<s>[INST] Proof on 3 equations crossing at one point $(a+b+c=0)$ WITHOUT determinant of the matrix So, as in title, I have to prove that if there are 3 linear equations:

*

*$ax + by + c = 0$

*$bx + cy + a = 0$

*$cx + ay + b = 0$
and those cross at one point, then $a + b + c = 0$. I know that's true because it is easy to do the proof using matrix determinants. However, in that case I was told that it is possible to do the proof using only linear algebra in matrix. I have no idea how to do it and I'm curious.
Thank you in advance for your effort.
edit:
I think I know the answer. You can correct me if I'm wrong. If we have a matrix of those equations it looks like this: r_1 = [a b c], r_2 =[b c a], r_3 = [c a b]. If we add r_1 and r_2 to r_3 we get a follownig equation: (a+b+c)x + (b+c+a)y + (c+a+b) = 0. The only way this equaton can be true is if a+b+c=0.
 [/INST] Let $A$ be the matrix of the coefficients and $A|B$ the augmented matrix.
Overdeterminate system like this have solution only if $\text{rank}(A)=\text{rank}(A|B)$.
As $\text{rank}(A)=2$ must be $\text{rank}(A|B)=2$ which means that
$$\det(A|B)=\left| 
\begin{array}{ccc}
 a & b & -c \\
 b & c & -a \\
 c & a & -b \\
\end{array}
\right|=0$$
which is $$a^3-3 a b c+b^3+c^3=0\to (a+b+c) \left(a^2-a b-a c+b^2-b c+c^2\right)=0$$
satisfied when $a+b+c=0$.
$$.......................................$$
Alternatively
$$
\begin{cases}
 ax + by + c = 0\\
bx + cy + a = 0\\
cx + ay + b = 0\\
\end{cases}
$$
Solve the first two equations
$$x=\frac{c^2-a b}{b^2-a c};\;y=\frac{a^2-b c}{b^2-a c}$$
plug in the third
$$\frac{a^3-3 a b c+b^3+c^3}{b^2-a c}=0\to a^3-3 a b c+b^3+c^3=0\\ (a+b+c) \left(a^2-a b-a c+b^2-b c+c^2\right)=0\to a+b+c=0$$
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