Poles of MIMO transfer function matrix.

Question

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.

Answer 1

A complex number $s_k$ is called a pole of a transfer function matrix $\boldsymbol{G}(s)$, when at least one element $G_{ij}(s)$ has $s_k$ as a pole. The multiplicity is given by the highest order of the pole $s_k$ for all the elements of the transfer function matrix.

Hence, the poles of the system are given by $\{-1,-2\}$.

Answer 2

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}