| k | 2k | |||
a2: −k*α1 −3α3 = 0 −−−> α3 = − | α1 = − | α2 | ||
| 3 | 3 |
| 1 | ||
a3: α2 + kα3 = 0 −−−> α3 = − | α2 | |
| k |
| 1 | 2k | 3 | √6 | |||||
tak więc musi zachodzić: − | = − | −−−> k2 = | −−> k = ± | |||||
| k | 3 | 2 | 2 |
| √6 | ||
i wtedy (sprawdzam tylko dla k = | , drugie sprawdź samodzielnie) | |
| 2 |
| √6 | ||
α3 = − | α2 ; α2 = α2 ; α1 = 2α2 | |
| 3 |
| √6 | √6 | |||
= α1(a1 − | a2) + α2(−2a1 + a3) + α3(−3a2 + | a3) = | ||
| 2 | 2 |
| √6 | √6 | √6 | ||||
= 2α2(a1 − | a2) + α2(−2a1 + a3) − | α2(−3a2 + | a3) = | |||
| 2 | 3 | 2 |
| √6*√6 | ||
= α2(2a1 −√6a2 −2a1 + a3 +√6a2 − | a3) = α2*0 = 0 | |
| 3*2 |