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Dear ladies and gentlemen,

over time I noticed I (and other) again and again have problems solving "systems of linear equations". It seems depending of the steps one chooses, we get different results!! How can that be? Should one not always get the same results no matter which path he goes down? What am I missing? Are there maybe rules I don't know and use even though I shouldn't?

I want to give you an example. The set of equations is from a state-price security calculation (see "State Preference Approach") which we shall solve using the Gaussian elimination:

I: 2P(1) + 2P(2) + 2P(3) =1,6

II: 3P(1) + 0P(2) + 1P(3) =1,0

III: 0P(1) + 2P(2) + 1P(3) =0,8

The solution shall be: P(1) = 0,2 P(2)=0,2 and P(3)=0,4

Not only got I different numbers on the first try that totally went in "into space", I want to write down for you my second approach to check for mistakes:

II-III = IV = 3P(1) - 2P(2) = 0,2 --> 2P(2) = 3P(1) - 0,2 (this far I'm with the solution) then I simply plug in the result in the following lines:

in III: 4P(1) = 1 --> P(1) = 0,25

in II: 0,75 + P(3) = 1,0 --> P(3) = 0,25

in I: 0,5 + 2P(2) + 0,5 = 1,6 --> P(2) = 0,3

But these results seem to differ. So it seems I clearly miss some important rule!

Must I not "plug in" results into other rows, as the "Gaussian" system seems to avoid? But how can there be a difference / can I be forced to avoid "plugging in" results?

This should normally be allowed, shouldn't it? Can you help me find my blind spot?

Thanks for your help in advance.

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    user6312 you are basically right, but we have to use this notation! :( But furthermore $I$ got the problem, that $I$ never come down to one variable! I also had this problem with later exercises. I can calculate around for 3$0$ minutes and don't come to an end / silver lining on the horizon!2011-07-12

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"Plugging in" or "substituting back" or whatever is equivalent to Gaussian elimination. You just made an error. You got $2P(2)=3P(1)-0{,}2$ all right. But when you plug that in equation III becomes: $3P(1)+P(3)=1{,}0$ and equation I becomes: $5P(1)+2P(3)=1{,}8$. Both of these still match the given solution.

IOW, you got variables mixed up. Not doing it systematically occasionally leads to a sleek solution, but you have to be careful.

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    @grunwald: Say that we have 3 unknowns+3eqs like in your example (similar thinking works with more). You solve one of them from some equation (no need to try those subtraction tricks, though they may make it more transparent). Set aside this equation, and plug in the solved variable into the other two equations. Now you have a pair of equations with two variables. Rinse. Repeat. IOW solve one of the remaining variables from one of those two eqs. Substitute to the remaining equation. One variable remaining, so can solve it. Substitute the solved value back to the equations that were set aside..2011-07-12
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You're not missing some important rule; you're just making mistakes, which we all do, and you have the good sense to check your results, so everything is in best order -- all you need to do is learn to find your mistakes :-)

Where you wrote "I-III", I think you meant "II-III", and where it says "in III", you added 3P(1) and 1P(3) to make 4P(1), but the second one was a P(3), not a P(1).

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    Oh yes you are indeed right. It's 1 P(3) and 3P(1) still. Indeed the notation is confusing, but we have to use it. But now I got the problem, that I never come down to one variable! I also had this problem with later exercises. I can calculate around for 30 minutes and don't come to an end / silver lining on the horizon!2011-07-12