A generalization of Liouville formula for the determinant of a system of ODE? Let $\Phi(t)$ be an $n\times n$ complex matrix whose columns are (independent) solutions of the  system of ordinary differential equations (ODE):
$\frac{d}{dt}y= A(t) y$, where $A(t)$ is a $t$-dependent $n\times n$ complex matrix.
Then, the Liouville formula provides a very simple relation between the determinant of $\Phi(t)$ at different $t$:
$\det(\Phi(t)) = e^{\int_0^t Tr A(s) ds} \det(\Phi(0))$.
Now, consider the alternative system of ODE:
$\frac{d}{dt}y= A(t) y + B(t) \bar{y}$, 
where $\bar{x}$ represents the complex conjugate of $x$. (Actually, I am interested only in the case where $A(t)$ is diagonal and $B(t)$ is skew-symmetric, if that helps.)
Does it exist any generalization of the Liouville formula applicable to this case? or any other expression for $\det(\Phi(t))$, which is simpler (to compute numerically) than finding the explicit solutions for all the starting conditions in $\Phi(0)$ and then computing the determinant explicitly?
The kind of manipulations used by the classical proof of Liouville formula do not seem very helpful to me, in this slightly modified case, unfortunately.
The solution of this problem would be very beneficial to the method described in arXiv:1205.3996, and to its very relevant physical applications.
 A: NB:  I'll slightly rearrange this for clarity: 
As I should have remarked at the beginning, 
writing the fundamental solution to your system in the form
$$
y(t) = \Phi(t)\,y(0),
$$
where $\Phi(0)=I$, is only possible when you allow $\Phi(t)$ to take values in $$\mathrm{Aut}_\mathbb{R}(\mathbb{C}^n,\mathbb{C}^n) \simeq \mathrm{GL}(2n,\mathbb{R}),
$$
i.e., you can't assume that $\Phi(t):\mathbb{C}^n\to\mathbb{C}^n$ is $\mathbb{C}$-linear, but only $\mathbb{R}$-linear.  Consequently, the very meaning of 'determinant' is probably not what you had in mind, since the homomorphism
$$
{\det}_\mathbb{R}:\mathrm{Aut}_\mathbb{R}(\mathbb{C}^n,\mathbb{C}^n)=\mathrm{GL}(2n,\mathbb{R})\to\mathbb{R}\setminus\{0\}
$$
takes (nonzero) values in $\mathbb{R}$, and there is no meaningful way to `lift' this to a 'determinant' that assumes all (nonzero) values in the complex numbers.
One should also note that, if $A:\mathbb{C}^n\to\mathbb{C}^n$ does happen to be $\mathbb{C}$-linear, then one has
$$
{\det}_\mathbb{R}(A) = \bigl|{\det}_\mathbb{C}(A)\bigr|^2.
$$
With that understood, there is a formula for $\det\bigl(\Phi(t)\bigr)$, it's just
$$
\det(\Phi(t)) 
= \exp\left(\int_0^t \mathrm{tr}\bigl(A(\tau)+\overline{A(\tau)}\bigr)\,d\tau\right).
$$
Here's how you can see this:  Since you know that $A(t)$ is diagonal, you can easily write $A(t) = -a(t)^{-1}a'(t)$ where $a(t)$ is a diagonal matrix satisfying $a(0)=I$ (just integrate and exponentiate each diagonal term separately).   
Hence you can regard $a(t)$ as known and write the equation in the form
$$
y'(t) = - a(t)^{-1}a'(t)y(t) + B(t)\,\overline{y(t)},
$$
which gives
$$
(a(t)y(t))' = a(t)B(t)\overline{(a(t)^{-1})}\,\,\overline{a(t)y(t)}.
$$
Set $z(t) = a(t)y(t)$ and write $L(t) = a(t)B(t)\overline{(a(t)^{-1})} C$, where $C:\mathbb{C}^n\to\mathbb{C}^n$ is the $\mathbb{R}$-linear map of conjugation.  Then the equation becomes the $\mathbb{R}$-linear system of ODE
$$
z'(t) = L(t) z(t).
$$
It's important to note that $\mathrm{tr}(L(t))$ vanishes identically.
Of course, $L(t)$, which can be regarded as known, can be written in the form $L(t) = -b(t)^{-1}b'(t)$ where $b(t)$ takes values in the matrix Lie group $\mathrm{Aut}_\mathbb{R}(\mathbb{C}^n,\mathbb{C}^n)$, and satisfies $b(0)=I$.  The general solution of this equation is  then
$$
z(t) = b(t)^{-1}z(0) \qquad \text{so}\qquad y(t) = a(t)^{-1}b(t)^{-1}y(0).
$$
Now, the Liouville formula yields 
$$
{\det}_\mathbb{C}(a(t)) = \exp\left(-\int_0^t \mathrm{tr}(A(\tau))\,d\tau\right)
$$
$$
{\det}_\mathbb{R}(b(t)) = \exp\left(-\int_0^t \mathrm{tr}(L(\tau)+\overline{L(\tau)})\,d\tau\right). = 1
$$
(After all the trace of $L(t)$ as a real endomorphism is identically zero.)
Thus, $y(t) = \Phi(t)y(0)$ where 
$$
{\det}_\mathbb{R}(\Phi(t)) 
= \exp\left(\int_0^t \mathrm{tr}\bigl(A(\tau)+\overline{A(\tau)}\bigr)\,d\tau\right).
$$
