When a system is in echelon form and contains no equation of the form 0=b, with b nonzero, every nonzero equation contains a basic variable with a nonzero coefficient. Either the basic variables are completely determined (with no free variables) or at least one of the basic variables may be expressed in terms of one or more free variables. In the former case, there is a unique solution; in the latter case, there are infinitely many solutions (one for each choice of values for the free variables).
These remarks justify the following theorem
Existence and Uniqueness Theorem
A linear system is consistent if and only if the rightmost column of the augmented matrix is not a pivot column — that is, if and only if an echelon form of the augmented matrix has no row of the form
[0 … 0 b ] with b nonzero
If a linear system is consistent, then the solution set contains either a unique solution, when there are no free variables, or infinitely many solutions, when there is at least one free variable.
The following procedure outlines how to find and describe all solutions of a linear system.
USING ROW REDUCTION TO SOLVE A LINEAR SYSTEM
Write the augmented matrix of the system.
Use the row reduction algorithm to obtain an equivalent augmented matrix in echelon form. Decide whether the system is consistent. If there is no solution, stop; otherwise, go to the next step.
Continue row reduction to obtain the reduced echelon form.
Write the system of equations corresponding to the matrix obtained in step 3.
Rewrite each nonzero equation from step 4 so that its one basic variable is expressed in terms of any free variables appearing in the equation.
—– from <<linear Algebra and its Applications 4E (Lay)>> CHAPTER 1, PAGE 20.