What to do with a zero column in a system of linear equations?

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A zero column in a system of linear equations indicates a variable that does not appear in any equation, which can be classified as a free variable. In a homogeneous system (Ax = 0), the zero column corresponds to a free variable, while in an inhomogeneous system (Ax = b), it may lead to a contradiction if the corresponding constant is non-zero. In the discussed case of a 2 x 3 system, the absence of an equation for variable x_3 means it remains a free variable, contributing to infinitely many solutions. The presence of a zero column does not necessitate testing for homogeneity if the equations do not yield a false statement. Thus, free variables like x_3 can still be included in the solution set.
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Hello, just started a LA course, and I have a small question.

If I get a zero column while solving for a system of linear equations, and that column corresponds to, say, variable x_{5}, do we just not include it in the solution set since it appears in no equation of the system? Or is it a free variable (that has no purpose)?
 
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That depends on what kind of system you have. If it's a homogeneous system (Ax = 0) then you just have 0=0, and x_5 is indeed just a free variable. However, if you have an inhomogeneous (Ax= b) then you have a possibility that 0=b_5 and b_5 =\=0. In this case, you have an obviously false statement, and the system has no solutions.
 
Thanks. We still haven't discussed homogeneity. I don't see, though, how you get 0=0. I have a 2 x 3 system of linear equations and I eventually get this (through Gauss-Jordan),

x_1 + 3x_4 - x_5 -x_6 = 32
x_2 +2x_4 -x_5 = 28

x_3 is missing in both equations, hence the zero column that I mention before appearing in the matrix. Do I still have to test homogeneity to determine the solution set here?
 
In this case, x_3 is just a free variable. Since you only have 2 equations, neither of which gave 0=0 then your system does have a solution(infinitely many solutions in fact). x_3 would still be included in the solution set as a free variable just like x_4-x_6. It just turns out that nothing depends on the value of x_3.
 
Thanks, Vid. :smile:
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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