$a_{n+1}=(r-2)a_{n}+(r-1)b_{n}$ and $b_{n+1}=a_{n}$ means that $a_{n+1}=(r-2)a_{n}+(r-1)a_{n-1}$ or $a_{n+1}-(r-2)a_{n}-(r-1)a_{n-1}=0$, it's character equation is $\lambda^2-(r-2)\lambda-(r-1)=0$.
More generally, character equation of type as $a_{n+1}+k_{1}a_{n}+k_{2}a_{n-1}=0$ is $\lambda^2+k_{1}\lambda+k_{2}=0$, if the two roots are $\lambda_{1}$ and $\lambda_{2}$, then $-k_{1}=\lambda_{1}+\lambda_{2}$ and $k_{2}=\lambda_{1}\lambda_{2}$.
Then the original equation is $a_{n+1}-(\lambda_{1}+\lambda_{2})a_{n}+\lambda_{1}\lambda_{2}a_{n-1}=0$, so we have
$\begin{align} a_{n+1}-\lambda_{1}a_{n}&=\lambda_{2}(a_{n}-\lambda_{1}a_{n-1})\\ a_{n}-\lambda_{1}a_{n-1}&=\lambda_{2}(a_{n-1}-\lambda_{1}a_{n-2})\\ &\vdots\\ a_{3}-\lambda_{1}a_{2}&=\lambda_{2}(a_{2}-\lambda_{1}a_{1}) \end{align}$
So we have $a_{n+1}-\lambda_{1}a_{n}=\lambda_{2}^{n-1}(a_{2}-\lambda_{1}a_{1})$, we can denote it as $a_{n+1}-\lambda_{1}a_{n}=d_{1}\lambda_{2}^{n}$.
Similarly, we can get $a_{n+1}-\lambda_{2}a_{n}=d_{2}\lambda_{1}^{n}$.
Consider above two equalities, by calculation, we can get $a_{n}$ have the form $a_n=c_{1}\lambda_{1}^{n}+c_{2}\lambda_{2}^{n}$