recurrence relations and closed form

Question

Let $a_n = 5a_{n-1} - 6a_{n-2}$. $a_0 = 12, a_1 = 27$. What is the closed form for this recurrence? What is $a_{10}$?

I have figured out what $a_n$ is up to $a_{10}$ Which is:

$a_0 = 12$, $a_1 = 27$, $a_2 = 63$, $a_3 = 153$, $a_4 = 387$, $a_5 = 1017$, $a_6 = 2763$, $a_7 = 7713$, $a_8 = 21987$, $a_9 = 63657$, $a_{10} = 186363$

However I am stuck on getting the closed form equation. I came up with this equation based on the example Image 2.4.6 below.

$x^2 - 5x + 6 = 0$

solving for x:

$(x-3) (x-2) = 0$

$x=3,2$

recurrence relation form:

$a_n = a3^n + b2^n$

plug in $n = 0, n = 1$

$12 = a3^0 + b2^0 = a + b$

$27 = a3^1 + b2^1 = 3a + 2b = 27$

This is the part that I am stuck on. In the example 2.4.6 I am not sure how they solved for this system and got $a = \frac{7}{3}$ and $b = - \frac{1}{3}$

Any insight will be much appreciated

Answer 1

You got $a+b=12$ and $3a+2b=27.$

Subtract twice the first equation from the second to get $a=3$.

Then solve for $b$ from the first equation.