Finding the recursive formula for calculating the solution of the linear equation system $A$x = b.

Question

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.

Answer 1

So,
a) solve the 1st as $x_2=b_1/p_1$;
b) solve the 2nd in terms of $x_3=(b_2-q_2x_1)/p_2$, the $x_1$ is unknown: keep it as such and continue;
c) all the following $x_{2k}$ will be a function of the precedent even-index and are therefore determined;
d) all the following $x_{2k+1}$ will be a linear function of the precedent odd-indexed and thus a linear function of $x_1$;
e) continue till the last two rows, which you are going to solve both in terms of $x_n$: one will refer to $x_{n-2}$ and one to $x_{n-1}$, thus one will contain the unknown and the other not, and you can solve for both $x_1$ and $p_n$;
f) replace the found value of $x_1$ in the precedent results depending on it.

Answer 2

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?