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Free variables in linear algebra?

by TheColorCute
Tags: free varibale matrix
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TheColorCute
#1
Jul7-10, 08:27 PM
P: 22
My questions is short and to the point: What exactly is a free variable (in a matrix, for example).
How do you know if a variable is free?

Here's a matrix that (apparently) has a free variable:

[1 4 -3 0 0]
[-2 -7 5 1 0 ]
[-4 -5 7 5 0 ]

Row reducing the matrix we end up with:

[1 4 -3 0 0 ]
[0 1 -1 1 0 ]
[0 0 6 -6 0 ]

Apparently this matrix has a variable. Where is it and how do I know that it's free?

Thanks!! :)
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vela
#2
Jul7-10, 08:44 PM
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Generally, if you have m unknowns and n independent and consistent equations, where n≤m, you have m-n free variables. Reducing a matrix allows you to find out how many equations you have that are actually independent. In your example, you have 4 variables and only 3 independent equations, so one variable is free.

Which variable is free is arbitrary. You get to choose, but when you have a reduced echelon matrix, it makes sense to use the variables that don't correspond to pivot columns, just because it makes the algebra more straightforward.
TheColorCute
#3
Jul7-10, 09:12 PM
P: 22
Ohhhh OK. So, to clarify, in this matrix we have 5 columns (but only four of which have constants in them), so we have four variables. Then we have three rows (each row equals an equation). So we have 4 variables minus 3 rows which is equal to 1 free variable. We don't want to choose any position in the pivot columns as free variables, so that eliminates columns 1, 2, and 3. So that leaves us with columns 4. So we could choose either x3 or x4 as our free variable? Or are they both free variables? Can we only have one?

vela
#4
Jul7-10, 10:20 PM
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Free variables in linear algebra?

The convenient choice would x4. In this particular case, since they turn out to be equal to each other, you could use either x3 or x4 pretty easily, but it's one or the other, not both. Generally, though since there's a pivot in the third column, you wouldn't use x3.
Mark44
#5
Jul8-10, 09:23 AM
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Quote Quote by TheColorCute View Post
Ohhhh OK. So, to clarify, in this matrix we have 5 columns (but only four of which have constants in them), so we have four variables.
All five columns have constants in them. The other posters in this thread are assuming that your matrix is an augmented matrix that represents the matrix equation Ax = 0. Augmented matrices usually have a dotted vertical line separating the coefficients of the matrix from the column of constants, something like this:
[1 4 -3 0 | 0]
[-2 -7 5 1 | 0 ]
[-4 -5 7 5 | 0 ]


The elements of A make up the first four columns. The elements of the zero vector are fifth column.
Quote Quote by TheColorCute View Post
Then we have three rows (each row equals an equation). So we have 4 variables minus 3 rows which is equal to 1 free variable. We don't want to choose any position in the pivot columns as free variables, so that eliminates columns 1, 2, and 3. So that leaves us with columns 4. So we could choose either x3 or x4 as our free variable? Or are they both free variables? Can we only have one?


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