<s>[INST] Solving system of $9$ linear equations in $9$ variables I have a system of $9$ linear equations in $9$ variables:
\begin{array}{rl}
-c_{1}x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} - c_{2}x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} - c_{3}x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} - c_{4}x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} + x_{4} - c_{5}x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} + x_{4} + x_{5} - c_{6}x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} - c_{7}x_{7} + x_{8} + x_{9} &= 0 \\ 
x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} - c_{8}x_{8} + x_{9} &= 0 \\ 
x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} - c_{9}x_{9} &= 0 
\end{array}
I want to find a general non-trivial solution for it. What would be the easiest and least time consuming way to find it by hand? I don't have a lot of background in maths, so I would very much appreciate if you actually found the solution and explained briefly.
Thanks in advance!
EDIT: Very important to mention is that always any $c_{i} > 1$ and any $x_{i} \geq 20$. Also it would be nice if someone posted how would a general non-trivial solution look in the form of $$S = \left \{( x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}, x_{8}, x_{9}\right )\}$$
 [/INST] obviously $x_i = 0$ is a solution.  If you want a non-trivial solution
Subtract any two lines and you get $(c_i+1)x_i =  (c_j+1)x_j = 0$
If any $c_j = -1$ we have all $(c_i + 1)x_i = 0$ so if $c_i \ne -1$ then $x_i = 0$ and if $c_i = -1$ then $x_i$ can be anything at all.
If none of the $c_i = -1$ then let $(c_i + 1)x_i = M \ne 0$.
Then $x_i = M/(c_i+1)$.  But is that possible?
Each line is $(\sum_{i= 0}^9 M/(c_i+1) ) - M = 0$
So $\sum_{i=0}^9 1/(c_i + 1) = 1$.
There are no non-zero solutions unless that very unlikely criteria is met.
So
(recap)
1)  $x_i= 0$ is a solution.
2) If any $c_i = -1$ then $x_i = \begin{cases} anything; c_i = -1 \\ 0; c_i \ne -1 \end{cases}$
3) If $c_i \ne -1 \forall i$ then $x_i = M/(c_i +1)$ will be a solution IF any of the lines add to 0, which will happen if and only if $\sum \frac 1{c_i + 1} = 1$.
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