What is the explicit formula for the general term of the sequence?

Question

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.

Answer 1

Hint: $$ A_n = \begin{bmatrix} a_n \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} 5a_{n-1} - 6a_{n-2} \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} 5 & -6 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n-1}\\ a_{n-2} \end{bmatrix} = \begin{bmatrix} 5 & -6 \\ 1 & 0 \end{bmatrix} A_{n-1}. $$

Answer 2

To compare with the matrix method a classic method is given here. Let $a_{n} = r^{n}$ to obtain \begin{align} a_{n} &= 5 \, a_{n-1} - 6 \, a_{n-2} \\ r^{n} &= 5 \, r^{n-1} - 6 \, r^{n-2} \\ r^{2} &= 5 \, r - 6 \\ r &\in \{2, 3\}. \end{align} From this it is given that $a_{n} = c_{0} \, 2^{n} + c_{1} \, 3^{n}$. Using $a_{0} = a_{1} = 1$ then the solution becomes $$a_{n} = 2^{n+1} - 3^{n}.$$