TITLE: Is the determinant of a matrix lower when all its elements are lower?
QUESTION [6 upvotes]: Problem
Consider a generic matrix $A$, we are going to think of a simple case by taking into consideration a $3 \times 3$ matrix:
$$
A = \begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3}\\
a_{2,1} & a_{2,2} & a_{2,3}\\
a_{3,1} & a_{3,2} & a_{3,3}\\
\end{pmatrix}
$$
Consider now having $A'$ as:
$$
A' = \begin{pmatrix}
a'_{1,1} & a'_{1,2} & a'_{1,3}\\
a'_{2,1} & a'_{2,2} & a'_{2,3}\\
a'_{3,1} & a'_{3,2} & a'_{3,3}\\
\end{pmatrix}
$$
The following holds:
$$a'_{i,j} \leq a_{i,j}$$
Question
I would like to know if the following:
$$|A'| \leq |A|$$
If it holds, can you prove it?
Another problem
What if we considered:
$$
a_{i,j} \leq 1, a'_{i,j} \leq 1
$$
Considering also that $A$ is a stochastic matrix?
This does not mean that both $A$ and $A'$ are stochastic. I am considering $A$ stochastic and $A'$ obtained as a reduced version of $A$ so that $A'$ is not stochastic but its values are all between 0 and 1.

REPLY [12 votes]: The claim is false. For instance $$A' = \begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{pmatrix}$$ and $$A = \begin{pmatrix}2 & 2 & 2\\2 & 2 & 2\\2 & 2 & 2 \end{pmatrix}$$
Clearly, $A'_{ij} \leq A_{ij}$, whereas $$\det(A') = 1 > 0 \det(A)$$
EDIT
If $A$ and $A'$ are both stochastic matrices, then $A'_{ij} \leq A_{ij}$ gives us $A'_{ij} = A_{ij}$ since $$1 = \displaystyle \sum_{j=1}^{3} A'_{ij} \leq \sum_{j=1}^{3} A_{ij} = 1, \,\forall i \in \{1,2,3\}$$
EDIT
If $A$ is stochastic, but $A'$ is not stochastic, then again it is false. For instance, $$A = \begin{pmatrix}\frac13 & \frac13 & \frac13\\\frac13 & \frac13 & \frac13\\\frac13 & \frac13 & \frac13 \end{pmatrix}$$
and $$A' = \begin{pmatrix}\frac13 & 0 & 0\\0 & \frac13 & 0\\0 & 0 & \frac13 \end{pmatrix}$$

REPLY [6 votes]: This is false. Let $A$ be the matrix all of whose entries are equal to $2$ and $A'$ be the identity matrix. Note that $\det(A) = 0$ since its columns are linearly dependent, while $\det(A') = 1$.