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linear algebra, free variables

 
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Dec9-11, 11:02 AM   #1
 

linear algebra, free variables


This is not a homework question. But a question on how to understand what my textbook does.

It is about choosing the free variables.

Let's say I have the system of equations:

x1-2x2+3x3+2x4+x5 =10
x3 +2x5=-3
x4 -4x5=7

Then my book says that we choose x2 and x5 to be free variables, since they are not leading. And we set
x2=t and x4=s and solve the system.

But why do we have to choose x2 and x5 just because they are not leading? I mean, Can't I say x2=t and x5=s instead of x4?
 
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Dec10-11, 08:22 PM   #2
 
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Quote by skwey View Post
But why do we have to choose x2 and x5 just because they are not leading? I mean, Can't I say x2=t and x5=s instead of x4?
As I understand the definition of "leading variable" you must have the matrix representing the system in row reduced echelon form before you can say which variables are "leading". When the matrix is in that form it is simplest to represent the general form of the solution by assign the non-leading variables to be arbitrary constants. As well as being simple, it is safe.

It's true that you often may be able to represent the general form of the solution by assigning arbitrary constants to some of the non-leading variables. But suppose you have the underdetermined system of equations:

x1 + x2 + x3 = 1
x2 = 0

You can't get the general form by assiging x2 to be an arbitrary constant.. Perhaps you can think of more complicated examples where the equations force some of the variables to have specific numerical values and leave the rest undetermined. It might not be obvious when you first look at the system of equations which variables are determined. So you aren't guaranteed that you can assign the variables of your choice arbitrary values.
 
Dec11-11, 08:11 AM   #3
 
Thanks!

And that was a good example!
 
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