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 http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum