The matrix
A=
\begin{pmatrix} A & B \\
C & D \end{pmatrix}
and the vector
\begin{pmatrix}X \\
Y \end{pmatrix}
define y' = A \cdot y + \begin{pmatrix}X \\
Y \end{pmatrix}
.
Determine {\color{red}X}
and {\color{blue}Y}
such that
{\color{orange}y_{\infty}} =
\begin{pmatrix} {\color{red}X} \\
{\color{blue}Y} \end{pmatrix}
is a stationary state of the inhomogeneous system.
\color{red} X
=
(-D*X+B*Y)/det
\color{blue} Y
=
(-A*Y+C*X)/det
We are looking for a solution \color{orange}y_{\infty}
with
{\color{orange}y'_{\infty}} = 0 = A \cdot {\color{orange}y_{\infty}} + \begin{pmatrix}X \\
Y \end{pmatrix}
.
Since A=
\begin{pmatrix} A & B \\
C & D \end{pmatrix}
is invertible, with \det(A) = det
and
A^{-1}=
\frac 1{det}\begin{pmatrix} D & -B \\
-C & A \end{pmatrix}
,
we can solve the equation
0 = A \cdot {\color{orange}y_{\infty}} + \begin{pmatrix}X \\
Y \end{pmatrix}
for \color{orange}y_{\infty}
.
It follows that - \begin{pmatrix}X \\
Y \end{pmatrix}= A \cdot {\color{orange}y_{\infty}}
and for multiplication with
A^{-1}
from the left
A^{-1} \left(- \begin{pmatrix}X \\
Y \end{pmatrix}\right)= {\color{orange}y_{\infty}}
.
Substituting this yields {\color{orange}y_{\infty}} =
\begin{pmatrix} {\color{red}SX} \\
{\color{blue}SY} \end{pmatrix}
.