The matrix
A=
\begin{pmatrix} A & {\color{red}b}\\
C & D \end{pmatrix}
defines a system y' = A \cdot y.
Determine the entry {\color{red}b} such that the system has a stationary state.
\color{red} b
=
A*D/C
For an eigenvector v with the corresponding eigenvalue \lambda is
y with y(t) = e^{\lambda \cdot t}\cdot v a solution of the system.
For \lambda =0 is e^{\lambda \cdot t} = e^{0} = 1 and a correspondung eigenvector
gives a stationary solution.
The eigenvalue zero exists if and only if \det(A) = 0.
It is \det(A) = negParens(A) \cdot negParens(D) -
{\color{red}b} \cdot negParens(C).
Setting this equal to zero and solving for {\color{red}b} gives:
{\color{red}b} =
fractionReduce(A*D,C).