Finding the recursive formula for calculating the solution of the linear equation system $A$x = b. Let $K$ be a field and n $\in$ $\mathbb{N}$. It's given that $A$ =  $
 \begin{bmatrix}
0 & p_{1} & 0 & 0 & \dots & 0 \\
q_{2} & 0 & p_{2} & 0 & \dots & 0 \\
0 & q_{3} & 0 & p_{3} & \dots & 0 \\
\vdots &  & \ddots & \ddots & \ddots & \vdots \\
0 & \dots & 0 & q_{n-1} & 0 & p_{n-1} \\
0 & \dots & 0 & 0 & q_{n} & p_{n}
\end{bmatrix}
$  $\in$ $K^{n,n}$, 
$b$ =  $
 \begin{bmatrix}
b_{1} \\
\vdots \\
b_{n} \\
\end{bmatrix}
$  $\in$ $K^{n,1}$ with $p_{i}$,$q_{i}$ $\neq$ 0 for all i = 1, $\dots$  ,n. 
Find a recursive formula for calculating the solution of the linear equation system $A$x = b. 
My ideas and thoughts:
I thought about writing the matrix as the following:$$
\left[\begin{array}{cccccc|c}
0 & p_{1} & 0 & 0 & \dots & 0 & b_{1} \\
q_{2}&0&p_{2}&0&\dots&0&\vdots\\
0&q_{3}&0&p_{3}&\dots&0&\vdots\\
\vdots&&\ddots&\ddots&\ddots&\vdots&\vdots \\
0 & \dots & 0 & q_{n-1}&0&p_{n-1}&\vdots \\
0&\dots&0&0&q_{n}&p_{n}&b_{n} \end{array}\right]
$$ so that \begin{equation}
    =
   \begin{cases}
     0 \cdot x_{1} + p_{1} \cdot x_{2} + 0 \cdot x_{3} + 0 \cdot x_{4} \dots 0 \cdot x_{n} = b_{1} \\
     q_{2} \cdot x_{1} + 0 \cdot x_{2} + p_{2} \cdot x_{3} + 0 \cdot x_{4} \dots 0 \cdot x_{n} = b_{2} \\
     0 \cdot x_{1} + q_{3} \cdot x_{2} + 0 \cdot x_{3} + p_{3} \cdot x_{4} \dots 0 \cdot x_{n} = b_{3} \\
     \vdots \\
     0 \cdot x_{1} \dots 0 \cdot x_{n} + q_{n-1} \cdot x_{n+1} + 0 \cdot x_{n+2} + p_{n-1} \cdot x_{n+3} = b_{n-1} \\
     0 \cdot x_{1} \dots 0 \cdot x_{n} + 0 \cdot x_{n+1} + q_{n} \cdot x_{n+2} + p_{n} \cdot x_{n+3} = b_{n}
   \end{cases}
\end{equation} Now I could calculate the solution for each row:
(1) $0 \cdot x_{1} + p_{1} \cdot x_{2} + 0 \cdot x_{3} + 0 \cdot x_{4} \dots 0 \cdot x_{n} = b_{1}$ 
$\Leftrightarrow$ $ p_{1} \cdot x_{2} = b_{1}$ $\quad$ $\mid$ $\div$ $p_{1}$
$\Leftrightarrow$ $x_{2}$ = $\frac{b_{1}}{p_{1}}$
(3) $ 0 \cdot x_{1} + q_{3} \cdot x_{2} + 0 \cdot x_{3} + p_{3} \cdot x_{4} \dots 0 \cdot x_{n} = b_{3}$
$\Leftrightarrow$ $q_{3} \cdot x_{2} + p_{3} \cdot x_{4} = b_{3}$
$\Leftrightarrow$ $q_{3} \cdot \frac{b_{1}}{p_{1}} + p_{3} \cdot x_{4} = b_{3}$  $\quad$ $\mid$ $- q_{3} \cdot \frac{b_{1}}{p_{1}}$
$\Leftrightarrow$ $p_{3} \cdot x_{4} = b_{3} - q_{3} \cdot \frac{b_{1}}{p_{1}}$$\quad$ $\mid$ $\div p_{3}$
$\Leftrightarrow$ $x_{4}$ = $\frac{b_{3}-q_{3} \cdot \frac{b_{1}}{p_{1}}}{p_{3}}$
$\Leftrightarrow$ $\frac{p_{1} \cdot b_{3}-q_{3} \cdot b_{1}}{p_{1} \cdot p_{3}}$
And so on $\dots$ 
I don't know how to continue at this point to find a recursive formula for calculating the solution of the linear equation system. My ideas did not really lead me anywhere (I don't even think that they are correct) and im not sure if I even understand the matrix correctly. So how should I start instead? 
Any hints guiding me to the right direction I much appreciate.

Here are $A^{-1}$ for $n$ from $1$ to $4$:
$$(1/p_1),\ \pmatrix{ -p_2/(p_1 q_1) & 1/q_1\cr 1/p_1 & 0\cr},\
\pmatrix{p_2 q_2/(p_1 q_1 p_3) & 1/q_1 & -p_2/(q_1 p_3) \cr 1/p_1 & 0 & 0\cr -q_2/(p_1 p_3) & 0 & 1/p_3},\
\pmatrix{-p_2 q_2 p_4/(p_1 q_1 p_3 q_3) & 1/q_1 & p_2 p_4/(q_1 p_3 q_3) & -p_2/(q_1 q_3)\cr
1/p_1 & 0 & 0 & 0\cr q_2 p_4/(p_1 p_3 q_3) & 0 & -p_4/(p_3 q_3) & 1/q_3\cr
-q_2/(p_1 p_3) & 0 & 1/p_3 & 0} $$
See a pattern?
