Thus the theorem: Equivalent system of linear equations have exactly the same solution.

Given the follow systems:

System A

$$x_1 - x_2 = 0$$ $$2x_1 + x_2 = 0$$

System B

$$3x_1 + x_2 = 0$$ $$x_1 + x_2 = 0$$

I want to write each equation of the system in terms of the other system. Then, I know that a solution for the first equation of the second system is:

$$3x_1 + x_2 = \frac{1}{3}(x_1-x_2) + \frac{4}{3}(2x_1 + x_2)$$

How do I get that result? The book I'm using tells us to use the Eliminated Method but is hard to figure out what the operations to apply between the equations to prove that one equation is a linear combination of the equations of the other system.

There is a another way?

  • 2
    $\begingroup$ System $B$ has no equations so far. System $A$ has the solutions $x_1=x_2=0$, provided $3\neq 0$. So no need to write it in terms of the other. Just compute the solutions for $B$. So both systems have only the trivial solution, hence exactly the same one. $\endgroup$ Jan 13 '19 at 20:57
  • $\begingroup$ @DietrichBurde Sorry, I wrote wrong the system. $\endgroup$
    – Arduin
    Jan 13 '19 at 21:28

I presume your second system was meant to be $$3x_1 + x_2 = 0\\ x_1 + x_2 = 0\ ,$$ and you want to know a systematic way of determining the coefficients $\frac{1}{3} \mbox{ and }\frac{4}{3}$ in your equation $$3x_1 + x_2 = \frac{1}{3}(x_1-x_2) + \frac{4}{3}(2x_1 + x_2)\ ,$$ and corresponding coefficients in a similar expression for your fourth equation in terms of the first two. But it's not at all clear to me why you would want to do this, and if that isn't your question, then you'll need to state it more clearly.

To obtain the coefficients in your expression, simply treat them as unknowns $\alpha_1$ and $\alpha_2$ , for which you can obtain two equations from the expression they have to satisfy. You want your third equation to be expressed as $\alpha_1$ times the first plus $\alpha_2$ times the second: $$3x_1 + x_2 = \alpha_1 (x_1-x_2) + \alpha_2(2x_1 + x_2)\ .$$ Now rearrange the equation on the left so you have $x_1$ and $x_2$ multiplied by a single coefficient: $$3x_1 + x_2 = (\alpha_1 + 2\alpha_2)\, x_1 + (-\alpha_1 + \alpha_2)\, x_2\ .$$ For this equation to be satisfied, you now want the coefficients of $x_1$ and $x_2$ on the left to have the same value as they do on the right. That is: \begin{eqnarray} 3 &=& \alpha_1 + 2\alpha_2 \ \mbox{, and}\\ 1 &=& -\alpha_1 +\alpha_2\ , \end{eqnarray} which you can now solve by the method of (Gaussian) elimination.


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