Poles of MIMO transfer function matrix. I am looking for the poles of the following transfer function matrix:
$$G_a(s)=\begin{bmatrix} \frac{1}{(s+1)(s+2)} & \frac{-1}{(s+1)(s+2)} \\ \frac{s^2+s-4}{(s+1)(s+2)} & \frac{2s^2-s-8}{(s+1)(s+2)} \\  \frac{s-2}{s+1} & \frac{2(s-2)}{s+1} \end{bmatrix}$$
There are, as I understand, a total of six $1 \times 1$ principle minors and two $2 \times 2$ principle minors.
The multiplicity of $s =-1$ in all $1 \times 1$ minors is $1$. The multiplicity of $s = -1$ in both $2 \times 2$ minors is $4$. So the multiplicity of $s = -1$ equals $4 - 1 =3$. Is this correct?
And what about $s+2$ because this term isn't present in all $1 \times 1$ principle minors?
I only found one file explaining this and it only covered really basic examples and not a $3 \times 2$ transfer funciton matrix. 

The poles can be found in the least common denominator of all principle minors. The minors of order $1$ are:
\begin{align}
M1_{11}=\frac{1}{(s+1)(s+2)}, \ M1_{12}=\frac{-1}{(s+1)(s+2)}, \ M1_{21}=\frac{s^2+s-4}{(s+1)(s+2)},\\
M1_{22}=\frac{2s^2-s-8}{(s+1)(s+2)}, \ M1_{31}=\frac{s-2}{s+1}, \ M1_{32}=\frac{2(s-2)}{s+1}
\end{align}
And the minors of order $2$ are:
\begin{align}
M2_{1}=\frac{3s(s-2)}{(s+1)^2(s+2)}, \ M2_{2}=M2_{3}=\frac{3(s-2)}{(s+1)^2(s+2)}
\end{align}
The least common denominator is: $(s+1)^2(s+2)$. The poles are thus:
\begin{align}
s=-1 \quad \text{(2x)} \\
s=-2
\end{align}
