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Beantwortet: Gleichungssystem loesen .

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0
0

0    -a      -b
a     0      -c      | *b
b     c       0      | *a  und 2. Zeile - 3. Zeile

0    -a         -b
0     -ac      -c     
b      c          0 

also b*x1  +c*x2 = 0    wenn also etwa  x1= t   dann

                            x2 = - b*t / c

dann wird aus 2. Zeile     -ac*(-bt/c)   -cx3 = 0

                                                              -c*x3 = abt

                                                                   x3 = abt/ -c  

Kontrolle mit 1. Zeile -a*(-bt/c) -b*abt/(-c) =0

                                           abt/c  +  ab^2t/(-c) = 0

                                        abt/c  *  (  1 -   b ) = 0

also klappt das nur für b=1  ???


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