I was preparing a worksheet about cofactor expansion of $3\times3$ determinants and decided to create only matrices of order 3 whose entries are among $\pm1,\pm2,\dots,\pm9$ such that the absolute value of the entries form the set $\{1,2,\dots,9\}$. For example: $$\begin{bmatrix}-9&7&-4\\-2&5&6\\1&3&8 \end{bmatrix}\mbox{ has determinant 0.}$$
I was very surprised when I realized that by using only these values, I can create matrices with determinants $0,1,2,\dots,9$ and about other 20 values smaller than 100. Of course, we can create negative determinants if we just permute a pair of rows in the matrix so I will focus only on the nonnegative values of the determinant.
Let $D$ be the set of possible values of the determinant of a matrix created in this way.
Question 1: Is there a way to find $D$?
Question 2: Can $D$ be described as a sequence of consecutive integers? If not, what is the largest sequence of consecutive integers included in $D$
There are $9!\times 2^9=185,794,560$ matrices of order 3 that can be computed in this way. This large number brings me hope that we can find a large sequence of consecutive integers included in $D$.
(this doesn't help in analytical solution, or gives any interesting properties, posting per question author request)
We can bruteforce all variants with simple python code:
The first number $921$ is maximum determinant we can get, the rest are values less than $921$ that we can't get. To speedup things a bit we assume that top left element is $\pm 1$.