I want to calculate the determinant a matrix given as $$\det(\mathbf{A}) = \det \begin{pmatrix} \dfrac{\partial \omega_1}{\partial x_1} & \dfrac{\partial \omega_1}{\partial x_2} & \dfrac{\partial \omega_1}{\partial x_3}\\ \dfrac{\partial \omega_2}{\partial x_1} & \dfrac{\partial \omega_2}{\partial x_2} & \dfrac{\partial \omega_2}{\partial x_3}\\ \dfrac{\partial \omega_3}{\partial x_1} & \dfrac{\partial \omega_3}{\partial x_2} & \dfrac{\partial \omega_3}{\partial x_3} \end{pmatrix}, $$ where each entry is of the form $$\frac{\partial \omega_{j}}{\partial x_i} = f_{1,j}a_{1,ij} + f_{2,j}a_{2,ij}.$$ Therefore, $\det(\mathbf{A})$ can be written as $$\det \begin{pmatrix} f_{1,1}a_{1,11} + f_{2,1}a_{2,11} & f_{1,1}a_{1,21} + f_{2,1}a_{2,21} & f_{1,1}a_{1,31} + f_{2,1}a_{2,31}\\ f_{1,2}a_{1,12} + f_{2,2}a_{2,12} & f_{1,2}a_{1,22} + f_{2,2}a_{2,22} & f_{1,2}a_{1,32} + f_{2,2}a_{2,32}\\ f_{1,3}a_{1,13} + f_{2,3}a_{2,13} & f_{1,3}a_{1,23} + f_{2,3}a_{2,23} & f_{1,3}a_{1,33} + f_{2,3}a_{2,33}\\ \end{pmatrix}$$
By means of expansions, collection and rearrangement in MatLab I was able to determine that $$ \det(\mathbf{A}) = \sum_{u=1}^2\sum_{v=1}^2\sum_{w=1}^2 f_{u,1}f_{v,2}f_{w,3}g_{uvw}, \text{ where}$$ $$ g_{uvw} = \det \begin{pmatrix} a_{u,11} & a_{u,21} & a_{u,31}\\ a_{v,12} & a_{v,22} & a_{v,32}\\ a_{w,13} & a_{w,23} & a_{w,33}\\ \end{pmatrix}. $$
This a fairly neat result, and it appears as if one should be able to derive it without "brute-forcing" the determinant expansion.
Is there an elegant way to derive this result? Can it be generalized to arbitrary $n\times n$ matrices?
Thanks in advance!
EDIT:
Thanks to @joriki for the observation of the linearity of each row. Following this idea, $\det(\mathbf{A})$ can be written as
\begin{align} \det(\mathbf{A})&=\det\begin{pmatrix} f_{1,1}\begin{bmatrix}a_{1,11} & a_{1,21} & a_{1,31} \end{bmatrix} + f_{2,1}\begin{bmatrix}a_{2,11} & a_{2,21} &a_{2,31}\end{bmatrix}\\ f_{1,2}\begin{bmatrix}a_{1,12} & a_{1,22} & a_{1,32} \end{bmatrix} + f_{2,2}\begin{bmatrix}a_{2,12} & a_{2,22} &a_{2,32}\end{bmatrix}\\ f_{1,3}\begin{bmatrix}a_{1,13} & a_{1,23} & a_{1,33} \end{bmatrix} + f_{2,3}\begin{bmatrix}a_{2,13} & a_{2,23} &a_{2,33}\end{bmatrix} \end{pmatrix}\\ &= \sum_{u=1}^2\sum_{v=1}^2\sum_{w=1}^2 \det\begin{pmatrix} f_{u,1}\begin{bmatrix}a_{u,11} & a_{u,21} & a_{u,31} \end{bmatrix} \\ f_{v,2}\begin{bmatrix}a_{v,12} & a_{v,22} & a_{v,32} \end{bmatrix} \\ f_{w,3}\begin{bmatrix}a_{w,13} & a_{w,23} & a_{w,33} \end{bmatrix} \end{pmatrix} \\ &= \sum_{u=1}^2\sum_{v=1}^2\sum_{w=1}^2 f_{u,1} f_{v,2}f_{w,3}\det\begin{pmatrix} a_{u,11} & a_{u,21} & a_{u,31} \\ a_{v,12} & a_{v,22} & a_{v,32} \\ a_{w,13} & a_{w,23} & a_{w,33} \end{pmatrix}.\end{align}
This is the result I initially achieved by numerical means.
Each of the rows is a linear combination of two row vectors. This is just the linearity of the determinant applied to the three rows.