设a=(1 5 2 ) b=(1 1 1) (这两个是矩阵!)a b是否等价,求p和q令paq=b.3 4 7 1 -1 1

1.因为a,b的秩都是2,所以它们等价
2.这类题目太麻烦了!
A=
1 5 2
3 4 7
r2-3r1 对应初等矩阵 E2(2,1(-3))
1 5 2
0 -11 1
c2-5c1,c3-2c1 E3(1,2(-5)) E3(1,3(-2))
1 0 0
0 -11 1
c2+11c3,c2c3 E(3,2(11)) E3(2,3)
1 0 0
0 1 0
所以 E2(2,1(-3)) A E3(1,2(-5)) E3(1,3(-2)) E(3,2(11)) E3(2,3) = 等价标准形
B =
1 1 1
1 -1 1
r2-r1 E2(2,1(-1))
1 1 1
0 -2 0
c2-c1,c3-c1 E3(1,2(-1)) E3(1,3(-1))
1 0 0
0 -2 0
c2*(-1/2) E3(2(-1/2))
1 0 0
0 1 0
所以 E2(2,1(-1)) B E3(1,2(-1)) E3(1,3(-1)) E3(2(-1/2)) = 等价标准形
E2(2,1(-3)) A E3(1,2(-5)) E3(1,3(-2)) E(3,2(11)) E3(2,3)
= E2(2,1(-1)) B E3(1,2(-1)) E3(1,3(-1)) E3(2(-1/2))
B = E2(2,1(1))E2(2,1(-3)) A E3(1,2(-5)) E3(1,3(-2)) E(3,2(11)) E3(2,3)E3(2(-2)) E3(1,3(1))E3(1,2(1))
P = E2(2,1(1)) E2(2,1(-3)) = [1 0;-2 1].
Q = E3(1,2(-5)) E3(1,3(-2)) E(3,2(11)) E3(2,3)E3(2(-2)) E3(1,3(1))E3(1,2(1))
= [1 5 -26;0 0 1;0 -2 11]
则有 PAQ = B.