# Clarifying the notion of basic/free variables in a system of equations

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

• Note that the order of the columns is arbitrary in a way. Feb 5 at 13:22

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

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

• 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.
– TEX
Feb 5 at 13:27
• @TEX They can all be free, that's the point. It doesn't matter. Feb 5 at 13:28
• 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.
– TEX
Feb 5 at 13:29
• @TEX you just use $x_3$ and get the full solution set. Why do more? Feb 5 at 13:31
• 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.
– TEX
Feb 5 at 13:32