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As pointed out in Robin Chapman's answer and Frederio Poloni's comment thereto, there is a one-to-one map between normalized $n$-by-$n$ $(-1,1)$ matrices and $(n-1)$-by-$(n-1)$ $(0,1)$ matrices under which the determinant gets multiplied by $2^{-n+1}$ and possibly a minus sign. Since normalizing a $(-1,1)$ matrix changes the determinant by at most a sign, the maximal determinant problem for $(-1,1)$ matrices of size $n$ is equivalent to the maximal determinant problem for $(0,1)$ matrices of size $n-1$.

An advantage to working with $(-1,1)$ matrices is that their rows are all vectors of the same magnitude, $\sqrt{n}$. If the rows are orthogonal, the matrix is Hadamard and the magnitude of the determinant is $n^{n/2}$ which is the best possible. It is not hard to show that if a Hadamard matrix of size $n$ exists, than $n$ is 1, 2, or a multiple of 4, but it is not known whether a Hadamard matrix exists for every multiple of 4.

If $n$ is not a multiple of 4, better upper bounds on the determinant of an $n$-by-$n$ $(-1,1)$ matrix $R$ can be obtained by analyzing the properties that the matrix $G=RR^{T}$ must have. An upper bound on the determinant for the class of matrices $\tilde G$ with these properties leads to an upper bound on $\det R$ by taking the square root. This idea was developed by Barba, Ehlich, and Wojtas, leading to the following bounds:

$n\equiv1\pmod{4}$: The optimal $\tilde G$ is $J+(n-1)I$ where $J$ is the all-ones matrix. Therefore $\det R\le\sqrt{2n-1}(n-1)^{(n-1)/2}$. Clearly this bound is attainable only if $2n-1$ is a perfect square. It is not known whether this is if-and-only-if, but that is conjectured to be the case.

$n\equiv2\pmod{4}$: The optimal $\tilde G$ is $2\binom{J \ \ 0}{0 \ \ J}+(n-2)I$ where $J$ and $0$ are $n/2$-by-$n/2$ matrices. This provides the bound $\det R \le (2n-2)(n-2)^{(n-2)/2}$. It can be shown that an $R$ such that $RR^T$ equals this optimal form exists only if $n-1$ is the sum of two squares. Once again, this is conjectured to be if-and-only-if.

$n\equiv3\pmod{4}$: The optimal $\tilde G$ is $-J+4B+(n-3)I$ where $B$ is a block-diagonal matrix with all-ones blocks along the diagonal. These all-ones blocks should be as nearly equal in size as possible and their number should be five if $n=7$, five or six if $n=11$, six if $15 \le n \le 59$, and seven if $n \ge 63$. The bound on $\det R$ \tilde G$ one gets from this is seldom a perfect square, and even when it is, Tamura showed that the existence of an $R$ such that $RR^T$ equals the optimal form $\tilde G$ is ruled out by the Hasse-Minkowski theorem in many cases. The smallest $n$ for which the bound might be attainable is 511. Nevertheless, there exist matrices with determinant quite close to the bound. For example, when $n=19$, a matrix whose determinant attains 97.5% of the bound is known. Using a lot of computer time, this particular determinant was recently shown to be best possible by Brent, Osborn, Zimmermann and myself.
show/hide this revision's text 2 added 3 characters in body

As pointed out in Robin Chapman's answer and Frederio Poloni's comment thereto, there is a one-to-one map between normalized $n$-by-$n$ $(-1,1)$ matrices and $(n-1)$-by-$(n-1)$ $(0,1)$ matrices under which the determinant gets multiplied by $2^{-n+1}$ and possibly a minus sign. Since normalizing a $(-1,1)$ matrix changes the determinant by at most a sign, the maximal determinant problem for $(-1,1)$ matrices of size $n$ is equivalent to the maximal determinant problem for $(0,1)$ matrices of size $n-1$.

An advantage to working with $(-1,1)$ matrices is that their rows are all vectors of the same magnitude, $\sqrt{n}$. If the rows are orthogonal, the matrix is Hadamard and the magnitude of the determinant is $n^{n/2}$ which is the best possible. It is not hard to show that if a Hadamard matrix of size $n$ exists, than $n$ is 1, 2, or a multiple of 4, but it is not known whether a Hadamard matrix exists for every multiple of 4.

If $n$ is not a multiple of 4, better upper bounds on the determinant of an $n$-by-$n$ $(-1,1)$ matrix $R$ can be obtained by analyzing the structure properties that the matrix $G=RR^{T}$ must have. An upper bound on the determinant for the class of matrices $\tilde G$ with this structure these properties leads to an upper bound on $\det R$ by taking the square root. This idea was developed by Barba, Ehlich, and Wojtas, leading to the following bounds:

$n\equiv1\pmod{4}$: The optimal $\tilde G$ is $J+(n-1)I$ where $J$ is the all-ones matrix. Therefore $\det R\le\sqrt{2n-1}(n-1)^{(n-1)/2}$. Clearly this bound is attainable only if $2n-1$ is a perfect square. It is not known whether this is if-and-only-if, but that is conjectured to be the case.

$n\equiv2\pmod{4}$: The optimal $\tilde G$ is $2\binom{J \ \ 0}{0 \ \ J}+(n-2)I$ where $J$ and $0$ are $n/2$-by-$n/2$ matrices. This provides the bound $\det R \le (2n-2)(n-2)^{(n-2)/2}$. It can be shown that an $R$ such that $RR^T$ equals this optimal form exists only if $n-1$ is the sum of two squares. Once again, this is conjectured to be if-and-only-if.

$n\equiv3\pmod{4}$: The optimal $\tilde G$ is $-J+4B+(n-3)I$ where $B$ is a block-diagonal matrix with all-ones blocks along the diagonal. These all-ones blocks should be as nearly equal in size as possible and their number should be five if $n=7$, five or six if $n=11$, six if $15 \le n \le 59$, and seven if $n \ge 63$. The bound on $\det R$ one gets from this is seldom a perfect square, and even when it is, Tamura showed that the existence of an $R$ such that $RR^T$ equals the optimal form is ruled out by the Hasse-Minkowski theorem in many cases. The smallest $n$ for which the bound might be attainable is 511. Nevertheless, there exist matrices with determinant quite close to the bound. For example, when $n=19$, a matrix whose determinant attains 97.5% of the bound is known. Using a lot of computer time, this particular determinant was recently shown to be best possible by Brent, Osborn, Zimmermann and myself.
show/hide this revision's text 1

As pointed out in Robin Chapman's answer and Frederio Poloni's comment thereto, there is a one-to-one map between normalized $n$-by-$n$ $(-1,1)$ matrices and $(n-1)$-by-$(n-1)$ $(0,1)$ matrices under which the determinant gets multiplied by $2^{-n+1}$ and possibly a minus sign. Since normalizing a $(-1,1)$ matrix changes the determinant by at most a sign, the maximal determinant problem for $(-1,1)$ matrices of size $n$ is equivalent to the maximal determinant problem for $(0,1)$ matrices of size $n-1$.

An advantage to working with $(-1,1)$ matrices is that their rows are all vectors of the same magnitude, $\sqrt{n}$. If the rows are orthogonal, the matrix is Hadamard and the magnitude of the determinant is $n^{n/2}$ which is the best possible. It is not hard to show that if a Hadamard matrix of size $n$ exists, than $n$ is 1, 2, or a multiple of 4, but it is not known whether a Hadamard matrix exists for every multiple of 4.

If $n$ is not a multiple of 4, better upper bounds on the determinant of an $n$-by-$n$ $(-1,1)$ matrix $R$ can be obtained by analyzing the structure that the matrix $G=RR^{T}$ must have. An upper bound on the determinant for the class of matrices $\tilde G$ with this structure leads to an upper bound on $\det R$ by taking the square root. This idea was developed by Barba, Ehlich, and Wojtas, leading to the following bounds:

$n\equiv1\pmod{4}$: The optimal $\tilde G$ is $J+(n-1)I$ where $J$ is the all-ones matrix. Therefore $\det R\le\sqrt{2n-1}(n-1)^{(n-1)/2}$. Clearly this bound is attainable only if $2n-1$ is a perfect square. It is not known whether this is if-and-only-if, but that is conjectured to be the case.

$n\equiv2\pmod{4}$: The optimal $\tilde G$ is $2\binom{J \ \ 0}{0 \ \ J}+(n-2)I$ where $J$ and $0$ are $n/2$-by-$n/2$ matrices. This provides the bound $\det R \le (2n-2)(n-2)^{(n-2)/2}$. It can be shown that an $R$ such that $RR^T$ equals this optimal form exists only if $n-1$ is the sum of two squares. Once again, this is conjectured to be if-and-only-if.

$n\equiv3\pmod{4}$: The optimal $\tilde G$ is $-J+4B+(n-3)I$ where $B$ is a block-diagonal matrix with all-ones blocks along the diagonal. These all-ones blocks should be as nearly equal in size as possible and their number should be five if $n=7$, five or six if $n=11$, six if $15 \le n \le 59$, and seven if $n \ge 63$. The bound on $\det R$ one gets from this is seldom a perfect square, and even when it is, Tamura showed that the existence of an $R$ such that $RR^T$ equals the optimal form is ruled out by the Hasse-Minkowski theorem in many cases. The smallest $n$ for which the bound might be attainable is 511. Nevertheless, there exist matrices with determinant quite close to the bound. For example, when $n=19$, a matrix whose determinant attains 97.5% of the bound is known. Using a lot of computer time, this particular determinant was recently shown to be best possible by Brent, Osborn, Zimmermann and myself.