<s>[INST] Recurrent equation system How to solve this recurrent equation system?
$$\begin{cases}a_{n+1}=a_n+2b_n\\
b_{n+1}=2a_n+b_n\end{cases}$$
The possible solutions should be
\begin{cases}a_n=\frac{1}{2}3^n+\frac{1}{2}(-1)^n\\
b_n=\frac{1}{2}3^n-\frac{1}{2}(-1)^n\end{cases}
 [/INST] Let $X_n =(a_n,b_n)$ with $X_0=(1,0)=(a_0,b_0)$
\begin{cases}a_{n+1}=a_n+2b_n\\
b_{n+1}=2a_n+b_n\end{cases} Then it is equivalent $$X_{n+1} = AX_n \Longleftrightarrow X_n =A^nX_0$$
where, $$A= \begin{pmatrix}1&2\\ 2&1\end{pmatrix}$$
Then prove by induction that,
$$A^n= \begin{pmatrix}\frac{1}{2}3^n+\frac{1}{2}(-1)^n &\frac{1}{2}3^n-\frac{1}{2}(-1)^n\\\frac{1}{2}3^n-\frac{1}{2}(-1)^n&\frac{1}{2}3^n+\frac{1}{2}(-1)^n\end{pmatrix}$$ 
Therefore you will get the desired result.
$$X_n= \begin{pmatrix}a_n\\b_n\end{pmatrix}= \begin{pmatrix}\frac{1}{2}3^n+\frac{1}{2}(-1)^n &\frac{1}{2}3^n-\frac{1}{2}(-1)^n\\\frac{1}{2}3^n-\frac{1}{2}(-1)^n&\frac{1}{2}3^n+\frac{1}{2}(-1)^n\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\\=\begin{pmatrix}\frac{1}{2}3^n+\frac{1}{2}(-1)^n\\
\frac{1}{2}3^n-\frac{1}{2}(-1)^n\end{pmatrix}$$
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