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A matrix containing a line such as this is invalid, right? $$ (0 \space0\space 0\space 0\space|\space1) $$ the matrix in question is this: $$\left( \begin{array}{rrrr|r} 1&0&1&1&2\\ 0&1&-4&0&-6\\ 0&0&0&0&1 \end{array} \right)$$

Edit: my original question was bad. I was asked to "solve the system of linear equations, or say there isn't one". My question is if this is a "no answer" case. Thank you.

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    Why do you think it is "invalid"? I see no problem with it at all.2012-11-08
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    @copper.hat I think the OP is referring to "row", when using "line"2012-11-08
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    In the context of row reduction, it just means it is not in the range of the lhs. But it is neither wrong nor right.2012-11-08
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    @amWhy: Thanks! I missed the point of the question until I read your answer. I thought it was a question about notation!2012-11-08
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    The question is ambiguous. What do you mean by "a line"? If you mean "a row", then we don't really understand what "invalid" means. And if by "line" you mean that thing between a $0$ and a $1$, then this line is there simply for convenience.2012-11-08
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    That's a pretty common way to represent a matrix which is, at heart, a combination of two matrices. If $A$ and $B$ are matrices with the same number of rows, we will often write $(A|B)$ for the matrix with the rows of $A$ and $B$ appended. The "actual" matrix is the object without the $|$ separator, the separator is just a way of remembering where the matrix came from2012-11-08
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    @ThomasAndrews I think the OP means "row", and not "line" as in the "vertical line"...2012-11-08
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    @amWhy Actually, I think the real confusion is the word "invalid." Lots of sloppy language in this question to go around.2012-11-08

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I wouldn't use the term "invalid".

The matrix with the row in question - or rather, (as @rschwieb points out below), the associated system of three equations in four unknowns - is inconsistent.

Edit:

Given your clarification, you are correct, there is no solution to the associated system of equations.

To see why, note that the bottom-most row of your matrix tells you that $$0\cdot x + 0\cdot y + 0\cdot z = 1.$$ There does not exist any solution. Can you see that whatever the values of $x, y, z$, we will never have the left-hand-side $=$ right-hand-side?

Note also that, in the following example (representing a system of $5$ equations in $4$ unknowns $w,x,y,z$):

$$\left( \begin{array}{rrrr|r} 1&0&0&0&2\\ 0&1&0&0&-6\\ 0&0&1&0&1\\ 0&0&0&1&1\\ 0&0&0&0&1\\ \end{array} \right),$$

where it appears that $w=2, x=-6, y=1, z=1$ is a solution, that "pseudo-solution" is incompatible with the equation associated with the $5^{th}$ row: $$0\cdot w +0\cdot x + 0\cdot y + 0\cdot z = 1.$$ Hence the entire associated system of equations is inconsistent (and thus no solution exists).

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    There is simply no solution to the associated system of linear equations.2012-11-08
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    Or rather than the matrix being inconsistent, one might say the *system of equations* is inconsistent.2012-11-08
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    Hello, my dear friend, @Babak.2013-08-11