I'm trying to understand a statement given by a book ("Analysis and Design of Autonomous Microwave Circuits", Almudena Suarez). It defines a linear dynamical system in the Laplace domain as $sX=AX+GU$ or equivalently $(sI-A)X=G(s)U(s)$ where $X$ is a vector of system states, $A$ is the system matrix, $U$ is a vector with inputs to the system, $G$ is a matrix that combines the inputs in some manner, and $s$ is the Laplace variable.
With zero input to the system ($U=0$), we can write $(sI-A)X=0$ from which we obtain the characteristic polynomial of the system, $P(s)=det(sI-A)$.
The textbook then states that some output of the system can be written as $$Y(s)=B(sI-A)^+$$ where the "+" denotes the transpose of the cofactor matrix and $B$ is some matrix. This is where I get confused. My understanding is that the transpose of the cofactor matrix is called the adjugate, but I'm not sure what significance the adjugate has in this context.
The book goes further to state in the case of a single-input, single-output system (ie $U$ and $Y$ are both scalar), then we can write the transfer function of this system as $$H(s)=\frac{Y(s)}{U(s)}=\frac{B(sI-A)^+G}{P(s)}$$
I'm having trouble understanding how the author goes from the definitions of $U$ and $Y$ to obtain the transfer function.
Thanks to @Semiclassical for providing a source that led to an answer.
Essentially, we can write the following: $$sX=AX+GU$$ $$Y=BX$$
Then we can solve for $X$ in the first equation: $X=(sI-A)^{-1}GU$
and substitute into the second equation: $Y=B(sI-A)^{-1}GU$
Finally, we write the transfer function as $$H=\frac{Y}{U}=B(sI-A)^{-1}G=\frac{B(sI-A)^+G}{det(sI-A)}=\frac{B(sI-A)^+G}{P(s)}$$
Regarding the invertibility of $(sI-A)$, we look at whether $P(s)$ is zero or not. Since $P(s)$ is the characteristic polynomial of $A$, it is clearly only zero when $s$ is an eigenvalue of $A$, or in this case a pole of the system. So this behavior is expected.