How to show inconsistency of a linear system of equations? [closed]

Question

Consider the (real) linear system of equations $A\mathbf{x}=\mathbf{c}$ of size $N$ as $$ \begin{bmatrix} a_{N}-a_{2}& a_{2}& 0 &\dots& 0 & -a_{N} \\-a_{1} & a_{1}-a_{3}&a_{3}& 0 & \dots&0\\ 0 & -a_{2} & a_{2}-a_{4}&a_{4}&0 & \dots \\\vdots&\ddots&\ddots&\ddots&\ddots\\ a_1& 0& \dots& 0&-a_{N-1}&a_{N-1}-a_1 \end{bmatrix}\begin{bmatrix} x_1\\x_2\\x_3\\\vdots\\x_N\end{bmatrix}=\begin{bmatrix} c\\c\\c\\\vdots\\c\end{bmatrix}. $$ So, the vector $\mathbf{c}$ is constant and $A$ has a periodic structure and its kernel (at least) contains constant vectors (since the sum of columns is zero).

How can one prove that

1. $rank(A)=N-1$
2. The system is inconsistent, unless $c=0$.

I suppose that these claims are valid. But any counter-example is also very appreciated.

Answer 1

For a counterexample to (1) with all $a_i$ nonzero, consider the $N=4$ case with $a_3 = a_1$, noting that $$ \left[\matrix{a_4-a_2 & a_2 & 0 & -a_4\cr -a_1 & 0 & a_1 & 0\cr 0 & -a_2 & a_2-a_4 & a_4\cr a_1 & 0 & -a_1 & 0\cr}\right] \left[\matrix{0 \cr a_4 \cr 0 \cr a_2}\right] = 0$$

For a counterexample to (2) with all $a_i$ nonzero, consider the $N=3$ case with $a_1=4, a_2=a_3=1$: $$ \left[\matrix{0 & 1 & -1\cr -4 & 3 & 1\cr 4 & -1 & -3\cr}\right] \left[ \matrix{1 \cr 2\cr 0\cr}\right] = \left[\matrix{2\cr 2\cr 2\cr}\right]$$