Solving system of $9$ linear equations in $9$ variables

Question

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 $$S = \left \{( x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}, x_{8}, x_{9}\right )\}$$

Answer 1

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 $T \ge 20 (1 + \max(c_1, \ldots, c_9))$.

Answer 2

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 $\sum \frac 1{c_i + 1} = 1$.

Answer 3

This is a linear algebra method. First, you would take the coefficients of each of your variables and put them in a matrix: $$ \begin{bmatrix}-c_1 &1&1&1&1&1&1&1&1\\1&-c_2&1&1&1&1&1&1&1\\1&1&-c_3&1&1&1&1&1&1\\1&1&1&-c_4&1&1&1&1&1\\1&1&1&1&-c_5&1&1&1&1\\1&1&1&1&1&-c_6&1&1&1\\1&1&1&1&1&1&-c_7&1&1\\1&1&1&1&1&1&1&-c_8&1\\1&1&1&1&1&1&1&1&-c_9\end{bmatrix}$$ Then put the variables into vector form: $$\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7\\x_8\\x_9\end{bmatrix}$$ Finally, take the constants that the equation is equivalent to and put those in vector form: $$\begin{bmatrix}0\\0\\0\\0\\0\\0\\0\\0\\0\end{bmatrix}$$ Then put it all together in an equation: $$\begin{bmatrix}-c_1 &1&1&1&1&1&1&1&1\\1&-c_2&1&1&1&1&1&1&1\\1&1&-c_3&1&1&1&1&1&1\\1&1&1&-c_4&1&1&1&1&1\\1&1&1&1&-c_5&1&1&1&1\\1&1&1&1&1&-c_6&1&1&1\\1&1&1&1&1&1&-c_7&1&1\\1&1&1&1&1&1&1&-c_8&1\\1&1&1&1&1&1&1&1&-c_9\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7\\x_8\\x_9\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\0\\0\\0\\0\\0\end{bmatrix}$$ This equation is of the form $Ax=b$ and I'll refer to each piece of the equation as $A$, $x$, or $b$ here on out. First, take $\det(A)$. If $\det(A)=0$ then check if $b$ is in the column space of $A$. If it is not, there are no solutions, if it is, there are an infinite number of solutions. If $\det(A) \neq 0$ then proceed to find the inverse of $A$. When this is completed, multiply each side by the inverse, and you will be left with $x=bA^{-1}$. Once multiplying the inverse by $b$, the variables will line up with the solutions. I'll be updating this to show how to do these manipulations.

Hope this helps!

Answer 4

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.