the minimum and maximum values of the determinant of order $3\times3$ matrix with entries $\{0,1,2,3\}$ The minimum and maximum values of the determinant of order $3\times3$ matrix with entries $\{0,1,2,3\}$.
Brute force enumeration gives a maximum of $54$ (and hence a minimum of $-54$ since $\det (-A) = - \det A$ for $n$ odd).

A maximising matrix would be $\begin{bmatrix} 0 &amp; 3 &amp; 3 \\
3 &amp; 0 &amp; 3 \\
3 &amp; 3 &amp; 0
\end{bmatrix}$.

In Python:

import numpy
import itertools

N = 3
cols = list(itertools.product(range(0, N+1), repeat=N))
mats = itertools.product(cols, cols, cols)
sup = 0
sup_p = None
for p in mats:
    d = numpy.linalg.det(p)
    if d &gt; sup:
        sup = d
        sup_p = p

print sup
print sup_p


Note: 

A slight improvement would notice that the problem can be written
as $\max \{ \det A \,|\,  [A]_{ij} \in \{0,1,2,3\} \}$ and since the cost is
affine in each matrix element, the maximiser must have extreme values
for all entries. Hence the problem is equivalent to 
$\max \{ \det A \,|\, [A]_{ij} \in \{0,3\} \}$.

Since the problem is also equivalent to 
$\max \{ |\det A| \,|\, [A]_{ij} \in \{0,3\} \}$, the order of rows (or columns)
doesn't matter and so we can choose the first row from
$\{(0,0,0), (3,0,0), (3,3,0), (3,3,3) \}$. This reduces the search space.

Since $(0,0,0)$  (or repeating a row) will result in a zero determinant,
the search space can be reduced further to at most
$3 \cdot \binom{7}{2} = 63$ possibilities, which is a little better
than the original $4^9 = 262,144$.

And finally, although irrelevant to the problem, we note that
this solves the problem $\max \{ |\det A| \,|\, [A]_{ij} \in [0,3] \}$,
the maximum volume of a box whose sides have $l_1$ length no greater than $3$.
I don't know anything about Hadamard's inequality...
But I have a better technique.
Here the smallest element is $0$ and the largest element is $3$
So, the determinant will be max when $0$ is the diagonal element and $3$ is the non-diagonal element, meaning that the answer is $54$.