Is the Linear System inconsistent?

Mr.Tibbs
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Hey guys, I'd assume that this will be a very easy answer. I need to clarify what is an inconsistent linear system. I've worked out my matrix and have achieved this form:

x+0y+z+0w=0
0x+0y+z+w=0
0x+0y+0z+w=0

I know that an inconsistent linear system has no solution i.e 0=1. This is throwing me off because of the y components being all zero which means I can make y what ever I want, but does this also make the system inconsistent?
 
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Mr.Tibbs said:
Hey guys, I'd assume that this will be a very easy answer. I need to clarify what is an inconsistent linear system. I've worked out my matrix and have achieved this form:

x+0y+z+0w=0
0x+0y+z+w=0
0x+0y+0z+w=0

I know that an inconsistent linear system has no solution i.e 0=1. This is throwing me off because of the y components being all zero which means I can make y what ever I want, but does this also make the system inconsistent?

x=y=z=w=0 is a solution. It will always be a solution when the RHS is all zeros. You have an infinite number of solutions, but it's not inconsistent. Inconsistent means no solutions.
 
Awesome! Thanks for clearing that up for me!
 
Mr.Tibbs said:
Hey guys, I'd assume that this will be a very easy answer. I need to clarify what is an inconsistent linear system. I've worked out my matrix and have achieved this form:

x+0y+z+0w=0
0x+0y+z+w=0
0x+0y+0z+w=0

I know that an inconsistent linear system has no solution i.e 0=1. This is throwing me off because of the y components being all zero which means I can make y what ever I want, but does this also make the system inconsistent?

"Inconsistent" does not apply to homogeneous linear systems (where the right-hand-sides are all zero) because setting all variables = 0 *does* give a solution. No: "consistency" and its opposite are relevant when (some of the) right-hand sides are *nonzero*. For example, the system x = 0 and 2*x = 0 is perfectly consistent, but the system x = 1 and 2*x = 3 is inconsistent.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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