Are there relation between $\mathbb{E}[ \det(A)]$ and $\det(\mathbb{E}[A])$? Let $A$ be an $n \times n$ matrix. 
Is there a relationship between $\mathbb{E}[\det(A)]$ and $\det(\mathbb{E}[A])$?
For example, for trace we have an equality relationship
\begin{align}
\operatorname{Tr}(\mathbb{E}[A])=\mathbb{E}[\operatorname{Tr}(A)]
\end{align}
 A: No. For the trace, you can use linearity of expectation. For the determinant, you have products, and unless you have strong independence assumptions between the entries of your random matrices, anything can happen.


*

*Take $A$ to be $0_n$ with probability $1/2$, and $4I_n$ with probability $1/2$.
Then $\mathbb{E}[\det A] = \frac{1}{2}\left( 0+4^n\right) = 4^{n-1}$, but
$\det \mathbb{E}[A] = \left(\frac{0+4}{2}\right)^n = 2^n$. In this case,
$$
\mathbb{E}[\det A] > \det \mathbb{E}[A] \tag{1}
$$

*Take $A$ to be $\begin{pmatrix}0&0\\0&2\end{pmatrix}$ with probability $1/2$, and $\begin{pmatrix}2&0\\0&0\end{pmatrix}$ with probability $1/2$.
Then $\mathbb{E}[\det A] = \frac{1}{2}\left( 0+0\right) = 0$, but
$\det \mathbb{E}[A] = \det\begin{pmatrix}1&0\\0&1\end{pmatrix}=1$. In this case,
$$
\mathbb{E}[\det A] < \det \mathbb{E}[A] \tag{2}
$$
A: A simple counter example, 
let $A = \begin{pmatrix}x&0\\0&y\end{pmatrix}$ 
$\mathbb{E}\{\det A\}=E\{xy\}$  whereas, $\det \mathbb{E}\{A\}=E\{x\}E\{y\}$ Unless $x$ and $y$ are independent you can't claim any relationships.
