How do you bound the determinant of an $n\times n$ matrix with values $-1, 0, 1$?

Question

How do you bound the determinant of an $n\times n$ matrix with values $-1, 0, 1$?

Answer 1

$$|A| := \sum_{\sigma \in S_n} sgn(\sigma) a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n\sigma(n)}$$

.

The triangle inequality tells us $|det(A)|$ is bounded by the the sum of $ |sgn(\sigma) a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n\sigma(n)}|$

There are $n!$ sums here each of which has an absolute value of at most $1$. Symbolically,

$$\bigg|\sum_{\sigma \in S_n} sgn(\sigma) a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n\sigma(n)}\bigg| \leq \sum_{\sigma \in S_n} \bigg|sgn(\sigma) a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n\sigma(n)}\bigg| \leq \sum_{\sigma \in S_n} 1 = n!$$