Question

线性代数,w^3=1,w可以是复数吗

Answer 1

解: 由已知 w^3-1=0

所以 (w-1)(w^2+w+1)=0

若w=1 行列式=0

若w≠1, 则 w^2+w+1 = 0. (w必为复数)

用Laplace定理按1,2行展开

D =

1 1

1 w

*

1 1 c1 1

1 w^2 c2 w

1 w c3 w^2

0 0 w 0

+ (-1)^(1+2+1+5)*

1 1

1 w^2

*

b1 1 1 1

b2 1 w^2 w

b3 1 w w^2

w^2 0 0 0

+ (-1)^(1+2+2+5)*

1 1

w w^2

*

a1 1 1 1

a2 1 w^2 w

a3 1 w w^2

1 0 0 0

= [(w-1)(-w) -(w^2-1)(-w^2) - (w^2-w)]*

1 1 1

1 w^2 w

1 w w^2

= [(w-1)(-w) -(w^2-1)(-w^2) - (w^2-w)]*(w^4-3w^2+2w)

= (w^4-3w^2+2w)(w^4-3w^2+2w)

= (-3w^2+3w)^2

= (-3(-w-1)+3w)^2

= (6w+3)^2

= 9(4w^2+4w+1)

= 9(4(-w-1)+4w+1)

= 9*(-3)

= -27.