We have the following LES $I$ \begin{align} 3x_1-2x_2+4x_3-x_4&=1\\ -\frac43x_2-\frac{13}3x_3+\frac13x_4&=-\frac{19}3 \end{align} with the solution set $\mathcal{L}_{I}$, which contains all solutions of $I$. Let $\mathcal{L}_2$ be some other solution, which is defined as the following: $$\mathcal{L}_2=\{(x_1,x_2,x_3,x_4)\mid(x_1,x_2,x_3,x_4)=(3,3,1,6)+\lambda_1(3,3,-1,-1)+\lambda_2(2,3,1,4),\;\lambda_1,\lambda_2\in\mathbb{R}\}$$

We want to determine whether $\mathcal{L}_2\subseteq \mathcal{L}_{I}$ or not.

Therefore, we will insert the tuples into $I$. If all equations hold for the inserted values $x_1,\cdots,x_4$, the solution set $\mathcal{L}_2$ is a subset of all solutions.: First equation holds. Moving on to the next one. We can simplify the term $$-\frac{4}3(3+3\lambda_1+3\lambda_2)-\frac{13}{3}(1-\lambda_1+\lambda_2)+\frac13(6-\lambda_1+4\lambda_2)=-\frac{19}3$$ to $$-\frac{19}3-7\lambda_2=-\frac{19}3$$ How do we go on now? Can we choose $\lambda_2=0$ and assume $\mathcal{L}_2\subseteq \mathcal{L}_I$ because all equations hold for $\lambda_2=0$ or is this in fact the sign for $\mathcal{L}_2\nsubseteq \mathcal{L}_I$?

  • $\begingroup$ What do you mean by "THE solution"? The "solution set'' for a given problem contains all solutions to the problem. $\endgroup$ – user247327 Apr 12 at 16:57
  • $\begingroup$ I wanted to express, that $\mathcal{L}_I$ is the solution set of the linear system $I$ while $\mathcal{L}_2$ is just one set which still needs to be analysed in order to know if $\mathcal{L}_2$ is a subset of the solution set of $I$ $\endgroup$ – Doesbaddel Apr 12 at 17:07
  • $\begingroup$ @user247327 I fixed some typos and changed the overall style a bit. I hope you understand my question better now. $\endgroup$ – Doesbaddel Apr 12 at 19:56

If you choose $\lambda_1=0,\lambda_2=-1$ you get $x \in \mathcal{L}_2$ of the form $(1, 0, 0, 2)$, which does not solve the second equation, therefore $\mathcal{L}_2\nsubseteq \mathcal{L}_I$.

  • $\begingroup$ Is it a necessary condition that if you choose fixed $\lambda_1,\lambda_2$ that both equations must hold in order to be a subset of all solutions $\mathcal{L}$? If we choose $\lambda_2=0,\lambda_1\in\mathbb{R}$ both equations hold for example. $\endgroup$ – Doesbaddel Apr 14 at 11:19
  • $\begingroup$ If we choose $\lambda_2=0,\lambda_1\in\mathbb{R}$ both equations hold for example. Therefore, $\mathcal{L}_2\subseteq\mathcal{L}$? I don't understand that $\endgroup$ – Doesbaddel Apr 14 at 11:25
  • $\begingroup$ $\mathcal{L}_I$ is a set of all solutions of the system of equations, which you have defined above, right? So any $x \in \mathcal{L}_2$ must also a solution of this system (i.e. of both equations) in order $\mathcal{L}_2 \subseteq \mathcal{L}_I$ to be true. $\endgroup$ – Vítězslav Štembera Apr 15 at 9:28
  • $\begingroup$ Yeah, that's right. That's why I can't pick a fixed $\lambda$ because that wouldn't be any $x\in\mathcal{L}_2$? $\endgroup$ – Doesbaddel Apr 15 at 13:41
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    $\begingroup$ If you want to prove $\mathcal{L}_2\subseteq \mathcal{L}_I$ you have to prove it for all $x \in \mathcal{L}_2$, if you want to prove $\mathcal{L}_2\nsubseteq \mathcal{L}_I$ you can pick only one $x \in \mathcal{L}_2$ and show that $x$ does not belong to $\mathcal{L}_I$. $\endgroup$ – Vítězslav Štembera Apr 15 at 14:05

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