The matrix
A=
\begin{pmatrix} A & B \\
fractionReduce(A*D,B) & D \end{pmatrix}
defines a system y' = A \cdot y.
Determine the entry {\color{red}X} such that
{\color{orange}y_{\infty}} =
\begin{pmatrix} {\color{red}X} \\
Y \end{pmatrix}
is a steady state of the system.
\color{red} X
=
- B/A * Y
We seek a solution \color{orange}y_{\infty} with
\color{orange}y_{\infty}' = 0, i.e. the entry
{\color{red}X} in {\color{orange}y_{\infty}}=
\begin{pmatrix} {\color{red}X} \\
Y \end{pmatrix}
satisfying A\cdot {\color{orange}y_{\infty}} = \begin{pmatrix} A & B \\
fractionReduce(A*D,B) & D \end{pmatrix} \cdot
\begin{pmatrix} {\color{red}X} \\
Y \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \end{pmatrix}
.
We compute \begin{pmatrix} A & B \\
fractionReduce(A*D,B) & D \end{pmatrix} \cdot
\begin{pmatrix} {\color{red}X} \\
Y \end{pmatrix} =
\begin{pmatrix} negParens(A) \cdot {\color{red}X} + negParens(B) \cdot negParens(Y) \\
negParens(fractionReduce(A*D,B)) \cdot {\color{red}X} + negParens(D) \cdot negParens(Y) \end{pmatrix} =
\begin{pmatrix} negParens(A) \cdot {\color{red}X} + B *Y \\
negParens(fractionReduce(A*D,B)) \cdot {\color{red}X} + D *Y \end{pmatrix}.
Both coordinates are zero for {\color{red}X} = fractionReduce(- B*Y,A).