Question

In terms of using matrices to solve linear equations, what is a 'free variable'?

Answer 1

In this augmented matrix i understand that there are 3 'free variables' as there are an infinte amount of solutions for them\[\left[\begin{matrix}1 & 2 & | & 6 \\ 0 & 0 & | & 0 \\ 0 & 0 & | & 0\end{matrix}\right]\]

Answer 2

my actual question is why is there 1 'basic' variable in the above augmented matrix? if it was x + 2y = 6....surely there are an infinite number of solutions

Answer 3

sorry, in my first message i meant to say that there are 2 free variables

Answer 4

I would call one free variable and one dependent variable.

Answer 5

How do you have three variables ? You only have two variables in the system.

You have only one free variable because you can choose to define x in terms of y or y in terms of x. Either way one variable determines the value of the other.

Answer 6

sorry, it should have looked something like this: \[\left[\begin{matrix}1 & 2 & 5 & | & 6 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & | & 0\end{matrix}\right]\]

Answer 7

i am told that there is one 'basic' variable (in contrast to a free variable)) in this system...I would like to know why

Answer 8

free variable = no of column - rank of matrix

Answer 9

the way I understand it is that I have: x + 2y + 5z = 6 as my only equation. Yet I have three variables. So I can't solve for any really so dont i have 3 free variables?

Answer 10

woops!! no of column = no of columns of square matrix

Answer 11

You can solve for x by choosing values for y and z. or solve for y by choosing values for x and z and so on.

Though this isn't a good explanation.

Answer 12

if you have free variables, then you have infinite number of solutions as you can freely choose the variables ... so try to get a system where there are no free variables ... you will have unique solutions.

Answer 13

Okay. But I am asking why the system i have given has only 2 free variables (because of all the zeroes) instead of 3 free variables, as having one equation (which i have) doesn't allow you to solve for any of the variables, therefore, shouldn't there be 3 free variables?

Answer 14

@hartnn

Answer 15

what is your idea of free variables? independent ?

Answer 16

x + 2y + 5z = 6 means we can freely assume 2 variables to get the 3rd , hence 2 free variables

Answer 17

and one dependent which we calculate, using assumed 2 other variables

Answer 18

couldn't you freely assume two to get the other. then freely assume another two of the 3 and get another and so on...?

Answer 19

so at a time there are only 2 free variables.