2/9 & 10/27 \\ 
11/27 & 2/9 & 10/27 \\ 
11/27 & 2/9 & 10/27\end{array}\right) ;\quad \det \mathbb{C}_{1}=0,  \label{C15}$$ $$\mathbb{C}_{2}=\left( 
\begin{array}{rrr}
{\ 0.304793} & {\ -0.544452} & {\ 0.239659} \\ 
{\ -0.424622} & {\ 0.758503} & {\ -0.333880} \\ 
{\ -0.326908} & {\ 0.143795} & {\ -0.063296}\end{array}\right) ;\quad \det \mathbb{C}_{2}=1.\,\allowbreak 076\,11\,\times 10^{-7}\text{,}  \label{C16}$$ $$\mathbb{C}_{3}=\left( 
\begin{array}{rrr}
{\ 0.287800} & {\ 0.322230} & {\ -0.610029} \\ 
{\ 0.017215} & {\ 0.019275} & {\ -0.036490} \\ 
{\ -0.326908} & {\ -0.366017} & {\ 0.692926}\end{array}\right) ;\quad \det \mathbb{C}_{3}=3.\,\allowbreak 245\,95\times 10^{-13}.
\label{C17}$$ While matrix $\mathbb{C}_{1}$ does not have an inverse, $\mathbb{C}_{2}$ and $\mathbb{C}_{3}$ do have, but they are not stochastic. The decay characteristic times associated with the modes $2$ and $3$ are calculated as $T_{k}=-\left( \ln \left\vert \lambda _{k}\right\vert \right) ^{-1}$, resulting in the values $T_{2}= 1.000178$ and $T_{3}=0.371878$, so mode $3$ decays faster than mode $2$, while $T_{1}=\infty $ for mode $1$. Thus we can write $$\mathbb{M}^{n}=\mathbb{C}_{1}+\mathbb{C}_{2}\exp \left( -\frac{n}{T_{2}}\right) +\mathbb{C}_{3}\exp \left( -\frac{n}{T_{3}}\right) .  \label{C18}$$ An initial state vector, for instance, $$\mathbb{U}^{\mathrm{T}(0)}=\left( 
\begin{array}{ccc}
100 & 160 & 300\end{array}\right) ,  \label{C18a}$$ will evolve as $$\mathbb{U}^{\mathrm{T}}\left( n\right) =\mathbb{U}^{\mathrm{T}}\left(
0\right) \mathbb{C}_{1}+\left( \mathbb{U^{\mathrm{T}}}\left( 0\right) 
\mathbb{C}_{2}\right) \exp \left( -\frac{n}{T_{2}}\right) +\left( \mathbb{U^{\mathrm{T}}}\left( 0\right) \mathbb{C}_{3}\right) \exp \left( -\frac{n}{T_{3}}\right) .  \label{C19}$$ Thereafter, one 