The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Leading Variables, Free Variables, and the Coefficient Zero

Date: 01/23/2010 at 15:58:29
From: Jon
Subject: Leading Variable

In linear algebra, I was taught that there were 2 kinds of variables: 
leading and free.  Leading variables have a leading 1 in front of 
them, and free variables do not.

If a variable has a 0 in front of it, does that make the variable 
leading or free?

Date: 01/23/2010 at 19:35:19
From: Doctor Vogler
Subject: Re: Leading Variable

Hi Jon,

Thanks for writing to Dr. Math.

These terms only apply *after* you triangulate your system of 
equations (or finish with Gaussian elimination, depending on which 
term your textbook/teacher used), so that you end up with a system of 
linear equations sort of like the following:

  x + 5y + 7z + 13w = 0
       y +  z +  2w = 0
                  w = 0

Note that it is actually not the coefficients that are important.  
The coefficients of the "leading" variables are all 1 only because 
you divided those equations by the coefficients of those variables.

The important thing is that each equation (read from the bottom up, or
from the one with the fewest variables to the most) adds one extra 
"leading" variable and all other new variables are "free."  
Because you have already done your Gaussian elimination, every 
equation adds at least one extra variable.  If an equation adds two or 
more new variables, then it really doesn't matter which one is 
"leading" and which ones are "free"; but it is 
customary to write the leading one first and the free ones next.

For example, the equations above could have been written as

  x + 7z + 5y + 13w = 0
       z +  y +  2w = 0
                  w = 0

and if the coefficient of z in the second equation hadn't been 1, 
then we could have divided the equation by its coefficient in order 
to make it 1.

Now, if the coefficient is 0, then what happens?  Well, if it is the 
coefficient of a variable that was already used in an equation below, 
then it doesn't matter; that variable was determined to be either 
"leading" or "free" by the lowest equation it appeared in.  
Otherwise, it's not a new variable, since it doesn't appear in this 
equation; so it will be determined "leading" or "free" in some other 
equation where it does appear.

But if it doesn't appear in *any* equation, then it's free.

The real difference between "free" variables and "leading" variables 
is that the free variables can be anything, and the leading variables 
are determined by solving the equations.  So in the example I gave 
above, w = 0 is determined.  Then y = -z, and x = -2z, so all 
solutions have the form

  (x, y, z, w) = (-2, -1, 1, 0)*z.

If we had made y the free variable, then we would have gotten z = -y, 
x = 2y, and

  (x, y, z, w) = (2, 1, -1, 0)*y,

which defines the same subspace of solutions.

Does that make sense?

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum 
Associated Topics:
High School Linear Algebra

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.