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We have two systems of linear equations $$a_1x_1+b_1x_2 = 0\\a_2x_1 + b_2x_2 = 0\\\qquad\qquad\vdots\\a_nx_1 + b_nx_2 = 0$$ and $$m_1y_1+n_1y_2 = 0\\m_2y_1 + n_2y_2 = 0\\\qquad\qquad\vdots\\m_ny_1 + n_ny_2 = 0$$ It will be enough to prove that a linear combination of the first system is equal to any equation from the second system as the same process can be repeated. Now if we multiply $n$-th equation by some $c_n$ then the linear combination will be $$x_1(a_1c_1 + a_2c_2 + \cdots a_nc_n) +x_2(b_1c_1 + b_2c_2 + \cdots b_nc_n) = m_1y_1 + n_1y_2$$ The two systems having the same solutions mean $x_1=y_1$ and $x_2=y_2$ so $a_1c_1 + a_2c_2 + \cdots a_nc_n = m_1 $ and $b_1c_1 + b_2c_2 + \cdots b_nc_n = n_1$. As $a_1,a_2,\cdots,a_n$ and $b_1,b_2,\cdots,b_n$ are constants we now have two equations with $n$ unknowns ($c_1,c_2,\cdots,c_n$), which is undetermined so there are infinitely many $c_1,c_2,\cdots,c_n$. This shows that as long as the solutions to the systems are the same, at least one (if there are two equations in each system) or infinitely many (more than two equations) ways to linearly combine the system, which in turn proves they are equivalent.
Is my reasoning correct, if not is there any way to correct it? Are there any simpler/more elegant proofs to this? In my textbook, definition of equivalency is this: If each equation in a system of equations is the linear combination of the equations in the other system, then the systems are equivalent.