Stuck for generalization of a proof on specific Linear System of Equations I have a specific linear system of $n$ equations with $n$ unknowns that clearly has a solution generalizable for any $n$ (as I tried several $n$ and see the pattern), but I cannot seem to find a proof that shows this is the case. Any hint that could help me to prove this in the general case would be greatly appreciated! The system of equations can be presentend by $n$ equations for each $i \in [1:n]$, $\rho \geq 0$ and $\rho \ne 1$:
$x_{i} + \rho \sum\limits_{\substack{j=1, \\ j \ne i}}^{n} x_{j} = a_{i}  $ 
Consider for instance for $n=5$, then we can solve the system using the Gaussian elimination method like so:
$ 
\left(\begin{array}{cccccc} 1 & \rho  & \rho  & \rho  & \rho  & a_{1}\\ \rho  & 1 & \rho  & \rho  & \rho  & a_{2}\\ \rho  & \rho  & 1 & \rho  & \rho  & a_{3}\\ \rho  & \rho  & \rho  & 1 & \rho  & a_{4}\\ \rho  & \rho  & \rho  & \rho  & 1 & a_{5} \end{array}\right)
$
$
\left(\begin{array}{cccccc} 1 & \rho  & \rho  & \rho  & \rho  & a_{1}\\ 0 & 1 & \frac{\rho }{\rho +1} & \frac{\rho }{\rho +1} & \frac{\rho }{\rho +1} & -\frac{a_{2}-a_{1}\,\rho }{\rho ^2-1}\\ 0 & 0 & 1 & \frac{\rho }{2\,\rho +1} & \frac{\rho }{2\,\rho +1} & \frac{a_{3}-a_{1}\,\rho -a_{2}\,\rho +a_{3}\,\rho }{-2\,\rho ^2+\rho +1}\\ 0 & 0 & 0 & 1 & \frac{\rho }{3\,\rho +1} & -\frac{a_{1}\,\rho -a_{4}+a_{2}\,\rho +a_{3}\,\rho -2\,a_{4}\,\rho }{-3\,\rho ^2+2\,\rho +1}\\ 0 & 0 & 0 & 0 & 1 & -\frac{a_{1}\,\rho -a_{5}+a_{2}\,\rho +a_{3}\,\rho +a_{4}\,\rho -3\,a_{5}\,\rho }{-4\,\rho ^2+3\,\rho +1} \end{array}\right)\\
$
And the solution is similar for other $n$, suggesting we can say in general that
$x_i = - \frac{a_i + \rho (n-2) a_i + \rho \sum_{j \ne i}^{n}a_j}{-(n-1)\rho^2 + (n-2)\rho + 1 }$
Now my question is, how should I proceed in trying to prove this formally? 
 A: The matrix of your system can be written in a very nice way: it is 
$$
(1-\rho)I+\rho E,
$$
where $E$ is the matrix with all entries equal to $1$. Since $E=nE$, the algebra generated by $E$ is equal to the span of $I$ and $E$; this suggestse that one can find the inverse explicitly: for $\rho\ne1$ and positive, if we assume that $\frac1{1-\rho}I+tE$ is an inverse for $(1-\rho)I+\rho E$, one can easily find what $t$ is, to get
$$
[(1-\rho)I+\rho E]^{-1}=\frac1{1-\rho}\,I-\frac\rho{(1-\rho)(1+(n-1)\rho)}E.
$$
So the system can be solved explicitly and 
$$
x_j=\frac{1+(n-2)\rho}{(1-\rho)(1+(n-1)\rho)}a_j-\frac\rho{(1-\rho)(1+(n-1)\rho)}\sum_{k\ne j}a_k,
$$
or, equivalently,
$$
\bbox[5px,border:2px solid green]{x_j=\frac{1+(n-1)\rho}{(1-\rho)(1+(n-1)\rho)}a_j-\frac\rho{(1-\rho)(1+(n-1)\rho)}\sum_{k=1}^na_k.}
$$
It is worth noting that the solution above works for any $\rho$ other than $1$ and $-1/(n-1) $, even if negative (or complex, even). 


*

*The case $\rho=1$: the matrix of system is $E$, which is rank-one. The range of $E$ consists of all vectors with equal entries; so in this case the system will have solution (and then, infinitely many) if and only if $a_1=a_2=\ldots=a_n$. 

*The case $\rho=-1/(n-1) $: now the matrix of the system is $\frac n {n-1}I-\frac1 {n-1}E $, which has rank $n-1$. This is a scalar multiple of $I-\frac1n E $, which is the projection onto the orthogonal complement of the subspace of vectors with all entries equal. So here the condition for the existence of a solution is that $\sum_{j=1}^na_j=0$.
