Isolating $a_n$ in a recursive formula I have three equations with three sequences, $a_n, b_n, c_n$.
I tried to isolate $a_n$ with no luck. 
$$a_n = 2b_{n-1}+c_{n-1}$$
$$b_n=2a_{n-1}+2b_{n-1}+c_{n-1}$$
$$c_n = 4a_{n-1}+4b_{n-1}$$
Is it even possible to get an expression based only on $a_n$ terms here?
 A: Notice that the expression $2b_{n-1}+c_{n-1}$, which is equal to $a_n$, appears in the expression for $b_n$. So we can rewrite that as 
$$b_n = 2a_{n-1}+a_n$$
Thus $$c_n = 4a_{n-1} + 4b_{n-1} = 4a_{n-1} + 8a_{n-2} + 4a_{n-1} = 8a_{n-1} + 8a_{n-2}$$
And, finally, $$a_n = 2b_{n-1}+c_{n-1} = 4a_{n-2} + 2a_{n-1} + 8a_{n-2} + 8a_{n-3} = 2a_{n-1} + 12a_{n-2} + 8a_{n-3}$$
A: Your coefficient matrix is
$$
M =
\left(
\begin{array}{ccc}
0 & 2 & 1 \\
2 & 2 & 1 \\
4 & 4 & 0 \\
\end{array}
\right)
$$
which satisfies (Cayley-Hamilton)
$$ M^3 - 2 M^2 - 12 M - 8I = 0 $$
$$  a_{n+3} = 2 a_{n+2} + 12 a_{n+1} + 8 a_n \; . $$
You also get
$$  b_{n+3} = 2 b_{n+2} + 12 b_{n+1} + 8 b_n \; , $$
$$  c_{n+3} = 2 c_{n+2} + 12 c_{n+1} + 8 c_n \; . $$
If we make the column vector
$$
x_n =
\left(
\begin{array}{c}
a_n \\
b_n \\
c_n \\
\end{array}
\right) \; ,
$$
we find $x_{n+1} = M x_n \; ,$ then $x_{n+2} = M x_{n+1} = M^2 x_n \; ,$ finally $x_{n+3}= M x_{n+2} = M^3 x_n.$ Cayley Hamilton says
$$  x_{n+3} = M^3 x_n = \left( 2M^2 + 12 M + 8 I \right)x_n =  2M^2 x_n + 12 M x_n + 8 I x_n = 2 x_{n+2} + 12 x_{n+1} + 8 x_n $$
