Example 10 In the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, $a_{1}=1, b_{1}=2$, and $\left\{\begin{array}{l}a_{n+1}=3 a_{n}-2 b_{n} \cdots \text { (1) } \\ b_{n+1}=2 a_{n}-b_{n} \cdots \text { (2) }\end{array}\left(n \in \mathbf{N}^{*}\right)\right.$, find $a_{n}$ and $b_{n}$, and their respective sums of the first $n$ terms.
Solution: From equations (1) and (2), we get
$$
\begin{aligned}
a_{n+2} & =3 a_{n+1}-2 b_{n+1}=3 a_{n+1}-2\left(2 a_{n}-b_{n}\right) \\
& =3 a_{n+1}-4 a_{n}+2 b_{n}=3 a_{n+1}-4 a_{n}+\left(3 a_{n}-a_{n+1}\right) .
\end{aligned}
$$

Thus, $a_{n+2}=2 a_{n+1}-a_{n}\left(n \in \mathbf{N}^{*}\right) \cdots$ (3)
Similarly, we get $b_{n+2}=2 b_{n+1}-b_{n}\left(n \in \mathbf{N}^{*}\right) \cdots(4)$
The characteristic equations of (3) and (4) are both $x^{2}=2 x-1$, solving which gives $x_{1}=x_{2}=1$.

Therefore, we can let $a_{n}=(A+B n) x_{1}^{n-1}=A+B n\left(n \in \mathbf{N}^{*}\right), b_{n}=\left(A^{\prime}+B^{\prime} n\right) x_{1}^{n-1}=A^{\prime}+B^{\prime} n$ $\left(n \in \mathbf{N}^{*}\right)$.
Since $a_{1}=1, b_{1}=2$, we have $a_{2}=3 a_{1}-2 b_{1}=-1, b_{2}=2 a_{1}-b_{1}=0$.
Thus, from $\left\{\begin{array}{l}A+B=1 \\ A+2 B=-1\end{array}\right.$ we get $A=3, B=-2$; from $\left\{\begin{array}{l}A^{\prime}+B^{\prime}=2 \\ A^{\prime}+2 B^{\prime}=0\end{array}\right.$ we get $A^{\prime}=4, B^{\prime}=-2$.
Therefore, $a_{n}=3-2 n\left(n \in \mathbf{N}^{*}\right), b_{n}=4-2 n\left(n \in \mathbf{N}^{*}\right)$.
It is easy to find that the sums of the first $n$ terms of $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are respectively
$$
\begin{array}{l}
S_{n}=3 n-2(1+2+\cdots+n)=3 n-2(n+1)=2 n-n^{2}\left(n \in \mathbf{N}^{*}\right), \\
S_{n}{ }^{\prime}=4 n-2(1+2+\cdots+n)=4 n-n(n+1)=3 n-n^{2}\left(n \in \mathbf{N}^{*}\right) .
\end{array}
$$