find the smallest positive integer $k$ equal to the determinant of a certain square matrix 
Let $n, m$ be two positive integers. Find the smallest positive integer k so that there is an $m\times m$ square matrix whose entries have absolute values $1,2,\ldots, n$ and that has determinant $k$.

Consider first the case where $n=1$. We claim that $k=2^{m-1}$, and to begin the proof we show that every $m\times m$ matrix with entries in $\{\pm 1\}$ must have a determinant divisible by $2^{m-1}$. To see this, subtract the first row from every other row. Then assuming there are no zero rows (otherwise the determinant is 0, which is divisible by $2^{m-1}$), we may factor out a $2$ from the m-1 rows other than the first row to get a matrix C with all entries in $\{\pm 1, 0\}.$ Then the determinant is $\det(C) 2^{m-1},$ which is a multiple of $2^{m-1}$, as required. Note that a matrix with determinant $2^{m-1}$ whose entries are $\pm 1$ can be described as follows: consider the matrix $B$ whose (i,j) entry is given by $B_{i,j} = \begin{cases} 1, &\text { if $i\leq j$}\\
-1,&\text{ otherwise }\end{cases}.$ Then by subtracting the $i-1$th row from the $i$ row for $n\ge i \ge 2$ (in that order), we get a matrix whose main diagonal has 1 as the first entry and zeroes elsewhere while the entries on the diagonal directly below the main diagonal are all $-2$'s. Then expanding along the last column gives that the desired determinant is $(-1)^{m+1} (-2)^{m-1} = 2^{m-1}.$
I'm not sure what to do for the general case; the method above breaks down because there are way too many possibilities to consider when performing those row operations (e.g. subtracting the top row from each other row). Suppose $n=2$ and let A be an m by m matrix all of whose entries are in the set $\{\pm 1,\pm 2\}.$ If $m=2$, clearly $k=1$ can be achieved. If $m = 3,$ there's the matrix $$\begin{pmatrix}1 & 2 & 2\\
2 & 1 & 1\\
1 & 2 & 1\end{pmatrix}$$
$-1-2(-2)+0 = 3.$ I'm not sure if $k=1$ can be achieved.
 A: The case when $n=1$ is solved in the question.
When $n=2$, the smallest positive such integer is $1$ for all $m$, as remarked by eyballdrog.

*

*If $m=1$, then the determinant of $[1]$ is $1$.


*Otherwise $m\ge2$. The determinant of the following $m\times m$ matrix is $1$.
$$\begin{bmatrix}
1 & 1 &1&\cdots&1\\
1&2&1&\cdots&1\\
1&1&2&\cdots&1\\
\vdots&\vdots&\vdots&\ddots&\cdots\\
1&1&1&\cdots&2
\end{bmatrix}$$

Suppose we require further that each of $1,2,\cdots, n$ must appear in the $m\times m$ matrix.
If $n\le m(m+1)/2$, then the answer remains $1$. Here is an illustration for the case of $n=11$ and a $6\times 6$ matrix with determinant $1$, where each entry labelled with "*" can be any nonzero number between $-11$ and $11$.
$$\begin{bmatrix}
1 & 2 &*&*&*&*\\
1&3&4&*&*&*\\
1&3&5&6&*&*\\
1&3&5&7&8&*\\
1&3&5&7&9&10\\
1&3&5&7&9&11
\end{bmatrix}$$
A: I interpret the problem posed as constructing a $m \times m$ matrix satisfying the following conditions:

*

*containing all entries from the set $S = \{1, 2, \cdots, n\}$ and

*using every element of the set $S$ at least once in the matrix and

*has determinant $k$.

If the condition #2 is relaxed then for the set
$$S_m = \{x_1, \overbrace{x_2, x_2^2, \cdots, x_2^{m-1}}, \overbrace{x_3, x_3^2, \cdots, x_3^{m-1}}, \cdots, \overbrace{x_{m-1}, x_{m-1}^2, \cdots, x_{m-1}^{m-1}}\}$$
we can construct a $m \times m$ Vandermonde matrix $V$ as:
$$V = \begin{bmatrix}
x_1 & x_1^2 & \cdots & x_1^{m-1} \\
x_2 & x_2^2 & \cdots & x_2^{m-1} \\
\vdots & \cdots & \ddots & \vdots \\
x_{m-1} & x_{m-1}^2 & \cdots & x_{m-1}^{m-1}
\end{bmatrix}
$$
with determinant
$$k = |V| = \prod_{1\le i \le j \le n} (x_j - x_i).$$
This lets us construct square Vandermonde matrices with determinant non-zero $k$ provided the $x_i$ are distinct.
Note: This is not an answer to the OP's question, but is a related response for constructing a matrix in the general case with one of the conditions relaxed.
