It is clear that if the system of linear equations
$$ \left\{ \begin{array}{c} x_1-x_3=c_1 \\ x_2-x_1=c_2 \\ x_3-x_2=c_3 \end{array} \right. $$
is solvable, then we have $c_1+c_2+c_3=0$.
How could we prove for the backward direction? That is, how could we prove that if $c_1+c_2+c_3=0$, then the above system of linear equations must be solvable without any corner cases?
The augmented form of our system is $$ \left[\begin{array}{rrr|r} 1 & 0 & -1 & c_{1} \\ -1 & 1 & 0 & c_{2} \\ 0 & -1 & 1 & c_{3} \end{array}\right] $$ Row-reducing gives \begin{align*} \left[\begin{array}{rrr|r} 1 & 0 & -1 & c_{1} \\ -1 & 1 & 0 & c_{2} \\ 0 & -1 & 1 & c_{3} \end{array}\right] \xrightarrow{R_2+R_1\to R_2}\left[\begin{array}{rrr|r} 1 & 0 & -1 & c_{1} \\ 0 & 1 & -1 & c_{1} + c_{2} \\ 0 & -1 & 1 & c_{3} \end{array}\right] \\ \xrightarrow{R_3+R_2\to R_3}\left[\begin{array}{rrr|r} \fbox{1} & 0 & -1 & c_{1} \\ 0 & \fbox{1} & -1 & c_{1} + c_{2} \\ 0 & 0 & 0 & \fbox{$c_{1} + c_{2} + c_{3}$} \end{array}\right] \end{align*} The first two boxed positions are guaranteed to be pivots in the reduced form of our system. The third boxed position is a pivot position if and only if $c_1+c_2+c_3\neq 0$. This third position is in the augmented column, so our system is solvable if and only if $c_1+c_2+c_3=0$.
