x - 4·y - 5·z = 1
x - y - z = 4
-x - 2·y - 3·z = -7
II - I ; III + I
3·y + 4·z = 3
- 6·y - 8·z = -6
II ist linear abhängig zu I und kann gestrichen werden.
3·y + 4·z = 3 --> y = 1 - 4/3·z
x - (1 - 4/3·z) - z = 4 --> x = 5 - z/3
[5 - z/3; 1 - 4/3·z; z] = [1; y; z] --> y = -15 ∧ z = 12
[5 - z/3; 1 - 4/3·z; z] = [x; 5; z] --> x = 6 ∧ z = -3