1
$\begingroup$

I have a fundamental confusion with regards to the notion of free/basic variables.

Consider the following linear system $$ \begin{pmatrix} 1 & 2 & -1 \\ 2 & -4 & 0 \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} 4\\ 5 \end{pmatrix} $$ This system has augmented matrix $$ \begin{pmatrix} 1 & 2 & -1 & 4 \\ 2 & -4 & 0 & 5 \end{pmatrix} $$ which row reduces to $$ \begin{pmatrix} 1 & 2 & -1 & 4 \\ 0 & -8 & 2 & -3 \end{pmatrix} $$ The last matrix is in echelon form. Columns 1 and 2 are pivot columns, so $x_1,x_2$ are basic variables. The third column is not a pivot column, so $x_3$ is a free variable. Finally, the last column is not a pivot column, so the system is consistent.


The above analysis tells me that $x_3$ is a free variable. I interpret this as saying "$x_1,x_2$ can be expressed as a function of $x_3$, while $x_3$ can be set equal to any number". For example, I can set $x_3=200$. In turn, $$ \begin{cases} x_1+2x_2-200=4 \\ 2x_1-4x_2=5 \end{cases}\Rightarrow x_2=\frac{403}{8}, x_1=\frac{413}{4} $$

Nevertheless, just by doing naive computations, I realised that we could also "express $x_1$, $x_3$ as a function of $x_2$, while setting $x_2$ equal to any number". For example, I can set $x_2=7$. In turn, $$ \begin{cases} x_1+14-x_3=4\\ 2x_1-28=5 \end{cases}\Rightarrow x_1=\frac{33}{2}, x_3=\frac{53}{2} $$ Alternatively, we could also "express $x_2$, $x_3$ as a function of $x_1$, while setting $x_1$ equal to any number". For example, I can set $x_1=0$. In turn, $$ \begin{cases} 2x_2-x_3=4\\ -4x_2=5 \end{cases}\Rightarrow x_2=-\frac{13}{2}, x_3=-\frac{5}{4} $$


Hence, my question: what is the relation between

  • the fact that $x_3$ is a free variable and $x_1,x_2$ are basic variables (from the row-reduced matrix)

  • and the fact that in the system above I am free to fix the value of anyone among $x_1,x_2,x_3$ and I will always be able to solve for the other two unrestricted variables

$\endgroup$
1
  • $\begingroup$ Note that the order of the columns is arbitrary in a way. $\endgroup$ Feb 5 at 13:22
0
$\begingroup$

You've got three column vectors, and any pair of them are independent and thus span $\Bbb R^2$. The solution set is one-dimensional (affine) (aka a line) so of the form $v+tw$, where $v,w$ are some vectors and $t$ is a free real variable. We can choose this to be $x_3$, or any of the other variables if you prefer.

You can then transform the equations to $$\begin{align}x_1+2x_2 &= 4+x_3\\ x_1 - 4x_2 &= 5\end{align}$$

and solve these in terms of $x_3$ uniquely. We get (add twice the first equation to the second and we get $3x_1 = 13+2x_3$ so $x_1 = \frac{1}{3}(13+2x_3)$, and subtraction the equations: $6x_2 = x_3-1$ and hence $x_2 = \frac{1}{6}(x_3 - 1)$ (we could also do another elimination step on this new system) so we can write the solution line as

$$\begin{pmatrix} \frac{13}{3}\\ -\frac16\end{pmatrix} + x_3 \begin{pmatrix} \frac{2}{3}\\ \frac16\end{pmatrix}$$

$\endgroup$
8
  • $\begingroup$ Thanks. (1) Can we say that in this specific case "$x_3$ is a free variable, $x_1,x_2$ are basic variables" or, equivalently, "$x_1$ is a free variable, $x_3,x_2$ are basic variables" or, equivalently, "$x_2$ is a free variable, $x_1,x_3$ are basic variables"? (2) Is there any way to realise from the echelon form that we are "free" to permute the columns? This somehow relates to my other question math.stackexchange.com/questions/4012647/… if you can help. $\endgroup$
    – TEX
    Feb 5 at 13:27
  • $\begingroup$ @TEX They can all be free, that's the point. It doesn't matter. $\endgroup$ Feb 5 at 13:28
  • $\begingroup$ Thanks. My doubt is: is there a formal way to understand from the echelon form (or any other form) when any variable can be set as the free variable? This is not always the case. $\endgroup$
    – TEX
    Feb 5 at 13:29
  • $\begingroup$ @TEX you just use $x_3$ and get the full solution set. Why do more? $\endgroup$ Feb 5 at 13:31
  • $\begingroup$ Because this relates to a more general result that I need to prove. I want to understand when I'm free to set the free variables and how I can detect that. $\endgroup$
    – TEX
    Feb 5 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.