Does $\det(A + B) = \det(A) + \det(B)$ hold? Well considering two $n \times n$ matrices does the following hold true:
$$\det(A+B) = \det(A) + \det(B)$$
Can there be said anything about $\det(A+B)$?
If $A/B$ are symmetric (or maybe even of the form $\lambda I$) - can then things be said?

Although the determinant function is not linear in general, I have a way to construct matrices $A$ and $B$ such that $\det(A + B) = \det(A) + \det(B)$, where neither $A$ nor $B$ contains a zero entry and all three determinants are nonzero:
Suppose $A = [a_{ij}]$ and $B = [b_{ij}]$ are 2 x 2 real matrices.  Then $\det(A + B) = (a_{11} + b_{11})(a_{22} + b_{22}) - (a_{12} + b_{12})(a_{21} + b_{21})$  and  $\det(A) + \det(B) = (a_{11} a_{22} - a_{12} a_{21}) + (b_{11} b_{22} - b_{12} b_{21})$.  
These two determinant expressions are equal if and only if 
$a_{11} b_{22} + b_{11} a_{22} - a_{12} b_{21} - b_{12} a_{21}
= $ $\det \left[ 
\begin{array}{cc}
a_{11} & a_{12}\\
b_{21} & b_{22}
\end{array} \right]$
+
$\det \left[ 
\begin{array}{cc}
b_{11} & b_{12}\\
a_{21} & a_{22}
\end{array} \right]$
= 0.
Therefore, if we choose any nonsingular 2 x 2 matrix $ A = [a_{ij}]$ with nonzero entries and then create $B = [b_{ij}]$ such that $b_{11} = - a_{21}, b_{12} = - a_{22}, b_{21} = a_{11},$ and $b_{22} = a_{12}$, we have solved our problem.  For example, if we take
$$A = 
\begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix}
\quad
\text{and}\quad
B =
\begin{bmatrix}
-3 & -4 \\
1 & 2\end{bmatrix}
,$$
then $\det(A) = -2, \det(B) = -2, $  and  $\det(A + B) = -4$, as required.
