Does a matrix $A$ of $\pm1$'s of order $11$ exist with $\det A >4000$?
Solution 1:
This is a $12\times12$ Hadamard matrix (from Neil Sloane's page).
+-----------
++-+---+++-+
+++-+---+++-
+-++-+---+++
++-++-+---++
+++-++-+---+
++++-++-+---
+-+++-++-+--
+--+++-++-+-
+---+++-++-+
++---+++-++-
+-+---+++-++
Its rows are pairwise orthogonal, and have squared length $12$, so the determinant of the whole thing is $12^6$. What is the maximum of the determinant of an $11\times11$ minor?
Well, $HH^T=12I_{12}$, so the minors all have absolute value $12^5=248832>4000$ (think cofactors in $H^{-1}=H^T/12$).
Solution 2:
Lets walk backwards. Pick any such matrix with the extra $$A=\begin{pmatrix} 1&1&1&1&1&1&1&1&1&1&1 \\ -1&*&*&*&*&*&*&*&*&*&* \\ -1&*&*&*&*&*&*&*&*&*&* \\ -1&*&*&*&*&*&*&*&*&*&* \\ -1&*&*&*&*&*&*&*&*&*&* \\ -1&*&*&*&*&*&*&*&*&*&* \\ -1&*&*&*&*&*&*&*&*&*&* \\ -1&*&*&*&*&*&*&*&*&*&* \\ -1&*&*&*&*&*&*&*&*&*&* \\ -1&*&*&*&*&*&*&*&*&*&* \\ -1&*&*&*&*&*&*&*&*&*&* \\ \end{pmatrix}$$
Adding the first row to the others, and getting a two from the other rows we get
$$\det(A)=2^{10} \det\begin{pmatrix} 1&1&1&1&1&1&1&1&1&1&1 \\ 0&*&*&*&*&*&*&*&*&*&* \\ 0&*&*&*&*&*&*&*&*&*&* \\ 0&*&*&*&*&*&*&*&*&*&* \\ 0&*&*&*&*&*&*&*&*&*&* \\ 0&*&*&*&*&*&*&*&*&*&* \\ 0&*&*&*&*&*&*&*&*&*&* \\ 0&*&*&*&*&*&*&*&*&*&* \\ 0&*&*&*&*&*&*&*&*&*&* \\ 0&*&*&*&*&*&*&*&*&*&* \\ 0&*&*&*&*&*&*&*&*&*&* \\ \end{pmatrix}=\\2^{10} \det\begin{pmatrix} *&*&*&*&*&*&*&*&*&* \\ *&*&*&*&*&*&*&*&*&* \\ *&*&*&*&*&*&*&*&*&* \\ *&*&*&*&*&*&*&*&*&* \\ *&*&*&*&*&*&*&*&*&* \\ *&*&*&*&*&*&*&*&*&* \\ *&*&*&*&*&*&*&*&*&* \\ *&*&*&*&*&*&*&*&*&* \\ *&*&*&*&*&*&*&*&*&* \\ *&*&*&*&*&*&*&*&*&* \\ \end{pmatrix}$$
where this time every $*$ is $0$ or $1$. Note that starting from this last $0,1$ matrix you can backtrack your row reduction and get a $-1,1$ matrix.
Thus the problem reduced to the following:
Find a $10\times 10$ matrix with all entries $0,1$ so that the determinant is at least $4$, which is easy to solve. An easy way to produce such a matrix is the following:
$$B=\begin{pmatrix} 0&1&1&1&1 \\ 1&0&1&1&1 \\ 1&1&0&1&1 \\ 1&1&1&0&1 \\ 1&1&1&1&0 \\ \end{pmatrix}$$
Then $B$ is invertible, and the determinant is a multiple of $4$ [just add all the rows to the first one.]
Construct now the $0,1$ matrix
$$\begin{pmatrix} B&0 \\ 0&I_5 \\ \end{pmatrix}$$
Backtracking back to, any $1$ in $\begin{pmatrix} B&0 \\ 0&I_5 \\ \end{pmatrix}$ stays a $1$, any $0$ becomes $-1$ and you add the first row and column of $A$.