The maximum value of determinant Suppose we have a matrix $A_{3\times3} = (a_{ij})$, where $a_{ij}\in \mathbb{Z}$ and $|a_{ij}|\le 9$ for $1\le i,j\le 3$.
What is the maximum value may take the value $\det (A)$?
I'm looking for a reasonable solution without the use of brute-force.
 A: This is by no means a concise answer, but we can at least avoid using brute force.
It is well known that for $[-1,1]$-matrices (i.e. real matrices with all its entries lying inside the closed interval $[-1,1]$), the maximal determinant is always attainable with a $\{-1,1\}$-matrix (i.e. an integer matrix whose entries are $-1$ or $1$). The reason is that if $A$ is a maximizer and some $a_{ij}$ lies in the interior of the interval, its corresponding minor must be zero. Therefore, it doesn't matter what the exact value of $a_{ij}$ is, and one may move it to the interval's boundary. Move the entries in the interior to the boundary one by one, you get a maximizer $A$ of which all entries lie on the boundary.
So, in your case, the maximizer is just $9$ times the maximizer of the determinant of all $\{-1,1\}$ matrices (and the determinant of the former is $9^3$ times the maximal determinant of the latter). Let $A$ be a maximizer of the latter. As the value of $|\det A|$ is unaffected by left/right multiplication of $A$ by diagonal matrices with diagonal entries taken from $\{-1,1\}$, we may assume that
$$
A=\pmatrix{-1&1&1\\ 1&a&b\\ 1&c&d}
$$
where $a,b,c,d\in\{-1,1\}$ and we want to maximize $|\det A|=|(a+1)(d+1)-(b+1)(c+1)|$. By swapping the last two columns of $A$ if necessary, we may further assume that
$$
\det A=(b+1)(c+1)-(a+1)(d+1)\ge0.
$$
Since all bracketed terms in the above are either $2$ or $0$, the maximal determinant occurs when $b+1=c+1=2$ and $(a+1)(d+1)=0$, i.e. when $b=c=1$ and $a$ or $d$ or both are equal to $-1$. Hence the maximum determinant is $4$ and the answer to your original problem is $9^3\times 4$.
