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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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    @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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    Hello, my dear friend, @Babak.2013-08-11