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Solving Linear Systems

Solving Linear Systems

Author: c o

To introduce the concept of free variables
To use reduced row echelon form to determine the solutions of linear systems
To use Gaussian elimination to find the solution of linear systems

First the notion of a reduced row echelon form matrix is reviewed before it is shown how such matrices may be used to determine what sort of solution a system has. Next a slide show presents the determination of a set of solutions in the case that a system contains free variables.

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Classifying Solution Sets

Reduced Row Echelon Form

You will recall that a matrix satisfying reduced row echelon form has a 1 in each row's pivot position and each row's pivot is the only non-zero entry in its column.  We are interested in this form because it allows us to quickly classify a linear system as having a single unique solution, as having multiple solutions, or as having no solutions.  For background about reduced row echelon form, and about Gaussian elimination, see this lesson.  

For what follows, we will examine reduced row echelon matrices in order to classify what sort of solutions they admit.

Unique Solution

A unique solution is easiest to spot.  If we have our reduced row echelon matrix, then the system has a unique solution if the only entries in each row are the pivot entry and the final entry.  Here is an example:

We can simply read the solution off by looking at the last column of the matrix.

Free Variables

Often, when a system contains more variables than equations, we end up with so-called free variables.  These can be identified in a reduced row echelon matrix, such as the following:

In the above matrix, the variable x4 is considered to be free.  In general, for a reduced row echelon matrix, a variable is free when its column does not contain a pivot entry.  Free variables are interesting because, when they are present, we have not one, but infinitely many solutions.  This is because the conditions imposed by the system are met regardless of the value of of the free variable.  

In the next section's slide show, we will see an example wherein we determine the set of solutions when a system contains a free variable. 

No Solution

Sometimes a system will not have a solution.  This results from a situation in which we discover some contradiction.  Here is an example:

Why doesn't this have a solution?  Simple, the final row expresses a contradiction.  Lets suppose that the variables of this system are x1,x2, and x3.  Then the final row stats that 0x1 + 0x2 + 0x3 = 1, which is plainly impossible because zero does not equal one.  In general, whenever a reduced row echelon matrix contains a row the pivot of which is the final entry, then it represents a linear system that has no solutions.

Example: Determining Solutions In the Presence Of Free Variables

Starting with a matrix in which we have more variables than equations, we apply Gauss-Jordan elimination to arrive at a matrix in reduced row echelon form. From there we identify free variables and use these to express the set of all vector solutions to the linear system the matrix represents.