"A system in echelon form can be inconsistent." Can someone give me an explanation as to why this is false?

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"A system in echelon form can be inconsistent." Can someone give me an explanation as to why this is false?

Mathematics
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any system in echelon form can be rewritten in row reduced echelon form
|dw:1439931493101:dw|
so u cannot have inconsistency

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Isn't a matrix in echelon form if it has a row of zeroes on the bottom? Like, is the following not in echelon form? \[\begin{bmatrix}1&&1&&1\\0&&0&&1\end{bmatrix}\](The system in this case would be \(x + y = 1\) and \(0x + 0y = 1\))
I just didn't see anything in my book's definition of echelon form that prevented this, but I'm probably wrong on that.
that has 3 variables x ,y ,z we dont know the constants
|dw:1439931839159:dw|
I meant for there to to be the line for an augmented matrix between columns 2 and 3, but I don't know how to do that in \(\LaTeX\) just yet.
|dw:1439931879525:dw|
this is not echelon form there is simply no echelon form for this
http://prntscr.com/866zua
actually wait i dont know, what does inconsistent mean again?
there is actually an example there where RREF has all 0s in the bottom soo
Yeah, that's what's confusing me.
i was thining inconsistent is same as dependant but thats not it
They said \[\begin{bmatrix}1~~~1~~~3\\0~~~0~~~2\end{bmatrix}\]isn't in ehcelon form but \[\begin{bmatrix}1~~~1~~~3\\0~~~0~~~1\end{bmatrix}\]is?\[\\\]And my book cites inconsistent as having no solution.
gotta have it as 1 i guess
Hmm...but if the second one is allowed, then is it assumed that the matrix has 3 unknowns, but we don't know the constant matrix since there is no divider?
what is the definition of inconsistent, that should clear this up
My book's definition of inconsistent is that there is no solution that satisfies all parts of the system simultaneously. "Inconsistent: No pair of numbers (s1,s2) satisfies all three equations simultaneously."
And to clarify, the question on my assignment was asking if the above statement was true or false. The answer ended up being false, and I just can't seem to figure out why.
Ohh okay
Is it just something weird based on definition or am I way off? Also, here's my book's definition of echelon form if it helps. "In each row of a system, the first variable with a nonzero coefficient is the row’s leading variable. A system is in echelon form if each leading variable is to the right of the leading variable in the row above it, except for the leading variable in the first row, and any all-zero rows are at the bottom."
They use those examples to show you how a matrix is called echelon. \(\left[\begin{matrix}1&1&3\\0&0&2\end{matrix}\right]\) is not an echelon matrix because the leading 1 of the second row is 2, not 1. But you can get the echelon form by divided row2 by 2 to get \(\left[\begin{matrix}1&1&3\\0&0&1\end{matrix}\right]\)
That is just NOTATION.
That shows you how "an echelon form can be inconsistent"
Ok, I'm pretty sure the leading one's thing is just differences in definition between texts (I cross checked multiple books, and some said they HAD to have leading one's and others didn't care.)
But the question answer was that the statement was false, indicating that it cannot be inconsistent...That's what I was confused on.
It's above my head!! :)
lol ok. I'm just as confused as you are, and this is the assignment from my first Linear Algebra class XD
I have to go, but @ganeshie8 , if you could take a look at this whenever you get online, that'd be great!
The question is about "system of equations", not about a "matrix" being inconsistent. We say a "system of equations" is consistent or inconsistent. But never say a matrix is inconsistent or inconsistent. If the system of equations is inconsistent, then after eliminations, the augmented matrix in echelon form will have at least one row in the bottom with all 0's except for the last entry. For example : \(\begin{bmatrix}1&2&5&|&5\\0&0&0&|&\color{red}{3}\end{bmatrix}\).
The given statement, "A system in echelon form can be inconsistent." means that the "starting" system of equations itself are in echelon form, for example : \[ \begin{array}{} x&+&3y&+&z&=&5\\ &&y&+&5z&=&3\\ &&&&z&=&8 \end{array}\] In other words, the given system of equations itself is in some kind of triangular/staircase form. Our task is to prove that any system like this can never be inconsistent..
So, when it said "in echelon form", it meant that the system was given in echelon form, not manipulated into it?
Exactly! the "given" set of equations are in "staircase" looking form
The given statement is false because the system of equations in that staircase form "always has a solution" by back substitution.
heard of "back substitution" before ?
Oh! That makes a lot more sense. I guess it was just the wording that I misinterpreted. And yes, I know what back substitution is. I've self-studied Linear Algebra before, but I never formally studied it in a classroom setting (which I am now doing).
Fact : you can never learn linear algebra fully, it is a vast subject and there is always something we never thought of..
haha seems like it. My teacher started off lecture stressing a similar point, bringing in lots of different fields where it is applied. Anyway, thanks so much for the help. :)

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