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,-

1个回答

  • (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.