# A matrix containing a line such as this is invalid, right? $(0 \space0\space 0\space 0\space|\space1)$

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. – Fly by Night Nov 8 '12 at 17:41
@copper.hat I think the OP is referring to "row", when using "line" – amWhy Nov 8 '12 at 17:42
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. – copper.hat Nov 8 '12 at 17:42
@amWhy: Thanks! I missed the point of the question until I read your answer. I thought it was a question about notation! – copper.hat Nov 8 '12 at 17:43
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 from – Thomas Andrews Nov 8 '12 at 17:49

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. – Hagen von Eitzen Nov 8 '12 at 17:43
Or rather than the matrix being inconsistent, one might say the system of equations is inconsistent. – rschwieb Nov 8 '12 at 17:43
+1 Nice to see u again, Amy :-) – S. Snape Aug 11 '13 at 0:22
Hello, my dear friend, @Babak. – amWhy Aug 11 '13 at 0:23