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?
 A: 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.
A: I think the simplest way to reduce $$A=M-\omega I$$ to a $3\times 3$ matrix is to use the Schur complement with respect to the $(\bar 2,\bar 2)$-elements of $A$,
\begin{align}
C = A/A_{\bar 2,\bar 2} = A_{2,2} - A_{2,\bar 2} A_{\bar 2,\bar 2}^{-1} A_{\bar 2,2}.
\end{align}
Here, $\bar 2$ denotes the index complement of $2$, i.e., $\bar 2\equiv\{1,3\}$. We get
\begin{align}
C = -\omega(N_1 - N_2 + N_3).
\end{align}
With
\begin{align}
\det A_{\bar 2,\bar 2}=\omega^4(\omega^2-1)
\end{align}
and $\det A=\det A_{\bar 2,\bar 2}\det C$ the result follows.
@Carlo: Thanks for posting the MMA notebook! Note that there is a small mistake in the definition of $N_3$, which needs to be transposed.
