Bounding the determinant of a matrix with bounded coefficients Suppose that $A$ is a real matrix of dimension $n \times n$ and that its coefficients are bounded by $c\ge0$ ($\vert a_{ij} \vert \le c$ for all $1\le i,j \le n$).
How to prove that
$$\vert \det A \vert \le c^n n^{n/2}$$
 A: If we consider $A/c$, we see that it is equivalent to show that for $c = 1$, we have $\det(A) \leq n^{n/2}$.  Moreover, since $|\det(A)| = \sqrt{\det(A^TA)}$, it suffices to note that $\det(A^TA) \leq n^n$.
With this in mind, suppose that $A$ is such that $|a_{ij}| \leq 1$.  We note that $M = A^TA$ satisfies
$$
|M_{ij}| = \left|\sum_{k=1}^n a_{ik}a_{kj}\right| \leq \sum_{k=1}^n |a_{ik}| \,|a_{kj}| \leq \sum_{k=1}^n 1 = n
$$
Moreover, $M$ is symmetric positive definite.  Let $\lambda_i$ denote the eigenvalues of $M$.  We have
$$
\operatorname{tr}(M) = \sum_i \lambda_i = \sum_{i} M_{ii} \leq n^2
$$
Now, consider the problem of maximizing $\prod_{i=1}^n \lambda_i$ under the constraint that we have $\lambda_i \geq 0$ and $\sum \lambda_i \leq n^2$.  We find that this product is maximized when
$$
\lambda_1 = \cdots = \lambda_n = n^2/n = n
$$
for a maximum product of $n^n$.  Or, if you prefer, the AM-GM inequality tells us that
$$
\prod_{i=1}^n \lambda_i = \left(\left[\prod_{i=1}^n \lambda_i\right]^{1/n}\right)^n \leq 
\left(\left[\sum_{i=1}^n \lambda_i/n\right]\right)^n \leq n^n
$$
Thus, it follows that
$$
\det(A^TA) = \det(M) = \prod_{i=1}^n \lambda_i \leq n^n
$$
As was desired.
A: Since 

is ture for every element in the matrix, let us consider a matrix where all of the coefficients are the value of c. 
Now factor out the c throughout the entire matrix. 

Using a formula we obtain

Which reduces to

Using another formula

Reduces to

Since all elements in the matrix where of value c then all of the elements left are valued as 1. The result will be that the determinant of a matrix times a matrix transpose will be zero. 

