Is the following statement true?

If two linear systems have the same solutions then 
the corresponding matrices are row equivalent

I attempted a simple example with the systems

5x + 2y = 23 
3x + 8y = 41 


2x + 3y = 18 
8x + 7y = 52

I was able to row reduce both of their augmented matrices into reduced row echelon form \begin{bmatrix}1&0\\0&1\end{bmatrix} Which I believe indicates row equivalence, however I don't know how I might prove the statement is true for all two same solution equivalent systems. From my research I believe the following Theorem is critical for my understanding of the question enter image description here


HINT.-Draw four straight lines concurring at a point $P = (x, y)$. This point is a solution of the system formed by two of these four lines. But also it is solution of the system formed by the other two lines which can be arbitrary respect of the two first ones. What can you deduce of this?

  • $\begingroup$ I'm not sure I follow your reasoning, would you care to elaborate? $\endgroup$ – 1forrest1 Apr 18 '18 at 16:13
  • $\begingroup$ Let $P(a,b)$ and four points $Q_i(c_i,d_i)$ You have the two systems: $$(1)\begin{cases}\dfrac{y-a}{x-b}=\dfrac{d_i-a}{c_i-b};\space i=1,2\end{cases}$$ $$(2)\begin{cases}\dfrac{y-a}{x-b}=\dfrac{d_i-a}{c_i-b};\space i=3,4\end{cases}$$ Write the matrix for $(1)$ and chose the matrix for $(2)$ as you want because you do have freedom to do it. $\endgroup$ – Piquito Apr 18 '18 at 16:51

Try to reformulate your question with some equations

If $ A_1 x = b_1 $ and $ A_2 x = b_2 $ then $ A_1 $ is row equivalent to $ A_2 $.

Now, what does it mean for $A_1$ to be row equivalent to $ A_2 $? They have the same row space. Or, equivalently, there is a sequence of elementary row transformations that changes one matrix into the other.

Somehow, I think there is something missing in the question...

  • $\begingroup$ What do you reckon is missing in the question? Is there a way to alter the statement to make it more general for all cases? $\endgroup$ – 1forrest1 Apr 18 '18 at 16:12
  • $\begingroup$ The question could be interpreted as "two systems have the same solution for all right-hand sides" (Pick a RHS, then the two solutions are equal). Anyway, Piquito clears it all up. Very nice answer. $\endgroup$ – Robert van de Geijn Apr 18 '18 at 16:46

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