5
$\begingroup$

I'm having trouble understanding this: The free variables are called free because they can take on any value; none of the equations relates any of them to each other. Why can we set free variables to an arbitrary s or t?

Example:

Say we have $A = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0 \end{bmatrix}$ and we are looking for solutions to $Ax = 0$. Here, we have column $1$ and $3$ as our pivot columns (and $x_1$, $x_3$ are pivot varibles) and column $2$ and $4$ as our free columns (and $x_2$, $x_4$ are free variables)[1].

To find a solution vector $x =\begin{bmatrix} x_1 \\ x_2 \\x_3 \\ x_4\end{bmatrix}$, we can set any arbitrary values to $x_2$ and $x_4$ which will lead to some values for $x_1$ and $x_3$, thus giving us our solution.

The confusion: setting arbitrary initial values to the "pivot variables" ($x_1$ and $x_3$) and then finding the values of "free variables" also leads to a solution. Then why do these "free variables" exist? And why are they called "free"?

[1] Page 80, Strang, Gilbert, Linear algebra and its applications, Boston, MA: Brooks/Cole, Cengage Learning (ISBN 978-0-534-42200-4). 496 p. (2007). ZBL1329.15004.

Lecture by Gilbert Strang illustrating this example: https://youtu.be/VqP2tREMvt0?t=322

$\endgroup$
2
  • 6
    $\begingroup$ You seem to be confused. It is kind of difficult to say what the source of confusion is without any source you may have read. $\endgroup$ Commented Mar 22, 2017 at 2:29
  • $\begingroup$ When a system of equations has multiple solutions, we need to describe the set of solutions somehow. Often, the goal when solving equations is to have solutions you can construct. Free variables are generally chosen in a way that lets you construct all the solutions, ideally without repetition. $\endgroup$ Commented Aug 16, 2023 at 22:54

1 Answer 1

Reset to default
8
$\begingroup$

Consider this system of equations: $$ \begin{cases} x+y+z = 2 \\ x-y-z = 0 \end{cases} $$ If we convert it to a matrix, we get $$ \begin{bmatrix} 1 & 1 & 1 & 2 \\ 1 & -1 & -1 & 0 \end{bmatrix} $$ which reduces to $$ \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & \boxed{1} & 1 \end{bmatrix} $$ which indicates that our original system can be rewritten $$ \begin{cases} x &= 1 \\ y + z &= 1 \end{cases} $$ with $z$ being "free".

Now recall what it means for a set of values $(x,y,z)$ to be a solution to our system: It means if we plug in those values for $x$, $y$, and $z$ all the equations are satisfied. Hence $(1,1,0)$ is a solution, whereas $(2,1,1)$ is not (because the second equation becomes false).

When we simplified our system down to $$ \begin{cases} x &= 1 \\ y + z &= 1 \end{cases} $$ we learn that any solution to the system must have $x=1$, but the only other restriction is that $y+z=1$. Any pair of $y$ and $z$ that add to $1$ will work as a solution (alongside $x=1$). How can we find such $(y,z)$ pairs? We can let one of the variables be "free" to be any value, and then solve for the other variable. In this case, we choose $z$ to be free. So if I choose $z=1$, then I can solve for $y=0$. If $z=2$, then $y=-1$. In general, if $z=t$, then $y = 1-t$.

So then we can characterize any solution to our system by saying it must have this form: $$ (x,y,z) = (1,\ 1-t,\ t) $$ where $t$ is free to be anything, and its value determines $y$. Using free variables allows us to write an explicit formula that can compute all the solutions of our system.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.