Question

a=(123,03-2,当t=时,r(a)=2

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?
这道题我计算的最后结果是当a=1时,R（A）=1,当a不等于1是,R（A）等于3,无论如何都不会等于2,最后结果是a=-1/2.

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