Number of words with no adjacent vowels I am faced with the following problem:

Given $C = \{B,C,D,F\}$ and $V = \{A, E, I, O, U\}$ find the number of 9-letter words with elements from $C$ and $V$ such that no two vowels (elements of V) are adjacent.

Following this answer about a very similar question I get that I should express the problem as a double recurrence:

$a_{n+1} = 5b_n$, $a_0 = 1$
$b_{n+1} = 4(a_n + b_n)$, $a_0 = 1$

where $a_n$ is the number of string starting by a vocal and $b_n$ is the number of string starting by a consonant.
Expressing it as a matrix I get $\begin{pmatrix} a_{n+1} \\ b_{n+1} \end{pmatrix} =  \begin{pmatrix} 0 & 5 \\ 4 & 4 \end{pmatrix} \begin{pmatrix} a_n \\ b_n \end{pmatrix}$
And I get that the eigenvalues of this matrix are $\{2+\frac{\sqrt{96}}{2}, 2-\frac{\sqrt{96}}{2}\}$. Is this the correct approach?
I don't really know where to go from here or how to solve the recurrence for any given $k$. Thank you in advance
 A: You boil your problem down to a linear recurrence relation, which is nice!  
There's a little issue with $a_0$ and $b_0$, though : how can the empty word start by anything? One should start the recurrence at $n=1$, we have $a_1 = 5$ and $b_1 = 4$.
Call 
$A = \begin{pmatrix} 0 & 5 \\ 4 & 4 \end{pmatrix}$, then 
$$\begin{pmatrix} a_n \\ b_n \end{pmatrix} = A^{n-1} \begin{pmatrix} 5 \\ 4 \end{pmatrix} $$
The point of computing the eigenvalues is to put $A$ in a nicer form so that $A^n$ is easy to compute. Since you have two distinct eigenvalues, you can diagonalize $A$ which will lead you to a nice expression of $a_n$ and $b_n$.
Diagonalization of $A$ :
Edit : in fact $96 = 4 \cdot 24$ and not $4\cdot 19$ !
One finds two distinct eigenvalues, $\lambda_1 = 2 + \sqrt{24}$ and $\lambda_2 = 2 - \sqrt{24}$.
Let $i=1,2$, an eigenvector $v_i$ corresponding to $\lambda_i$ is a nonzero solution to the equation $$(A - \lambda_i \operatorname{Id}) v_i = 0$$
Write $v_i = \begin{pmatrix} x \\ y \end{pmatrix}$, we get
$$\left\{\begin{matrix} -\lambda_i x + & 5 y = & x \\
4 x + & (4-\lambda_i)y = & y \end{matrix} \right.$$
The first line implies $y = \frac{\lambda_i}{5} x$, after which the second line vanishes. Now chose an arbitrary $x \neq 0$, let's take $x = 5$ for convenience.
Now $v_1 = \begin{pmatrix} 5 \\ 2 + \sqrt{24} \end{pmatrix}$ and $v_2 = \begin{pmatrix} 5 \\ 2 - \sqrt{24} \end{pmatrix}$ form a basis of eigenvectors. Take 
$$P = \begin{pmatrix} 5 & 5 \\ 2 + \sqrt{24} & 2 - \sqrt{24} \end{pmatrix}$$
One has 
$$ P^{-1}AP = \begin{pmatrix} 2 + \sqrt{24} & 0 \\ 0 & 2 - \sqrt{24}\end{pmatrix} $$
Hence, 
$$ A^k = P \begin{pmatrix} (2 + \sqrt{24})^k & 0 \\ 0 & (2 - \sqrt{24})^k\end{pmatrix} P^{-1}$$
