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this is what I'm trying to demonstrate

With the following system of $n \times n$ equations

$G_1B_1+G_2B_2+\cdots+G_nB_n=G_1A_1+G_2A_2+\cdots+G_nA_n$

$B_1-B_2=A_2-A_1$

$B_2-B_3=A_3-A_2$

$\vdots \hspace{1cm} \vdots \hspace{1cm} \vdots \hspace{1cm} \vdots $

$B_{n-1}-B_{n}=A_n-A_{n-1}$

Prove that in general any $B_{V}$ with $V \in \{1,2,\cdots,n\}$ is equal to:

$B_V=\left(\Gamma_1A_1+\Gamma_2A_2+\cdots+\Gamma_nA_n\right)-A_V$

With $\Gamma_V=\dfrac{2G_V}{G1+G_2+\cdots+G_n}$

This is my first time solving a problem like that, so the first thing that I tried, was to write the system in a matrix form

$\left( \begin{array}{cccccc} G_1 & G_2 & G_3 & G_4 & \cdots & G_{n-1} & G_n \\ 1 & -1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & -1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & -1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & -1 \\ \end{array} \right)$$\left( \begin{array}{c} B_1 \\ B_2 \\ B_3 \\ B_4 \\ \vdots \\ B_{n-1}\\ B_n \end{array} \right)$=$\left( \begin{array}{cccccc} G_1 & G_2 & G_3 & G_4 & \cdots & G_{n-1} & G_n \\ -1 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & -1 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & -1 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & -1 & 1 \\ \end{array} \right)$$\left( \begin{array}{c} A_1 \\ A_2 \\ A_3 \\ A_4 \\ \vdots \\ A_{n-1}\\ A_n \end{array} \right)$

But it gets me nowhere and I think that it isn't the right approach to achieve that demonstration, This is not a book exercise, is just a formula that I found in a paper and that I don't understant how was obtained.

This is the paper https://ccrma.stanford.edu/~jingjiez/portfolio/gtr-amp-sim/pdfs/Wave%20Digital%20Filters%20Theory%20and%20Practice.pdf , equation (29)

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First you should note that $$ A_1+B_1= A_2+B_2= \dots = A_n + B_n. $$

Now starting from the first given equation we have $$ 0= \sum_i G_i (A_i -B_i)= -\sum_i G_i (A_i +B_i) +\sum_i G_i (A_i +B_i) +\sum_i G_i (A_i -B_i) $$ so that $$ 0= -\sum_i G_i (A_i +B_i) +2\sum_i G_i A_i, $$ or $$ \sum_i G_i (A_i +B_i)= 2\sum_i G_i A_i. $$

Now recall that $A_i+B_i$ does not depend on $i$; we can replace them all by $A_j+B_j$ for a fixed $j$ and get: $$ \left(\sum_i G_i\right) (A_j +B_j)= 2\sum_i G_i A_i $$ which is what you wanted to prove.

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