Compressing two recurrences

Question

I have two recurrences

$a_n = 9a_{n-1} + b_{n-1}$

$b_n = 9b_{n-1} + a_{n-1}$

Is there a way to combine these two so it's only in terms of $a_n$?

$a_1 = 9, b_1 = 1$, if this information is needed.

Answer 1

From the first equation, $b_{n-1}=a_n-9a_{n-1}$.
Put that into the second equation, twice, to get rid of the $b_k$ and leave an equation in $a_k$.

Answer 2

Let us make the problem more general $$a_n = \alpha\, a_{n-1} +\beta \,b_{n-1}$$ $$b_n = \gamma\, b_{n-1} +\delta \,a_{n-1}$$ So, from the first $$b_{n-1}=\frac 1\beta \, a_n-\frac \alpha\beta \, a_{n-1}$$ Then the second equation becomes $$\frac 1\beta \, a_{n+1}-\frac \alpha\beta \, a_{n}=\gamma\left(\frac 1\beta \, a_n-\frac \alpha\beta \, a_{n-1}\right) +\delta a_{n-1}$$ which, after simplifications, write $$a_{n+1}-(\alpha+\gamma)\,a_n+(\alpha\gamma-\beta\delta)\,a_{n-1}=0$$ to be solved in the usual manner using the condition $a_1=A$; so the expression of $a_n$ still contain a constant the value of which being obtained solving for it $$b_1=\frac 1\beta \, a_2-\frac \alpha\beta \, a_{1}=B$$

Answer 3

$$a_n = 9a_{n-1} + b_{n-1}$$ $$b_n = 9b_{n-1} + a_{n-1}$$

There is another way to "combine" $a$ and $b$ into 1 variable. Let $V = \begin{bmatrix} a \\ b \end{bmatrix}$, then

$$V_n = \begin{bmatrix} 9 & 1 \\ 1 & 9 \end{bmatrix} V_{n-1}$$

And since $\begin{bmatrix} 9 & 1 \\ 1 & 9 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 10 & 0 \\ 0 & 8 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1}$, then

$$V_{n+1} = \begin{bmatrix} 9 & 1 \\ 1 & 9 \end{bmatrix}^n V_{1}$$ $$V_{n+1} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 10^n & 0 \\ 0 & 8^n \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}^{-1} V_{1}$$