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I am dealing with the following example:

$$ \pmatrix{ -1 & -2& -3 \\ 3 & 2 & 1 \\ 1& 1 & 1 \\ 2 & 4& 1 }* \pmatrix{ x_1\\ x_2 \\ x_3 \\ }= \pmatrix{ -6\\ 6 \\ 3 \\ 7 \\ }$$

I know I should solve that matrix with gauß' elimination, but my problem is that there are only $x_1 $ to $x_3 $, so I do not know how to interprete this example?

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Solve it like you normally would, and the last equation should change to $0x_1+0x_2+0x_3=0$, otherwise the system is inconsistent. –  Clayton Jan 16 '13 at 13:53
    
Check this Wiki link for a little more information on inconsistent or over-determined systems : en.wikipedia.org/wiki/Overdetermined_system –  MSEoris Jan 16 '13 at 13:55
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@Clayton so you would kill the last row and then have 3 equations? –  Le Chifre Jan 16 '13 at 14:00
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@maximus: It happens naturally (unless the system is inconsistent). This is because the basis only consists of three elements, so the fourth one has to be a linear combination of the first $3$. –  Clayton Jan 16 '13 at 14:01

1 Answer 1

up vote 3 down vote accepted

As @Clayton noted do the G.E., as you always do for any matrix. Here you have $$A:=\left( \begin{array} ~-1 & -2& -3&~~ -6 \\ 3 & 2 & 1&~~~~~~~ 6 \\ 1& 1 & 1&~~~~~~~ 3 \\ 2 & 4& 1&~~~~~~~ 7 \\ \end{array} \right)$$ which by doing some process became: $$\sim\left( \begin{array} 11 & 1& 1&~~~~~~~ 3 \\ 0 & 1 & 2&~~~~~~~ 3 \\ 0& 0 & -5&~~~ -5 \\ 0 & 0& 0&~~~~~~~ 0 \\ \end{array} \right)$$ This means that the forth equation in the original system was dependent to other 3 equations. It is clear now that your solution $(x_1,x_2,x_3)$ is unique and can be found easily.

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@DavidMitra: Yes of course. I assumed the OP already knows that in G.E. we do the same rules on the constant matrix as well as we are doing on the coefficient matrix. Honestly, I wanted to add more for the OP that the system shows 4 plane in $R^3$ which has no a vivid single intersection as well. I should say that Clayton started to tell the OP the main point. I wrote some of the way he wanted to say here. Thanks. –  Babak S. Jan 16 '13 at 14:22
    
Or maybe something to do with \begin{array} usually has formatting like \begin{array}{cccc} ...? –  adam W Jan 16 '13 at 15:53
    
@adamW: You are right. I fixed it. there was a gap and if you wanted to see that, you would do that. I add a ~ for doing it right. ;-) –  Babak S. Jan 16 '13 at 15:55
    
I don't want to stray off topic too much so I will delete these soon but - I've always wanted to do a dashed line instead of solid as with \begin{array}{ccc|c}, I end up using \phantom{-}1 or something for alignment. Never knew of using ~ is that supposed to be like a single space? –  adam W Jan 16 '13 at 16:00
    
@adamW:Thanks. It is very kind of you but, you know, I am interested to know how can I write $<A\mid b>$ when we know the matrix $A$ and $b$ in a system with a segment between them in Latex. You see through my answer that I made a big gap there to build this. –  Babak S. Jan 16 '13 at 16:05

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