The matrix
A =
\begin{pmatrix} A & B\\
C & D \end{pmatrix}
defines a system y' = Ay.
Compute the entry {\color{red}B}, such that the solution of the system starts with
y(0) = \begin{pmatrix} Y*Z \\
{\color{red}B} \end{pmatrix}
and that \displaystyle \lim_{t \to \infty} y(t) = \begin{pmatrix} 0 \\ 0 \end{pmatrix} holds.
\color{red} B
=
X*Z
At first, compute eigenvalues {\color{blue}\lambda_{1,2}} as zeroes:
\lambda^2 - T \cdot \lambda + S =
\left(\lambda - {\color{blue}negParens(L)} \right) \cdot
\left(\lambda - {\color{blue}L2}\right) =
\left(\lambda - {\color{blue}\lambda_1} \right) \cdot \left(\lambda - {\color{blue}\lambda_2} \right) .
With corresponding eigenvectors v_1 und v_2,
we get the general solution
y: \mathbb R \to \mathbb R^2 mit
y(t) =
C_1 \cdot e^{ {\color{blue}L} \cdot t} \cdot v_1 +
{\color{orange}C_2}\cdot e^{{\color{blue}\lambda_2} \cdot t} \cdot v_2.
Thus we must find {\color{red}B} such that
y(0) = \begin{pmatrix} Y*Z \\ {\color{red}B} \end{pmatrix}
and in addition \color{orange}C_2 = 0, as {\color{blue}\lambda_2} = L2 \geq 0.
Dies garantiert die Konvergenz
\displaystyle
\lim_{t \to \infty} y(t) = \lim_{t \to \infty} C_1 e^{ {\color{blue}L} \cdot t} \cdot v_1
= \begin{pmatrix} 0 \\ 0 \end{pmatrix}.
With the given numbers this yields to:
y(0) = \begin{pmatrix} Y*Z \\ {\color{red}B} \end{pmatrix} = C_1 \cdot e^{ {\color{blue}L}
\cdot 0} \cdot v_1 = C_1 \cdot v_1.
To proceed further, we need to determine the eigenvectors:
From A v_1 = {\color{blue}\lambda_1} v_1 we conclude that the eigenvectors are of the form
\alpha\cdot \begin{pmatrix} Y \\ X \end{pmatrix} with
0 \neq \alpha \in \mathbb R.
The 1st coordinate give
Y*Z = C_1 \cdot Y and hence
Z = C_1.
This Z = C_1 in the 2nd coordinate equation leads to
{\color{red}B} = Z*X.