TITLE: Can we say that $\det(A+B) = \det(A) + \det(B) +\operatorname{tr}(A) \operatorname{tr}(B) - \operatorname{tr}(AB)$.
QUESTION [5 upvotes]: Let $A,B \in M_n$.
Is this formula true?
$$\det(A+B) = \det(A) + \det(B) + \operatorname{tr}(A) \operatorname{tr}(B) - \operatorname{tr}(AB).$$

REPLY [4 votes]: $n=2$, $\newcommand{\tr}{\operatorname{tr}}$
$$\det(A+B) = \det(A) + \det(B) + \tr(A) \tr(B) - \tr(AB).$$
$n=3$, letting $c(X) = (\tr(X)^2 – \tr(X^2)) / 2$,
\begin{align*}
 \det(A + B) ={}& \det(A) + \det(B) – \tr(AB)\tr(A) – \tr(AB)\tr(B) +{} \\
 &{}+ c(A)\tr(B) + \tr(A)c(B) + \tr(AAB) + \tr(ABB)
 \end{align*}
$n>3$, a formula with  $2^n$ terms should be obtainable from the 1987 Reutenauer and Schützenberger's "A formula for the determinant of a sum of matrices"