已知向量组
a1=
(1
1
0
-1),
a2=
(-2
1
0
0),
a3=
(-1
2
0
1),
又β满足3(a1-β)+2(a3+β)=5(a2+β),求β
3a1-3β+2a3+2β=5a2+5β, 6β=3a1-5a2+2a3, 所以β=1/2a1-5/6a2+1/3a3=1/2 (1 1 0 -1) -5/6 (-2 1 0 0) +1/3 (-1 2 0 1) = (11/6 1/3 0 -1/6)
已知向量组
a1=
(1
1
0
-1),
a2=
(-2
1
0
0),
a3=
(-1
2
0
1),
又β满足3(a1-β)+2(a3+β)=5(a2+β),求β
3a1-3β+2a3+2β=5a2+5β, 6β=3a1-5a2+2a3, 所以β=1/2a1-5/6a2+1/3a3=1/2 (1 1 0 -1) -5/6 (-2 1 0 0) +1/3 (-1 2 0 1) = (11/6 1/3 0 -1/6)
