A Question about the free variable

[The Student]

In the following system of equations, I am confused with how we can know that z is the free variable and not y. I know that in this example it is obviously z, but I can't figure a general rule.

x + 3y + z = −1
3y + 4z = 1
0y + 0z = 0

 

[pka]

In the following system of equations, I am confused with how we can know that z is the free variable and not y. I know that in this example it is obviously z, but I can't figure a general rule.
x + 3y + z = −1
3y + 4z = 1
0y + 0z = 0

First, do you really mean that? If so, it contributes nothing to the question.

Then what is meant by a free variable? How do you know that "in this example it is obviously z"?

 

[DrPhil]

In the following system of equations, I am confused with how we can know that z is the free variable and not y. I know that in this example it is obviously z, but I can't figure a general rule.

x + 3y + z = −1
3y + 4z = 1
0y + 0z = 0

Since the 3rd equation you wrote is a tautology, what you have is a system of two equations in three unknowns. There is no "obvious" choice for the free variable - any one of them will do. The other two variables can be expressed as functions of the one you chose to be "free." If you want it to be z, you can rewrite your two equations like this:

x + 3y = -1 - z
.....3y = 1 - 4z

 

[The Student]

Since the 3rd equation you wrote is a tautology, what you have is a system of two equations in three unknowns. There is no "obvious" choice for the free variable - any one of them will do. The other two variables can be expressed as functions of the one you chose to be "free." If you want it to be z, you can rewrite your two equations like this:

x + 3y = -1 - z
.....3y = 1 - 4z

The question was originally

x + 3y + z = −1
2x + 9y + 6z = −1
−3x − 3y + 5z = 5

Then row operations made it

x + 3y + z = −1
3y + 4z = 1
6y + 8z = 2

then

x + 3y + z = −1
3y + 4z = 1
0y + 0z = 0

Are you saying that y could be the free variable? I thought that it was suppose to be obvious because my notes do not explain why they chose z to be the free variable.

 

[DrPhil]

The question was originally

x + 3y + z = −1
2x + 9y + 6z = −1
−3x − 3y + 5z = 5

Then row operations made it

x + 3y + z = −1
3y + 4z = 1
6y + 8z = 2

then

x + 3y + z = −1
3y + 4z = 1
0y + 0z = 0

Are you saying that y could be the free variable? I thought that it was suppose to be obvious because my notes do not explain why they chose z to be the free variable.

OK - that makes more sense for the statement of the question. You started with three equations, BUT you discovered that one was a linear combination of the others, so you only have two equations that are linearly independent.

Yes, I am saying y (or x) could be the free variable. With only two linearly independent equations, you can't solve for all three variables, but you can find any two of them as functions of the other one.

 

[The Student]

OK - that makes more sense for the statement of the question. You started with three equations, BUT you discovered that one was a linear combination of the others, so you only have two equations that are linearly independent.

Yes, I am saying y (or x) could be the free variable. With only two linearly independent equations, you can't solve for all three variables, but you can find any two of them as functions of the other one.

Ok, thanks, that makes sense.

 

[pka]

Ok, thanks, that makes sense.

@The Student,
I hope you learn from this experience.
You could have saved all of us a great deal of time had posted the entire original question.
Instead, you put us in the middle of work you had done and expected us to know what you were thinking.