# Functional Determinant of a matrix of operators

How would we calculate the functional determinant of a matrix with both continuous and discrete indices; such as

$$$$O =\begin{pmatrix} a(t) & \frac{d}{dt} +b(t) & c(t) \\ \frac{d}{dt}+ b(t) & 0 & d(t) \\ c(t) & d(t) & 1 \end{pmatrix}$$$$

subject to some boundary conditions e.g. $$\vec{\phi}(0)=\vec{\phi}(1)=0$$. Where $$\vec{\phi}$$ is a vector of functions?

I know that for a simpler problem e.g. $$O =\frac{d}{dt} + b(t)$$ the solution is just to find the product of the eigenvalues of $$O$$ (again with some boundary conditions) using the eigenvalue equation $$O \psi = \lambda \psi$$ and then divide by the products of the eigenvalues of $$O'=\frac{d}{dt}$$ with matching boundary conditions. My guess is that I'll eventually need to do something like this

$$$$\text{Det}[O]=\frac{\text{det} \begin{pmatrix} a(t) & \frac{d}{dt} +b(t) & c(t) \\ \frac{d}{dt}+ b(t) & 0 & d(t) \\ c(t) & d(t) & 1 \end{pmatrix}}{\text{det}\begin{pmatrix} 0 & \frac{d}{dt} & 0 \\ \frac{d}{dt} & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}}$$$$

Which means I will need to calculate the eigenvalues for both infinite-dimensional matrices.

I have tried to calculate $$$$\text{det}\begin{pmatrix} 0 & \frac{d}{dt} & 0 \\ \frac{d}{dt} & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$$$ with some boundry condition of the form $$\vec{\phi}(0) =0, \vec{\phi}(1)=1$$, by using the eigenvalue equation, getting a pair of differential equations and solving for them, but I find that this doesn't give me a discrete set of eigenvalues to multiply together? Should I be dividing by a different determinant, or just using a different method completely?

Any help or resource recommendations would be appreciated (I have so far not been able to find any online resources dealing with this type of, presumably quite common, problem?)