# Why can we use unrelated variables in solution vectors?

Given the system:

$$x_1$$ + $$2x_2$$ + $$x_5$$ + $$x_6$$ = $$0$$

$$x_1$$ + $$2x_2$$ + $$2x_3$$ - $$x_5$$ + $$x_6$$ = $$2$$

$$x_4$$ + $$2x_5$$ - $$x_6$$ = $$2$$

I get the solution vector $$\begin{pmatrix} -2r-s-t\\ r\\ s-t+1\\ 2-2s+t\\ s\\ t \end{pmatrix}$$ for $$x_1, x_2, x_3, x_4, x_5, x_6$$ corresponding to each row in the solution vector respectively.

My question is how should I interpret r, s, and t. If I were to write the reduced row echelon matrix for example, I could conclude that $$2x_2 = -x_1 - x_5 + x_6$$. Yet for simplification, we call $$x_2$$ 'r.' How are we able to perfectly encapsulate the behavior or $$x_2$$ with just the variable 'r' even though $$x_2$$ could depend on other variables in the matrix. Likewise, how can we assume the same for $$x_5$$ and $$x_6$$. Or is it simply that r, s, and t are equivalent to saying 'for all real numbers?'

• The set of all solutions to this linear system of equations is $S = \{ \begin{bmatrix} -2r -s - t \\ r \\ s- t + 1 \\ 2 - 2s + t \\ s \\ t \end{bmatrix} \mid r, s, t \in \mathbb R \}$. Jan 1, 2022 at 4:27
• One way to start to tease out the complexity you're correctly perceiving is to rewrite your solution as $\left\{ (0\ 0\ 1\ 2\ 0\ 0)^T + r(-2\ 1\ 0\ 0\ 0\ 0)^T + s(-1\ 0\ 1\ -2\ 1\ 0)^T + t(-1\ 0\ -1\ 1\ 0\ 1)^T | r, s, t \in \mathbb{R}\right\}$ which shows that the solutions are a three dimensional hyperplane sitting in six dimensional space. Jan 1, 2022 at 4:42
• As @user24142 suggested, you are trying to imagine something rather complex. I would suggest you look at very simple problems where the relationships between the variables can be seen without much effort. Then just apply the same logic to the more complex problems. Jan 2, 2022 at 3:39

In this case, after obtain rref of the matrix we choose $$r=x_2, s=x_5, t=x_6$$