Is it true that for matrices where all entry are lower than 1, determinant is lower than 1 as well? Generic square matrix with positive 1 bounded entries
Considering a matrix $A=(a_{i.j})$ where $0 \leq a_{i,j} < 1 \forall i,j$. It is important to consider that all entries are strictly lower than 1 and positive.
Rows sum to a number lower than 1
Let us consider that the sum of all entries of matrix $A$'s rows is lower than 1: $\sum_{j=1}^{n}a_{i,j} < 1$. Sorry, maybe I did not specify it, only wrote in the formula, I talk about rows. Rows sum to a number lower than 1.
Determinant...
Let us consider $\det(A)$ (determinant).
Is it true that $\det(A)<1$???
Or maybe $|\det(A)| < 1$???
 A: Note that $\sum_j a_{ij}^2 < \sum_j a_{ij} < 1$. So the magnitude of each row, viewed as a vector in ${\mathbb R}^n$, is less than one. The absolute value of the determinant of $A$ is the volume of the parallelopiped spanned by the rows, which is at most the product of the magnitudes of the row vectors, and therefore is less than one in this case. 
If you want to do it algebraically, you can prove it by induction on the dimension, the $1$ by $1$ case being trivial. Then you can do a cofactor expansion along any $i$th row, getting that 
$$det(A) = \sum_j (-1)^{i + j} a_{ij} \,det(A_{ij})$$
Note that each matrix $A_{ij}$ also satisfies the conditions of the problems, so each $|det(A_{ij})| < 1$ by induction hypothesis. You then get
$$|det(A)| < \sum_j |a_{ij}||det(A_{ij})|$$
$$< \sum_j |a_{ij}|$$
$$< 1$$
A: EDIT: Taking into account the condition that the sum of the entries be less than one, the determinant is a sum of $n!$ terms, each of which is at most $n^{-n}$, so the determinant is bounded by $n!/n^n$, which is certainly less than 1 (for $n\gt1$). Each term is les than $n^{-n}$ because it's a product of $n$ numbers that add up to less than 1, and you maximize the product by taking all the number to equal $1/n$. 
(Never mind --- I just saw the part about the sum of all the entries, or maybe it's the sum of all the entries in each row, being less than 1.)
Are we only talking about $2\times2$ matrices? If not, then $$\pmatrix{a&b&0\cr0&c&d\cr e&0&f\cr}$$ will have determinant $acf+bde$ which can certainly exceed 1 even if all the variables stand for numbers between zero and one. 
