TITLE: Prove: if two homogeneous systems of linear equations in two unknowns have the same solutions, then they are equivalent.
QUESTION [5 upvotes]: From Exercise 6 of Sec 1.2 of Linear Algebra by K.Hoffman and R.Kunze.
Equivalence is defined as follows:

Two systems of linear equations are equivalent if each equation in
  each system is a linear combination of the equations in the other
  system.

I did some research, but most answers I found on math.stackexchange.com either use some serious math like ranks or null spaces, or are merely "intuitive". I'm curious how to formalize the proof, with concepts previously defined.  
Thx.
More specifically --
$A_{11}x_1+A_{12}x_2=0$, $A_{21}x_1+A_{22}x_2=0$;
$B_{11}x_1+B_{12}x_2=0$, $B_{21}x_1+B_{22}x_2=0$;
We are asked to find $c_{11},c_{12},c_{21},c_{22}$ such that:
$A_{11}c_{11}+A_{21}c_{12}=B_{11}$;
$A_{12}c_{11}+A_{22}c_{12}=B_{12}$;
$A_{11}c_{21}+A_{21}c_{22}=B_{21}$;
$A_{12}c_{21}+A_{22}c_{22}=B_{22}$; (Sorry about the mess...I don't know how to type systems of equations...)
I can't convince myself in formal language the existence of such $c$s.
Since this problem appears at the beginning of the book, I assume there's a rather rookie proof. I suppose I have to express the $c$s in terms of the given quantities. But how?

REPLY [5 votes]: This time I finally get it done without using the words 'determinant', 'rank' and so on. Special thanks to @orangeskid, and thank you all.

6. Prove that if two homogeneous systems of linear equations in two unknowns have the same solutions, then they are equivalent.
[Proof] A proof by contraction.
Assuming the negation of the conclution, suppose systems $\bf{A}$ and $\bf{B}$ are not equivalent. Accoording to the symmetry of the problem, this is the same as asserting that the first equation of $\bf{A}$: $$A_1x_1+A_2x_2=0$$ is not a linear combination of the equations of $\bf{B}$: $$B_{i1}x_1+B_{i2}x_2=0, (i=1,2,...,k).$$
Case 1. $A_1\neq0,(\forall i)\ B_{i1}=0$. In this case, $(x_1,x_2)=(x_1\neq0,0)$ solves $\bf{B}$ but not $\bf{A}$.
Case 2. $A_1=0,A_2\neq0,(\forall i)\ B_{i1}=B_{i2}=0$. In this case, $\bf{B}$ yields $(x_1,x_2)=(x_1\in\mathbb{R},x_2\in\mathbb{R})$ while $\bf{A}$ yields $(x_1,x_2)=(x_1\in\mathbb{R},0)$.
Case 3. $A_1\neq0,(\exists i)\ B_{i1}\neq0,(\exists j)\ B_{j2}\neq0$. In this case, there exist $c_1,c_2,...,c_k$ such that $$A_1=c_1B_{11}+c_2B_{21}+...+c_kB_{k1};$$ and $$A_2\neq c_1B_{12}+c_2B_{22}+...+c_kB_{k2}.$$
Noticing that $A_1x_1+(c_1B_{12}+c_2B_{22}+...+c_kB_{k2})x_2=0$ and $A_1x_1+A_2x_2=0$ gives $$(A_2-(c_1B_{12}+c_2B_{22}+...+c_kB_{k2}))x_2=0$$ which means $x_2=0$. A closer examination of the $j$th equation of $\bf{B}$:$$B_{j1}x_1+B_{j2}x_2=0$$ suggests that for both systems to have exactly the same solutions, $B_{j1}x_1=0$ must happen to be true. However, realizing that $B_{j2}\neq0$, we can adjust $c_j$ so that $$A_2= c_1B_{12}+c_2B_{22}+...+c_kB_{k2}$$ also holds.
Thus, the statement is proved.