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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?'

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    $\begingroup$ The set of all solutions to this linear system of equations is $S = \{ \begin{bmatrix} -2r -s - t \\ r \\ s- t + 1 \\ 2 - 2s + t \\ s \\ t \end{bmatrix} \mid r, s, t \in \mathbb R \}$. $\endgroup$
    – littleO
    Jan 1, 2022 at 4:27
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    $\begingroup$ 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. $\endgroup$
    – user24142
    Jan 1, 2022 at 4:42
  • $\begingroup$ 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. $\endgroup$ Jan 2, 2022 at 3:39

1 Answer 1

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

We can do this because after reduce this matrix to its reduced row echelon form we can see that there are 3 variables what depends on the other 3 variables(the free variables). Technically you can choose the variables that you want for free variables. In this case the author chose the above variables.

This variables are related with the dimension of the Null space of the matrix.

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  • $\begingroup$ This is not an answer to the question. If I'm not mistaken, the author is asking how to "physically" interpret the variables, not what they are. $\endgroup$ Jan 2, 2022 at 3:34
  • $\begingroup$ I edited my comment. Thank you $\endgroup$
    – AngelGuale
    Jan 2, 2022 at 3:55

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