The matrix
A=
\begin{pmatrix} A & {\color{red}b} \\
{\color{teal}c} & D \end{pmatrix}
defines a system y' = A \cdot y.
Determine entries {\color{red}b} and {\color{teal}c} such that
{\color{orange}y_{\infty}}=
\begin{pmatrix} X \\
Y \end{pmatrix}
is a steady state of the system.
\color{red} b
=
B
\color{teal} c
=
C
We seek a solution \color{orange}y_{\infty} with
\color{orange}y_{\infty}' = 0, i.e. {\color{red}b} and {\color{teal}c}
with A\cdot {\color{orange}y_{\infty}} = \begin{pmatrix} A & {\color{red}b} \\
{\color{teal}c} & D \end{pmatrix} \cdot
\begin{pmatrix} X \\
Y \end{pmatrix} = {\color{blue}
\begin{pmatrix} 0 \\ 0 \end{pmatrix}}
.
We compute \begin{pmatrix} A & {\color{red}b} \\
{\color{teal}c} & D \end{pmatrix} \cdot
\begin{pmatrix} X \\
Y \end{pmatrix} =
\begin{pmatrix} negParens(A) \cdot negParens(X) + {\color{red}b} \cdot negParens(Y) \\
{\color{teal}c} \cdot negParens(X) + negParens(D) \cdot negParens(Y) \end{pmatrix} =
{\color{blue}
\begin{pmatrix} A*X \cdot + negParens(Y) \cdot {\color{red}b} \\
negParens(X) \cdot {\color{teal}c} + D *Y \end{pmatrix}
}.
Thus we get:
{\color{red}b} = fractionReduce(- A*X,Y) and {\color{teal}c} = fractionReduce(- D*Y,X).