Find a matrix that represents the dynamic system

Let $h_t$ and $m_t$ denote the number of healthy and sick fish in year number $t$ respectively.

According to the provided information, $70$% of the healthy fish duplicate, $20%$% become sick, and $10$% die in one year. This implies $70$% of the healthy fish in year $t$ remain alive and healthy for year $t+1$. Taking into account the $3$ additional healthy fish that migrate into the lake from abroad, we determine the following formula for $h_{t+1}$: $$h_{t+1}=(0.7)h_t+(0.7)h_t+3=1.4h_t+3$$

Since $20$% of the healthy fish in year $t$ become sick in year $t+1$ while all the sick fish in year $t$ die off, we see that $m_{t+1}=0.2h_t$ and so $$\begin{pmatrix}h_{t+1}\\ m_{t+1}\\ 1\end{pmatrix}=\begin{pmatrix}1.4h_t+3\\ 0.2h_t\\ 1\end{pmatrix}=\begin{pmatrix}1.4&0&3\\ 0.2&0&0\\ 0&0&1\end{pmatrix}\begin{pmatrix}h_t\\ m_t\\ 1\end{pmatrix}$$

Diagonalizing yields a more attractive dynamical system. $$\begin{pmatrix}h_{t+1}\\ m_{t+1}\\ 1\end{pmatrix}=\begin{pmatrix}0&-15&7\\ 1&-3&1\\ 0&2&0\end{pmatrix}\begin{pmatrix}0&0&0\\ 0&1&0\\ 0&0&\frac{7}{5}\end{pmatrix}\begin{pmatrix}-\frac{1}{7}&1&\frac{3}{7}\\ 0&0&\frac{1}{2}\\ \frac{1}{7}&0&\frac{15}{14}\end{pmatrix}\begin{pmatrix}h_t\\ m_t\\ 1\end{pmatrix}$$ Taking $\begin{pmatrix}h_0\\ m_0\\ 1\end{pmatrix}=\begin{pmatrix}100\\ 20\\ 1\end{pmatrix}$ reveals $$\begin{eqnarray*}\begin{pmatrix}h_t\\ m_t\\ 1\end{pmatrix} &=& \begin{pmatrix}0&-15&7\\ 1&-3&1\\ 0&2&0\end{pmatrix}\begin{pmatrix}0&0&0\\ 0&1&0\\ 0&0&\left(\frac{7}{5}\right)^t\end{pmatrix}\begin{pmatrix}-\frac{1}{7}&1&\frac{3}{7}\\ 0&0&\frac{1}{2}\\ \frac{1}{7}&0&\frac{15}{14}\end{pmatrix}\begin{pmatrix}100\\ 20\\ 1\end{pmatrix}\\ &=&\begin{pmatrix}0&-15&7\\ 1&-3&1\\ 0&2&0\end{pmatrix}\begin{pmatrix}0&0&0\\ 0&1&0\\ 0&0&\left(\frac{7}{5}\right)^t\end{pmatrix}\begin{pmatrix}-\frac{83}{7}\\ \frac{1}{2}\\ \frac{215}{14}\end{pmatrix} \\ &=&\begin{pmatrix}0&-15&7\cdot \left(\frac{7}{5}\right)^t\\ 0&-3&\left(\frac{7}{5}\right)^t\\ 0&2&0\end{pmatrix}\begin{pmatrix}-\frac{83}{7}\\ \:\frac{1}{2}\\ \:\frac{215}{14}\end{pmatrix} \\ &=& \begin{pmatrix}-\frac{15}{2}+\frac{215}{2}\cdot \left(\frac{7}{5}\right)^t\\ -\frac{3}{2}+\frac{215}{14}\cdot \left(\frac{7}{5}\right)^t\\ 1\end{pmatrix}\end{eqnarray*}$$ Take $t\longrightarrow +\infty$ to see the result.