What do the leading variables and free variables in a matrix mean? I have the system below and am trying to understand which are which. I searched a lot for this, please help me ! $$w + x + y + z = 6 \qquad w + y + z = 4 \qquad w + y = 2$$

Have you seen my answer? Any further questions? – Gerry Myerson Mar 22 '14 at 22:42
2 Answers
The terms "leading variable" and "free variable" are usually defined for the matrix representing a system, and only when the matrix is in rowechelon form.
The augmented matrix for your system is $$ \left( \begin{array}{c c c cc} 1&1&1&1&6\\ 1&0&1&1&4\\ 1&0&1&0&2\\ \end{array}\right) $$ Notice that each column corresponds to a variable.
Do you know how to use elementary row operations to bring this matrix to rowechelon form?
In rowechelon form, the first nonzero entry in each row (if there is one) is a $1$, and the column it is in corresponds to a leading variable; the columns that don't have that kind of $1$ correspond to the free variables. Essentially, columns that don't have a leading variable, have a free variable.
So for example if a rowechelon form is given by $$ \left( \begin{array}{c c c c cc} 1&2&3&4&5&6\\ 0&0&1&7&8&9\\ 0&0&0&1&\pi&\sqrt2\\ \end{array}\right) $$ then the 1st, 3rd, and 4th variables are leading variables while the 2nd and 5th variables are free variables.
 111
 4
 168,500
 12
 196
 359

1Thats good when thought of in the context of matrix representation. But what they mean in (homogeneous/nonhomogeneous) linear systems context: Is it just that **the leading variables are bound/constrained by the given equations and free variables are parametric/not bound by the equations**? Or are there more alternative interpretations / significance / facts / observations associated with it in the context of linear system? – Maha Feb 05 '15 at 14:55

3@awe, that would depend on how the systems are written. If they are written as, say, $a=3b7d+11,c=12d+44, e=92$, then you can say $a,c,e$ are leading and $b,d$ are free. But if they are written $w+x+y+z=6,w+y+z=4,w+y=2$ then there is no sensible way to distinguish some of the variables as leading and others as free. The distinction only arises during the process of solving the system. – Gerry Myerson Feb 05 '15 at 22:37

`the first nonzero entry in each row (if there is one) is a 1` This condition is not really needed for echelon form. It does does not have to be 1 and it is still echelon form. – Nasser Sep 29 '20 at 03:01
If a set of linear equations can be expressed as let's say
a = 3x + 4y + 5z  12
b = 2x + 8y + z  11
c = 9x + 7y z  15
where
 The left hand variables don't appear on the right side and vice versa.
 On the left side, there is only one variable.
Then,
x, y and z can take values of any combination and are called free variables.
a, b and c are dependent on the above free variables (x, y and z) and cannot be any combination. a, b and c are called pivot or leading variables.
One more thing, depending upon how we form the above equations the pivot and free variable might be changed. For ex.,
a = x + y + z ==> a is pivot => x, y, z are free variables.
The same equation can be expressed as
x = a  y  z ==> x is pivot => a, y, z are free variables.
I thought of explaining the free/leading variables in a nonmatrix way, since the original query was in a nonmatrix way. Thanks to Gilbert Strang for getting me again started with Linear Algebra.
 131
 3