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 $$S = \left \{( x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}, x_{8}, x_{9}\right )\}$$
 A: If $T = x_1 + x_2 + \ldots + x_9$, you can write this as
$$ T - (1+c_1) x_1 = T - (1 + c_2) x_2 = \ldots = T - (1+c_9) x_9 = 0$$
If any $c_i = -1$, then $T = 0$, and any $x_j$ for which $c_j \ne -1$ must be $0$, while those for which $c_j = -1$ must add to $0$.
On the other hand, if all $c_i \ne -1$,
then each $x_i = T/(1+c_i)$, and then
$$T = x_1 + \ldots + x_9 = T \left( \dfrac{1}{1+c_1} + \ldots + \dfrac{1}{1+c_9}\right) $$
Since you want a nontrivial solution, you don't want $T=0$ which would make all $x_i = 0$.
So then you need $$ \dfrac{1}{1+c_1} + \ldots + \dfrac{1}{1+c_9} = 1$$ and $T$ can be anything.
EDIT: The added condition that all $c_i > 1$ rules out the case where some $c_i = -1$, so you need $1/(1+c_1) + \ldots + 1/(1+c_9) = 1$.  You want 
$x_i \ge 20$, and since $x_i = T/(1+c_i)$ that says $T \ge 20 (1+c_i)$.
So now the solutions are 
$$ (x_1, \ldots, x_9) = \left(\dfrac{T}{1+c_1}, \ldots, \dfrac{T}{1+c_9}\right)$$
where $1/(1+c_1) + \ldots + 1/(1+c_9) = 1$ and $T \ge 20 (1 + \max(c_1, \ldots, c_9))$. 
A: 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 $\sum \frac 1{c_i + 1} = 1$.
A: Let
$$\mathrm A := 1_n 1_n^T - \mbox{diag} (1 + c_1, \dots, 1 + c_n)$$
where $c_i \neq -1$ for all $i \in \{1,2,\dots,n\}$. Using the matrix determinant lemma,
$$\det (\mathrm A) = \left( 1 - \sum_{i=1}^n \frac{1}{1 + c_i} \right) (-1)^n \left( \prod_{i=1}^n (1+c_i)\right)$$
We want the homogeneous linear system $\mathrm A \mathrm x = \mathrm 0_n$ to have non-trivial solutions. Thus, we impose the equality constraint $\det (\mathrm A) = 0$, or, equivalently,
$$\sum_{i=1}^n \frac{1}{1 + c_i} = 1$$
If this constraint is satisfied, using visual inspection, we conclude that all points on the line
$$\left\{ \gamma \begin{bmatrix} \frac{1}{1 + c_1}\\ \vdots\\ \frac{1}{1 + c_n}\end{bmatrix} : \gamma \in \mathbb R \right\}$$
are solutions to the aforementioned homogeneous linear system.
