116 CHAPTER 7. SYSTEMS OF EQUATIONS
gives x+4− s+ t = 3 showing that x =−1+ s− t,y = 2,z = s, and w = t. It is customaryto write this in the form
xyzw
=
−1+ s− t
2st
. (7.10)
This is another example of a system which has an infinite solution set but this timethe solution set depends on two parameters, not one. Most people find it less confusingin the case of an infinite solution set to first place the augmented matrix in row reducedechelon form rather than just echelon form before seeking to write down the description ofthe solution. In the above, this means we don’t stop with the echelon form 7.9. Instead wefirst place it in reduced echelon form as follows. 1 0 −1 1 −1
0 1 0 0 20 0 0 0 0
.
Then the solution is y = 2 from the second row and x = −1+ z−w from the first. Thusletting z = s and w = t, the solution is given in 7.10.
The number of free variables is always equal to the number of different parametersused to describe the solution. If there are no free variables, then either there is no solutionas in the case where row operations yield an echelon form like 1 2 3
0 4 −20 0 1
or there is a unique solution as in the case where row operations yield an echelon form like 1 2 2 3
0 4 3 −20 0 4 1
.
Also, sometimes there are free variables and no solution as in the following: 1 2 2 30 4 3 −20 0 0 1
.
There are a lot of cases to consider but it is not necessary to make a major production ofthis. Do row operations till you obtain a matrix in echelon form or reduced echelon formand determine whether there is a solution. If there is, see if there are free variables. In thiscase, there will be infinitely many solutions. Find them by assigning different parametersto the free variables and obtain the solution. If there are no free variables, then there willbe a unique solution which is easily determined once the augmented matrix is in echelonor row reduced echelon form. In every case, the process yields a straightforward way todescribe the solutions to the linear system. As indicated above, you are probably less likely