The question, with no simplifications or motivation:
Let $A$ and $B$ be square matrices of the same size (with real or complex coefficients). What is the most reasonable formula one can find for the determinant $\det((1-t)A + tB)$ as a function of $t \in [0,1]$? If no reasonable formula exists, what can we say about these determinants?
So we're taking a line between two matrices $A$ and $B$, and computing the determinant along this line. When $A$ and $B$ are diagonal, say $A = \operatorname{diag}(a_1,\ldots,a_n), B = \operatorname{diag}(b_1,\ldots,b_n),$ then we can compute this directly: $\begin{aligned} \det((1-t)A + tB) &= \det \operatorname{diag}((1-t)a_1 + tb_1, \ldots, (1-t)a_n + tb_n) \\ &= \prod_{j=1}^n ((1-t)a_j + tb_j). \end{aligned}$ I'm not sure if this can be further simplified, but I'm sure someone can push things at least a tiny bit further than I have.
I'm most curious about the case where $A = I$ and each $(1-t)A + tB$ is assumed to be invertible. Here's what I know in this case: writing $D(t) = \det((1-t)I - tB),$ we can compute that $ \dot{D}(t) = D(t) c(t)$ where $c(t) := \operatorname{trace}(((1-t)I + tB)^{-1}(B-I))$ (a warning: I am not 100% sure this formula holds). Thus we can write $D(t) = \exp\left(\int_0^t c(\tau) \; d\tau\right)$ since $D(0) = 1$. I have no idea how to deal with the function $c(\tau)$ though. Any tips?