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I would like to find the general solution to the following system of equations:

$$ x_1 + k_1 + \sum_{i=1}^N A_{1,i}x_i=0 $$ $$ x_2 + k_2 + \sum_{i=1}^N A_{2,i}x_i=0 $$ $$\vdots$$ $$ x_N + k_N + \sum_{i=1}^N A_{N,i}x_i=0 $$

where $k_j$ and $A_{j,i}$ are known constants, and $x_j$ are the unknown variables. This system has $N$ equations with $N$ unknown variables. However, $N$ in my case is at least $50$. I want to find the pattern for solving this system numerically in Matlab using recursion and loops. I have a feeling an algorithm can be constructed, but I can't find the pattern. Here is what I did.

I solved this problem for $N=1$, $N=2$ and $N=3$. For $N = 1$ the solution is:

$$ x_1 = -{k_1 \over 1 + A_{1,1}} \tag 1$$

Also, for $N = 2$ the solution is:

$$ x_1 = { -(1 + A_{2,2}) k_1 + A_{1,2} k_2 \over (1 + A_{1,1}) (1 + A_{2,2}) - A_{1,2} A_{2,1}} \tag 2 $$

$$ x_2 = { A_{2,1} k_1-(1 + A_{1,1}) k_2 \over (1 + A_{1,1}) (1 + A_{2,2}) - A_{1,2} A_{2,1} } \tag 3$$

And the solution for $N = 3$ is:

$$ x_1 = {-\big( (1 + A_{2,2}) (1 + A_{3,3}) - A_{2,3} A_{3,2} \big) k_1 + (A_{1,2} (1 + A_{3,3}) - A_{1,3} A_{3,2}) k_2 + (A_{1,3} (1 + A_{2,2}) - A_{1,2} A_{2,3}) k_3 \over C} \tag 4$$

$$ x_2 = { (A_{2,1} (1 + A_{3,3}) - A_{2,3} A_{3,1}) k_1 -\big( (1 + A_{1,1}) (1 + A_{3,3}) - A_{1,3} A_{3,1} \big) k_2 + (A_{2,3} (1 + A_{1,1}) - A_{1,3} A_{2,1}) k_3 \over C} \tag 5$$ $$ x_3 = { (A_{3,1} (1 + A_{2,2}) - A_{2,1} A_{3,2}) k_1 + (A_{3,2} (1 + A_{1,1}) - A_{1,2} A_{3,1}) k_2 -\big( (1 + A_{1,1}) (1 + A_{2,2}) - A_{1,2} A_{2,1} \big) k_3 \over C} \tag 6$$

where:

$$ C = (1 + A_{1,1}) (1 + A_{2,2}) (1 + A_{3,3}) - (1 + A_{1,1}) A_{2,3} A_{3,2} - (1 + A_{2,2}) A_{1,3} A_{3,1} - (1 + A_{3,3}) A_{1,2} A_{2,1} + A_{1,2} A_{2,3} A_{3,1} + A_{3,2} A_{2,1} A_{1,3} \tag 7$$

Unfortunately, I can't seem to notice the general pattern. The only thing I do notice is that there might be two separate patterns involved. One for the numerator and one for the denominator. I would like to ask for help.

In summary, I would like to numerically solve this system of equations, but I need to construct an algorithm for that based on the pattern I can't seem to notice. My question is, what is the general solution for $x_n$, where $n \in \{1, 2, ... N\}$, and $N \in \Bbb{N}$? If there is no general solution, how can I construct the algorithm I need using recursions and loops (if it can be done at all)?

NOTE: I also tried to solve for $N=4$, but the solution for one variable got two pages long, so I couldn't see the right way to group the terms in a way that would help find the general pattern. I will write it down here, but I doubt it is of any use:

$$x_1 = {Q_1 k_1 + Q_2 Q_2 + Q_3 k_3 + Q_4 k_4 \over D} \tag 8$$

$$ Q_1 = - \Big( (1 + A_{3,3}) (1 + A_{4,4}) - A_{3,4} A_{4,3} \Big)f_1 \tag 9$$

$$ Q_2 = \Big ( (1 + A_{3,3}) (1 + A_{4,4}) - A_{3,4} A_{4,3} \Big) f_3 \tag {10}$$

$$ Q_3 = \Big( (1 + A_{4,4}) A_{1,3} - A_{1,4} A_{4,3} \Big)f_1 - \Big( (1 + A_{4,4}) A_{2,3} - A_{2,4} A_{4,3} \Big) f_3 \tag {11}$$

$$ Q_4 = \Big( (1 + A_{3,3}) A_{1,4} - A_{1,3} A_{3,4} \Big)f_1 - \Big( (1 + A_{3,3}) A_{2,4} - A_{2,3} A_{3,4} \Big)f_3 \tag {12}$$

$$ f_1 = (1 + A_{2,2}) (1 + A_{3,3}) (1 + A_{4,4}) - (1 + A_{2,2}) A_{3,4} A_{4,3} - (1 + A_{3,3}) A_{2,4} A_{4,2} - (1 + A_{4,4}) A_{2,3} A_{3,2} + A_{2,3} A_{3,4} A_{4,2} + A_{2,4} A_{4,3} A_{3,2} \tag{13} $$

$$ f_2 = (1 + A_{1,1}) (1 + A_{3,3}) (1 + A_{4,4}) - (1 + A_{1,1}) A_{3,4} A_{4,3} - (1 + A_{3,3}) A_{1,4} A_{4,1} - (1 + A_{4,4}) A_{1,3} A_{3,1} + A_{1,4} A_{4,3} A_{3,1} + A_{1,3} A_{3,4} A_{4,1} \tag{14} $$

$$ f_3 = (1 + A_{3,3}) (1 + A_{4,4}) A_{1,2} - (1 + A_{3,3}) A_{1,4} A_{4,2} - (1 + A_{4,4}) A_{1,3} A_{3,2} - A_{1,2} A_{3,4} A_{4,3} + A_{1,3} A_{3,4} A_{4,2} + A_{1,4} A_{4,3} A_{3,2} \tag {15} $$

$$f_4 = -(1 + A_{3,3}) (1 + A_{4,4}) A_{2,1} + (1 + A_{3,3}) A_{2,4} A_{4,1} + (1 + A_{4,4}) A_{2,3} A_{3,1} + A_{2,1} A_{3,4} A_{4,3} - A_{2,3} A_{3,4} A_{4,1} - A_{2,4} A_{3,1} A_{4,3} \tag{16}$$

$$D = f_1 f_2 + f_3 f_4 \tag {17}$$

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If we denote
$$\begin{align} &a_{i,j} = \cases{A_{i,j} &if $i=j$ \\ A_{i,j} +1 &if $i\ne j$ } \hspace{1cm} &\forall 1\le i,j \le N \\ &b_{i} =-k_i &\hspace{1cm} \forall 1\le i \le N \end{align}$$ then the problem is simply equivalent to solve a system of linear equations: $$\sum_{i=1}^N a_{i,j}x_j=b_i \hspace{1cm} \forall 1\le i \le N \tag{1}$$

Without any further information about $(A_{i,j})_{1\le i,j \le N}$, only general solutions are available.

Let us denote $\mathbf{a} := (a_{i,j})_{i,j} \in \mathbb{R}^{N\times N}$ the invertible matrix, $\mathbf{x} := (x_1,..,x_N)' $ and $\mathbf{b} := (b_1,..,b_N)' $ then the system of linear equations $(1)$ can be expressed in matrix form as follows: $$\mathbf{a}\cdot \mathbf{x} = \mathbf{b}$$ As the matrix $\mathbf{a}$ is invertible, mathematically, the solution $\mathbf{x}$ is equal to $$\mathbf{x} = \mathbf{a}^{-1}\cdot \mathbf{b} \tag{2}$$

Remark: numerically, there are several methods for solving the system of linear equations. With only available information of $N$ (and not $\mathbf{a}$ or $\mathbf{b}$) , I think this answer is useful.

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  • $\begingroup$ I just noticed in your link that I can construct a matrix form of my problem $[A] \vec x=\vec b$. My matrix is square, and full rank, so I can just use its inverse to solve my problem: $\vec x = [A]^{-1} \vec b$. Can you demonstrate this in your answer so I can accept it? Thank you a lot! $\endgroup$ Jan 13 at 23:16
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    $\begingroup$ @NikolaRistic I add some other information. Hope that helps! $\endgroup$
    – NN2
    Jan 13 at 23:45

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