Aufgabe 2.24 ii)
2008-09-28 23:55:29

0 1 2 3 | 0  (1)
1 2 3 4 | 0  (2)
2 3 4 5 | 0  (3)
3 4 5 6 | 0  (4)
-> homogen

(1) und (2) tauschen

1 2 3 4 | 0  (1)
0 1 2 3 | 0  (2)
2 3 4 5 | 0  (3)
3 4 5 6 | 0  (4)

(3) + (-2) * (1) => 0 -1 -2 -3 | 0  (3)
(4) + (-3) * (1) => 0 -2 -4 -6 | 0  (4)

1  2  3  4 | 0  (1)
0  1  2  3 | 0  (2)
0 -1 -2 -3 | 0  (3)
0 -2 -4 -6 | 0  (4)

(4) + 2 * (2) => 0 0 0 0 | 0  (4)
(3) + (2)     => 0 0 0 0 | 0  (3)

Setze x3 = t1 und x4 = t2

x1 + 2x2 + 3t1 + 4t2 = 0  (1)
      x2 + 2t1 + 3t2 = 0  (2)

=> x2 = -2t1 -3t2

Einsetzen in (1)

x1 + 2(-2t1 -3t2) + 3t1 + 4t2 = 0
x1 -t1 - 2t2                  = 0
x1 = t1 + 2t2

Lösung:

x1 =   t1  + 2t2
x2 = - 2t1 - 3t2
x3 =   t1
x4 =   t2