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.


marked as duplicate by amd, José Carlos Santos linear-algebra Jul 10 '18 at 20:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ To my mind, two systems of equations having the same solutions is the definition of equivalence. What is your definition? $\endgroup$ – amd Jul 10 '18 at 18:44
  • $\begingroup$ @the post already explains the definition of equivalency according to the book I am using $\endgroup$ – user500668 Jul 11 '18 at 7:44
  • $\begingroup$ @José Carlos Santos I don't see how this is a duplicate, while I know there are already questions about this particular problem, what I asked is if my proof is correct, if not how to correct it, and maybe point out some better solutions $\endgroup$ – user500668 Jul 11 '18 at 7:54