<s>[INST] Reducing $9\times9$ determinant to $3\times3$ determinant Consider the $9\times 9$ matrix
$$M = \begin{pmatrix} i e_3 \times{} & i & 0 \\ 
-i & 0 & -a \times{} \\
0 & a \times{} & 0 \end{pmatrix}$$
for some vector $a \in \mathbb R^3$, where $\times$ is the cross product.
It is claimed in Fu and Qin - Topological phases and bulk-edge correspondence of magnetized cold plasmas that this can be reduced to the determinant of a $3\times3$ matrix, meaning
$$ \det(M-\omega ) = \det(N_1-N_2+N_3)=0.$$
Here
$$N_1 = \frac{aa^t}{\omega^2}, N_2 =\frac{a^ta}{\omega^2}\operatorname{id},\text{ and }N_3 = \begin{pmatrix} 1-\frac{1}{\omega^2-1} & i\frac{1}{\omega(\omega^2-1)} & 0 \\ -i\frac{1}{\omega(\omega^2-1)} &1-\frac{1}{\omega^2-1} & 0 \\ 0& 0 & 1-\frac{1}{\omega^2}  \end{pmatrix}.$$
This determinant is stated as equation (1) versus the original $9\times9$ matrix is equation (11).
How can we derive the determinant of the $3\times 3$ matrix from the determinant of the $9\times 9 $ matrix without first expanding the determinant?
 [/INST] The formula's in the OP contain an error: the $\omega$ in the denominator of $N_1$ and $N_2$ should be $\omega^2$, so
$$N_1 = \frac{aa^t}{\omega^2}, N_2 =\frac{a^ta}{\omega^2}\operatorname{id}.$$
Then it works out:
$${\rm det}\,(M-\omega)=-\omega^9+2 \omega^7 \left(a_1^2+a_2^2+a_3^2+2\right)-\omega^5 \left(a_1^4+2 a_1^2 \left(a_2^2+a_3^2+3\right)+\left(a_2^2+a_3^2\right)^2+6 a_2^2+6 a_3^2+4\right)+\omega^3 \left(2 a_1^4+a_1^2 \left(4 a_2^2+4 a_3^2+3\right)+2 \left(a_2^2+a_3^2\right)^2+3 a_2^2+4 a_3^2+1\right)-\omega a_3^2  \left(a_1^2+a_2^2+a_3^2\right)$$
$$=\omega^7 \left(1-\omega^2\right)\,{\rm det}\,(N_1-N_2+N_3).$$
Link to the Mathematica notebook.
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