Linear algebra on Fibonacci number Consider the sequence $\{a_n\}_{n\ge 0}$ given by the recurrence relation
$$a_0=1,\ a_1=-1,\ a_{n+1}=3a_n+10a_{n-1}\ \ \text{for } n\ge2$$
And I am asked to work out the closed form expression for an in the same fashion as the proof for Fibonacci numbers by using linear algebra way
Thanks all for the help!

We can use the generating function $f(x)$ of $\{a_n\}$ to solve. Let $f(x)=\sum_{n=0}^\infty a_nx^n$. Then
\begin{eqnarray*}
f(x)&=&\sum_{n=0}^\infty a_nx^n=1-x+\sum_{n=2}^\infty a_nx^n\\
&=&1-x+\sum_{n=1}^\infty a_{n+1}x^{n+1}\\
&=&1-x+\sum_{n=1}^\infty (3a_{n}+10a_{n-1})x^{n+1}\\
&=&1-x+3\sum_{n=1}^\infty a_{n}x^{n+1}+10\sum_{n=1}^\infty a_{n-1}x^{n+1}\\
&=&1-x+3x\sum_{n=1}^\infty a_{n}x^{n}+10x^2\sum_{n=1}^\infty a_{n-1}x^{n-1}\\
&=&1-x+3x(f(x)-1)+10x^2f(x)
\end{eqnarray*}
from which we obtain the following function equation
$$ f(x)=1-x+3x(f(x)-1)+10x^2f(x). $$
Hence
\begin{eqnarray*}
f(x)&=&\frac{4x-1}{10x^2+3x-1}\\
&=&\frac{6}{7}\frac{1}{1+2x}+\frac{1}{7}\frac{1}{1-5x}\\
&=&\frac{6}{7}\sum_{n=0}^\infty (-2)^nx^n+\frac{1}{7}\sum_{n=0}^\infty 5^nx^n
\end{eqnarray*}
and so
$$ a_n=\frac{6}{7}(-2)^n+\frac{1}{7}5^n. $$
