<s>[INST] Prove the condition for system of linear equations to be solvable 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?
 [/INST] 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 $c_1+c_2+c_3=0$.
</s>