According to this answer, there exists a relationship between determinant and integral, as follows:
$$ \frac{1}{\operatorname{vol} [M]}\int _M \det(B+\nabla \phi)=\det B $$
Then it goes to say that this applies in the case "[...] if $A$ is a constant $B$ plus the Jacobian matrix of a compactly supported field $\phi$ (for instance if $M$ has no boundary), then the integral and the determinant do commutte".
Can anyone unpack this relation, perhaps with an example or a link with more information? For instance, what is $M$, what is the volume of $M$? Does it work with any matrix $B$? What does $B$ being a constant plus the Jacobian matrix and does it relate to the expression $B+\nabla \phi$?
Edit:
My interest is to attack the problem in the other direction as that of the linked question; that is to start with the determinant of a matrix $B$ and then convert it to an equivalent "integral-form" using the equality above. Can $B$ be an arbitrary $n\times n$ matrix, or does it have special restrictions?
Then, it appears that it suffices that I introduce an arbitrary field $\nabla \phi$ and define the integral with respect to this field, correct?
Consider a sphere $M$ with standard metric and spherical coordinate chart ($\varphi$, $\theta$). Let's have a field $\phi=(\cos\theta, 0)$, then its Jacobian matrix: $$ \nabla\phi = \begin{pmatrix}0&-\sin\theta\\0&0\end{pmatrix}. $$
Take some constant $B=\begin{pmatrix}1&1\\-1&2\end{pmatrix}$, then $\det(B+\nabla\phi)=3-\sin\theta$.
Consider integral: $$ \int_M \det(B+\nabla\phi) = \int_{-\pi}^{\pi}\int_{-\pi/2}^{\pi/2}(3-\sin\theta)\cos\theta d\theta d\varphi = 12\pi = \underbrace{\mathop{\mathrm{vol}}(M)}_{4\pi}\underbrace{\det B}_{3}. $$
where $\mathrm{vol}(M)=\int_M1$ is a total volume (in that case area) of a manifold $M$.