Question

设三阶矩阵A,α1=（a,a,1),α2=(a,1,a),α3=(1,a,a),当a=多少时,R(A)=2?

Answer 1

a=1,R(A) =1

det A

=|α1 α2 α3|

|a a 1|

=|a 1 a|

|1 a a|

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

=-2a^3+3a^2-1

det A =0

2a^3-3a^2 +1 =0

(a-1)(2a^2-a -1 )=0

(a-1)(2a+1)(a-1)=0

a=1 or -1/2

ie

a=-1/2 R(A)=2