We know that we can apply EROs to any augmented matrix into REF.

Suppose the system has $n$ equations and $m$ variables, and let $k$ be the number of non-zero rows in REF. Also suppose the system is consistent: then the REF has no row of the form $[0~0~0~\dots~1]$.

• $k\le n$, because there are only $n$ rows in the whole matrix
• $k$ is precisely the number of leading variables. So $k$ is no bigger $m$, the total number of variables; in symbols, we have $k\le m$.
• All the other variables are free variables, so $$ \text{$m-k$ is the number of free variables.} $$

What does this tell us about the set of solutions? For example, how many solutions are there?

Observation 1: free variables and the number of solutions

For consistent systems, this shows that:

• either $k=m$;
  • so $m-k=0$
  • there are no free variables
  • the system has one solution and no more
  • We say it has a unique solution.
• or $k<m$
  • so $m-k>0$
  • there is at least one free variable
  • so the system has infinitely many solutions (one for each value of each free variable)
  • The number of free variables, $m-k$, is called the dimension of the solution set.

Observation 2: systems with fewer equations than variables

For consistent systems where $n<m$ (fewer equations than variables):

• $k\le n < m$, so $k<m$.
• So there is at least one free variable.
• So in this situation we always have infinitely many solutions.