$
$\begin{vmatrix} 1&-1&1&1&1&1 \\ 1&1&1&-1&-1&-1 \\ 1&1&1&-1&1&1\\ -1&-1&-1&1&1&-1\\ 1&1&1&1&-1&-1\\ -1&-1&1&1&1&1\\ \end{vmatrix}=32$
And so on...
CRUX -
Optimization- The determinant being optimized to maximum value always yields a number which follows up a geometric progression , Why isn't it even a bit higher or lower than it , to remove our this specific qualm , it's always saturated to a G.P. and strictly follows the same algorithm . Next thing is the Result of the Optimized Determinant - $D_n$ $=$ Always a G.P. , how does it strictly follow up being a G.P. , suppose that we frame it algebraically , then will it show the same behavior ?! It's very necessary for algebraic frameworks .
Configuration- Where to plot the elements so that in order to optimize it ?! The order increases ... So does the difficulty. Does it holds true for even more than $2$ elements ?! And will it also yield G.P. as output ? Can we also have negative numbers and later non-integral Reals & which also follows the same G.P. tradition ?! Configuration also includes the order of configuration of the determinant itself , so can we have $n\times n$ type determinant with at most $2$ distinct elements which follows the same above argument ?!
Till now we saw that as the order of the determinant increases , the difficulty in optimization & it's respective element's configuration also increases ...
Generalization- Generalizing it & framing it to a well finished conjecture ! Yes , but there always exists the Qualm about optimization & configuration issues as per mentioned above ... I did tried distinct elements - $1$ & $2$ this time , and surprisingly it came out to generate Geometric progressive numbers as output ! Holy Jesus ! Before generalizing one should keep all the above issues in mind & Is it a NP-HARD /COMPLETE problem which I end up with? ...
Proof Positive- The ultimate explanation for this Enigmatic behavior of such determinant's optimization and it also includes the strict algorithm for the configuration of elements and generation of the final & the absolute proof dealing with all the above arguments.
My dear colleague mrtaurho has made significant efforts towards the conjecture and I'll phrase his words here , and that'll count in the latest progress made in my conjecture over the sphere of internet.
[ Source - Aops community Forum named - Mathematics Research Station ]
mrtaurho's article -
As brilliantly found by Alphatrion there are some matrices of size $n\times n$ of the given type with a determinant precisely $2^{n-1}$. However, this value is far away from being maximal. I have not found a version of $D_3$ with determinant $\geqslant4$ but for all values $n>3$ (we may distinguish between odd and even $n$) it is worth considering matrices of the type $$D_{2k+1}^*=\begin{pmatrix}-1&\dots&1\\1&\ddots&\vdots\\1&\dots&-1\end{pmatrix}~~~D_{2k}^*=\begin{pmatrix}1&\dots&-1\\1&...&\vdots\\-1&\dots&1\end{pmatrix}$$ For those we gain $D^*_4=2^4=16,~D^*_5=s^4\cdot3=48,~D^*_6=2^7,~D^*_7=2^6\cdot5,~D^*_8=2^8\cdot3,~D^*_9=2^8\cdot7,\dots$ from where one can guess that in general we have $D^*_n=(n-2)2^{n-1}$ which is clearly greater than $2^{n-1}$ alone. This seems to work out but I have no idea how to prove this relation.
[Small Remark: sticking to the type $D^*_n$ given for odd $n$ the formula changes to $D_n^*=(-1)^{n-1}(n-2)2^{n-1}$; so the alternate form for even $n$ is just for aesthetics].
I have found the values using this online calculator Another observation worth being mentioned: $|D_n|\leqslant|D_{n+1}|$. This is quite simple by using block matrices as we have that $$|\det(D^\star_n)|=\left|\begin{pmatrix}\mathbf{1}&\mathbf{0}\\\mathbf{0}&\mathbf{D_{n-1}}\end{pmatrix}\right|=|1\cdot\det(D_n)|=|\det(D_n)|$$ And given $D_n^*$ we know that there are configurations for $D_n$ whose determinant exceeds $D_{n-1}$.
To end this post I would like to mention Leibniz Formula for Determinants which states that $$\det(A)=\sum_{\tau\in S_n}\operatorname{sign}(\tau)\prod_{i=1}^na_{i,\tau(i)}$$ Here $\tau$ is a permutation of the symmetric group (on $n$ letters) $S_n$ and $\operatorname{sign}(\tau)$ returns the sign of the permutation $\tau$. Note that there are precisely $|S_n|/2=n!/2=|A_n|$ even and thus $n!/2$ odd permutations. However, I do not see how one may apply this formula here but I thought it may be good to have it linked here.
So the above article ends up here and the contribution of user Zatanna on Aops ( which he gave values of $D_5$ & $D_6$ have been merged ) . I choose $\pm1$ to initially explain my observations because it's easier than picking up any other set of elements available to us, The results holds true for other specific cases too. Source - Mathematics Research Station ( MARS )