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I have the following matrix (called BRM):

$ BRM = \begin{bmatrix} -A & A & \mathbb{0}_{3\times3} & B & \mathbb{0}_{3\times1} & \mathbb{0}_{3\times1} \\ -C & \mathbb{0}_{3\times3} & C & D & \mathbb{0}_{3\times1} & \mathbb{0}_{3\times1}\\ \mathbb{0}_{3\times3} & -E & E & \mathbb{0}_{3\times1} & F & \mathbb{0}_{3\times1}\\ G & \mathbb{0}_{3\times3} & -G & \mathbb{0}_{3\times1} & \mathbb{0}_{3\times1} & H \end{bmatrix}\in\mathbb{R}^{12\times12}$

Where $A,C,E,G$ are $3\times3$ matrices, while $B,D,F,H$ are $3\times1$ matrices.

All those matrices depend from an angle ($\psi$) and by doing some numerical computations I found out that the determinant of the matrix $(BRM^TBRM)$ does NOT depend on this angle. I would like to find out how to proof, in an analytical way, that this happens. I want to find that the determinant of the matrix does not depend on the angle.

Without giving you the expression of every block, could you tell me a procedure you will adopt to proof this?

I was looking into the determinant of block matrices: https://en.wikipedia.org/wiki/Determinant#Block_matrices

But I didn't find yet a solution to my problem.

Let me know if it is clear what I want to do and if I need to put more details.

Thanks in advance.

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$\det(BRM)=0$ (add the first three block columns).

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