Proving a system of n linear equations has only one solution I have been given the system:
$-x_1 + 2x_2 + ... + 2x_{n-1} + 2x_n = 1$
$2x_1 - x_2 + ... 2x_{n-1} + 2x_n = 2$
$.$
$.$
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$2x_1 + 2x_2 + ... + 2x_{n-1} - x_n = n$
And the assignment to prove that it has only one solution. I am aware of the existence of discriminants and their role of solving system of linear equations, although I have no idea on how to use that to prove this system has only one solution. (And I currently lack materials on infinite determinants and recurrent relations, any links would be appreciated).
Could anyone provide suggestions or hints on how to prove this system has exactly one solution?
 A: Your equation is of the form $Ax = b$ where $$A = \begin{bmatrix}-1 & 2 & 2 & 2 & 2 & \cdots & 2 & 2\\ 2 & - 1 & 2 & 2 & 2 &\cdots & 2 & 2\\ 2 & 2 & -1 & 2 & 2 &\cdots & 2 & 2\\ 2 & 2 & 2 & -1 & 2 &\cdots & 2 & 2\\ 2 & 2 & 2 & 2 & -1 &\cdots & 2 & 2\\ \vdots & \vdots & \vdots & \vdots & \vdots &\ddots & \vdots & \vdots\\ 2 & 2 & 2 & 2 & 2 &\cdots & -1 & 2\\ 2 & 2 & 2 & 2 & 2 &\cdots & 2 & -1 \end{bmatrix}$$
We then have
$$A = -3I_{n \times n} + 2 \begin{bmatrix} 1\\ 1\\ 1\\ 1\\ 1\\ \vdots\\ 1\\ 1\end{bmatrix} \begin{bmatrix} 1& 1& 1& 1& 1& \cdots& 1& 1\end{bmatrix}$$
By Sherman-Morrison, we have
$$A^{-1} = -\dfrac13 I_{n \times n}-\dfrac2{3(3-2n)}\begin{bmatrix} 1\\ 1\\ 1\\ 1\\ 1\\ \vdots\\ 1\\ 1\end{bmatrix}\begin{bmatrix} 1& 1& 1& 1& 1& \cdots& 1& 1\end{bmatrix}$$
Clearly, $A^{-1}$ exists for all $n$.
A: Symmetrize!  Add up all the equations. You will find an expression for $x_1+\cdots+x_n$. 
Now it is easy to write down the solution.   
Details: Let $s=x_1+\cdots+x_n$. Adding up the equations, we get 
$$(2n-3)s=1+2+\cdots+n=\frac{n(n+1)}{2}.$$ 
The $k$-th equation can be written as $2s-3x_k=k$. It follows that
$$x_k=\frac{1}{3}\left(\frac{n(n+1)}{2n-3}-k\right).$$
A: The system can  be written in matrix form $Ax=b$ where $A$ is the matrix 
$$
A=2J-3I
$$
with J the matrix with all entries 1 and $I$ the identity matrix.
Show that the matrix $A$ is  invertible (equivalently all eigenvalues of $A$ are non zero) therefore the system has the unique solution $A^{-1}b$.
This answer is relevant. 
A: There is a very nice way to see this using the algebra of circulant matrices. Define
$$\rm X = \left[\begin{array}{}
0 & 1 & 0 & 0 &\cdots& 0\\
0 & 0 & 1 & 0 &\cdots& 0\\
0 & 0 & 0 & 1 &\cdots& 0\\
 & & & & \cdots &\\
0 & 0 & 0 & 0 &\cdots& 1\\
1 & 0 & 0 & 0 &\cdots& 0\end{array}\right]\ \Rightarrow\  
X^2\! = \left[\begin{array}{}
0 & 0 & 1 & 0 &\cdots& 0\\
0 & 0 & 0 & 1 &\cdots& 0\\
 & & & & \cdots &\\
0 & 0 & 0 & 0 &\cdots& 1\\
1 & 0 & 0 & 0 &\cdots& 0\\
0 & 1 & 0 & 0 &\cdots& 0\\\end{array}\right]\ \Rightarrow\ \cdots\ \Rightarrow\ X^n = 1$$
Your circulant matrix is $\rm\:M = f(X)\, =\, -1\!+\!2X\!+\!2X^2\!+\,\cdots\,+2X^{n-1}.\,$ One easily checks (see below) that $\rm\:f(x)\:$ is coprime to $\rm\:x^n-1,\:$ so by the Euclidean algorithm there is a Bezout identity
$$\rm a(x)\ f(x) + b(x)\ (x^n\!-1)\, =\ 1$$
Since $\rm\:X^n=1,\:$ evaluating above at $\rm\: x = X\:$ yields $\rm\:a(X)\,f(X)\, =\, 1,\:$ so $\rm\ M = f(X)\:$ is invertible.
For coprimality, let $\rm\:g = 1\!+\!x\!+\cdots+\!x^{n-1}.\:$ Then $\rm\:f = 2g\!-\!3\:$ is coprime to $\rm\:g\:$ since $\rm\:2g-f = 3,\:$ and $\rm\:f\:$ is coprime to $\rm\:x\!-\!1\:$ (else $\rm\:x\!-\!1\mid f\:\Rightarrow\:0=f(1) = 2n\!-\!3).\:$ Therefore, by Euclid's Lemma, we deduce that $\rm\:f\:$ coprime to both $\rm\:g\:$ and $\rm\,x\!-\!1\:\Rightarrow\:f\:$ coprime to their product $\rm\:(x\!-\!1)g = x^n\!-\!1.$
A: Note that  the system can be written as $(2J-3I)x=(1,2,\ldots ,n)^t$, where $x=(x_1,\ldots ,x_n)$, $J$ is the matrix with all entries $1$ and $I$ is the identity matrix. Now the eigen values of $J$ are $n$ with mutiplicity $1$ and $0$ with multiplicity $n-1$ (this follows from the fact that rank of $J$ is $1$, which is also the number of non-zero eigen values, and the sum of the eigen values is $n$). Hence the eigen values of $2J-3I$ are $2n-3$ with mutiplicity $1$ and $-3$ with multiplicity $n-1$, so all the eigen values are non-zero and hence the rank of $2J-3I$ is $n$. Therefore the system has a unique soloution.
