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

Question

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.

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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, $$ \begin{cases} 2x_2-x_3=4\\ -4x_2=5 \end{cases}\Rightarrow x_2=-\frac{13}{2}, x_3=-\frac{5}{4} $$

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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

Answer 1

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

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