# A zero column while solving

1. Jun 11, 2008

### JinM

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)?

Last edited: Jun 11, 2008
2. Jun 11, 2008

### Vid

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.

3. Jun 11, 2008

### JinM

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?

4. Jun 11, 2008

### Vid

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.

5. Jun 11, 2008

Thanks, Vid.