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### A nice symmetric system of linear equations

Suppose we want to solve the following system of linear equations in which $a$ and $b_i$ are real parameters.

$$ax_1 + ax_2 + \dots + ax_{n-1} + (a+1)x_n = b_1$$
$$ax_1 + ax_2 + \dots + (a+1)x_{n-1} + ax_n = b_2$$
$$\dots$$
$$(a+1)x_1 + ax_2 + \dots + ax_{n-1} + ax_n = b_n$$

We can proceed as follows.

Denote

$$T = \sum_{i=1}^n b_i$$
$$S = \sum_{i=1}^n x_i$$

Then the given system of equations can be rewritten as

$$aS + x_n = b_1$$
$$aS + x_{n-1} = b_2$$
$$\dots$$
$$aS + x_2 = b_{n-1}$$
$$aS + x_1 = b_n$$

Then it follows that

$$x_1 - b_n = x_n - b_1$$
$$x_2 - b_{n-1} = x_n - b_1$$
$$\dots$$
$$x_{n-1} - b_2 = x_n - b_1$$
$$x_n - b_1 = x_n - b_1$$

Let's call these equations the differences equations

When we sum up these equations we get that

$S-T = nx_n - nb_1$

So it follows that

$aS-aT = anx_n - anb_1$
$(b_1 - x_n)-aT = anx_n - anb_1$
$b_1(1 + an)-aT = (1 + an)x_n$

Note that in this equation only $x_n$ is unknown. Also, we can get analogical equations for all unknowns $x_i$ ($i=1,2,\dots,n)$. They look like this

$$(1 + an)b_{n+1-i}-aT = (1 + an)x_i \tag{*}$$

We note here that the equations $(*)$ (for $i=1,2,\dots,n$) are just a subsequence of the original system. They are not necessarily equivalent to the original system.

Now based on $(*)$ we can look at the possible cases for the parameters $a$ and $b_i$ (where $i=1,2,\dots,n$).

Case (1)

$1+an \ne 0$ i.e. $a \ne -1/n$

In this case from $(*)$ we find that

$$x_1 = b_{n} - \frac{aT}{1+an}$$
$$x_2 = b_{n-1} - \frac{aT}{1+an}$$
$$\dots$$
$$x_n = b_{1} - \frac{aT}{1+an}$$

A direct check (substituting the $x_i$ unknowns in the original system with these values) shows that these values are indeed a solution. So in this case the original system has a unique solution given by the formulas above.

Case (2)
$1+an = 0$ i.e. $a = -1/n$ and $aT \ne 0$
Note that since $a=-1/n$ the condition $aT \ne 0$ is in fact equivalent to $T \ne 0$.
So this case is the case when $a=-1/n$ and $T \ne 0$.

Obviously the equations $(*)$ cannot be satisfied in this case.
So in this case the system has no solutions.

Case (3)
$1+an = 0$ i.e. $a = -1/n$ and $aT = 0$
Note that since $a=-1/n$ the condition $aT = 0$ is in fact equivalent to $T=0$.
So this case is the case when $a=-1/n$ and $T = 0$.

Now $(*)$ is always satisfied. Going back to the differences equations we see that if we let $x_n=p$ we get

$$x_1 = p - b_1 + b_n$$
$$x_2 = p - b_1 + b_{n-1}$$
$$\dots$$
$$x_{n-1} = p - b_1 + b_2$$
$$x_{n} = p$$

Again, a direct check (substituting the $x_i$ unknowns in the original system with these values) shows that these values are indeed a solution. So in this case the system has infinite number of solutions which depend on a single parameter $x_n = p$.

No other cases are possible logically. So this completes the solution of the given system of linear equations.