106 次のベクトルを,3つのベクトルa=(1,2,3),(0, 2,5),(1,3,1) と適当な実数 s,
tu を用いて, Sa + to + uc の形に表せ。
→例題 15
(1) p=(03, 12)
S(1₁23)+2(0.2, 5) + ((. 3.1)
(stu, 25+2t+3u, 3s +5t+α)
(0₁ 3₁ (12) = (stu, 2s+ 2t+3u. 3 st 5t+u)
S+u = 0
2s+2t+3u = 3
3s+5t+u=12
P=a²+2b-c
(2)* q=(-2, 2, 9)
S+U= -2
→
2st2u= -4
-2s-2t-3u=-2
2s+2t+3u=23stSt+u=9
65+6+ +9u=6
-65-10t-20= - 18
-2t - U=-69
-14t-qu=-42
-4t+7u=-12
-18t=-54
t=3
S = 1₁ t = 2₁α = − 1 2+2+-3 = 3
22=4
t=2
-4t+70²= -12...
-25-2α=0
25+3u+24=3
-4-u=-6
- U= -2
U=2
-65-6t-qu=-9
65 + 10 + 20 = 24
utzt = 30
5+2= 2
S = - 4
4t-qu=15--0
-4t-2α=-6
-qu = 9
u = -1
S = 1
q=²-4a²+ 3 b² +₂0² q = -2ã+36