帮我解个线性代数题吧β=(1,1,b+3,5)T,α1=(1,0,2,3)T,α2=(1,1,3,5)T,α3=(1,-

(α1,α2,α3,α4,β) =
1 1 1 1 1
0 1 -1 2 1
2 3 a+2 4 b+3
3 5 1 a+8 5
r3-2r1,r4-3r1
1 1 1 1 1
0 1 -1 2 1
0 1 a 2 b+1
0 2 -2 a+5 2
r1-r2,r3-r2,r4-2r2
1 0 2 -1 0
0 1 -1 2 1
0 0 a+1 0 b
0 0 0 a+1 0
当a=-1且b≠0时,β不能由α1,α2,α3,α4线性表示
当a=-1且b=0时
(α1,α2,α3,α4,β)-->
1 0 2 -1 0
0 1 -1 2 1
0 0 0 0 0
0 0 0 0 0
β=0α1+α2+0α3+0α4.
当a≠-1时
(α1,α2,α3,α4,β)-->
1 0 2 -1 0
0 1 -1 2 1
0 0 a+1 0 b
0 0 0 a+1 0
-->
1 0 2 -1 0
0 1 -1 2 1
0 0 1 0 b/(a+1)
0 0 0 1 0
-->
1 0 0 0 -2b/(a+1)
0 1 0 0 (a+b+1)/(a+1)
0 0 1 0 b/(a+1)
0 0 0 1 0
β=[-2b/(a+1)]α1+[(a+b+1)/(a+1)]α2+[b/(a+1)]α3+0α4.