A sequence $\{a_n\}_{n \ge 1}$ is defined recursively by
$$a_0 = 1, a_1 = 1$$
$$a_n = 5a_{n-1}-6a_{n-2}, \text{ for } n \ge 2 $$
Find an explicit formula for the general term $A_n$.
So, I want to let the vector
$
\begin{equation*}
A_n = \begin{bmatrix} a_n\\
a_{n-1}\\
\end{bmatrix}
\end{equation*}
$
Then if I can find a relation between $A_n$ and $A_{n-1}$ using a $2 \times 2$ matrix I believe I'm close. Once the relation is found I could use eigenvector diagonalization to possibly find the explicit formula for $A_n$?
There should be a relation, I believe, between the characteristic polynomial of a $2 \times 2$ matrix to the coefficients of
$a_n – 5a_{n-1} + 6a_{n-2} = 0$
Thank you for the help.
Best Answer
Hint: $$ A_n = \begin{bmatrix} a_n \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} 5a_{n-1} - 6a_{n-2} \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} 5 & -6 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n-1}\\ a_{n-2} \end{bmatrix} = \begin{bmatrix} 5 & -6 \\ 1 & 0 \end{bmatrix} A_{n-1}. $$