Relationships between $\det(A+B)$ and $A+B$ When computing $\det(A+B)$ we notice that there is no relation between $\det A + \det B$.
However does the $\det(A+B)$ have any relation to the matrices $A+B$ as they stand?
 A: Here is something that can be said in $M_n(\mathbb{C})$, in the special case where the matrices commute.
If $AB=BA$, then it can be shown that there exists an invertible matrix $P$ such that $PAP^{-1}$ and $PBP^{-1}$ are both upper-triangular. 
Then the diagonal of $PAP^{-1}$ is $\{\lambda_1,\ldots,\lambda_n\}$ where the $\lambda_j$', are the eigenvalues of $A$ with multiplicities, and $PBP^{-1}$ has diagonal $\{\mu_1,\ldots,\mu_n\}$ whre the $\mu_j$'s are the eigenvalues of $B$ with multiplicities.
Now observe that $P(A+B)P^{-1}$ is upper triangular with diagonal $\{\lambda_1+\mu_1,\ldots,\lambda_n+\mu_n\}$.
Note that $\det PCP^{-1}=\det P\det C \det P^{-1}=\det C$ and that the determinant of an upper triangular matrix is the product of its diagonal terms.
It follows that 
$$
\det (A+B)=(\lambda_1+\mu_1)\cdots(\lambda_n+\mu_n).
$$
This yields $\det A +\det B$, plus $2^n-2$ mixed terms of the form $\lambda_{i_1}\cdots\lambda_{i_k}\cdot \mu_{i_{k+1}}\cdots\mu_{i_{n}}$ which could give anything.
A: There are formulas for $\det(A+B)$ in terms of $A$ and $B$, but they are not nicely self-contained, as one needs to refer to compositions that can't be expressed in terms of basic matrix operations.  For example, see Theorem 4.1 in U. Prells, M. I. Friswell and S. D. Garvey, “Use of geometric algebra: compound matrices and the determinant of the sum of two matrices”, Proc. R. Soc. Lond. A (2003) 459, pp. 273–285.  Here's a PDF link.
To get a sense of what I mean by "compositions" here is a nice formula that works for $2\times 2$ matrices:
$$\det(A+B) = \det(A) + \det(B) + \text{tr}(A^{\dagger} B),$$ where $A^{\dagger}$ is the adjugate of $A$ (essentially $\det(A) A^{-1}$ but still meaningful when $A$ is singular).  The series acquires more terms as the size of the matrices increases.
A: There is a formula of sorts, which I have actually had occasion to use. 
Assume and $A$ and $B$ are $n\times n$ and let $M$ denote the set of $2^n$ 
matrices we get by replacing, in turn, each subset of columns of $A$ with the
corresponding columns of $B$. Then
$$
  \det(A+B) = \sum_{C\in M} \det(C).
$$
If you take $B=xI$, you can derive Laplace's formula for the coefficients of the characteristic polynomial of $A$. 
A: Assuming everything is defined well and stuff. ($A$, $B$ are square matrices of same size)
The determinant of $A+B$ is simply the determinant of the sum of the matrices $A$ and $B$. However, it is NOT the sum of determinants. 
In general,
$$\det(A+B)\neq\det(A)+\det(B)$$
A: I can provide the following insight:
for every $A,B \in F^{n\times n}$
if we define $W=\left\{ M\in \left( \mathbb{Z} _{2}\right) ^{n\times n}| M \ is \ diagonal \right\} $ then the formal way of describing it is:
$$ det\left( A+B\right) =\sum _{M\in W}det\left( A\cdot M+B\cdot \left( I-M\right) \right) $$
in words, the sum of determinants above is the sum of all the determinants of the matrices that result from either removing the column from $B$ or from $A$ for each column in $A+B$ , the following is an example where $n = 2$ to better understand the idea:
$$
\left| A+B\right| =\begin{vmatrix}
a_{11}+b_{11} & a_{12}+b_{12} \\
a_{21}+b_{21} & a_{22}+b_{22}
\end{vmatrix}=\begin{vmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{vmatrix}+\begin{vmatrix}
b_{11} & a_{12} \\
b_{21} & a_{22}
\end{vmatrix}+\begin{vmatrix}
a_{11} & b_{12} \\
a_{21} & b_{22}
\end{vmatrix}+\begin{vmatrix}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{vmatrix}
$$
we can notice that:
$$
\left| A+B\right| =det(A)+\begin{vmatrix}
\vert & \vert \\
B_{1} & A_{2}  \\
\vert & \vert 
\end{vmatrix}+\begin{vmatrix}
\vert & \vert \\
A_{1} & B_{2} \\
\vert & \vert
\end{vmatrix}+det(B)
$$
so in a way there is a relation between $det(A+B)$ and $det(A)+det(B)$ .
