求向量a1=(1,2,1,1),a2=(0,-2,1,1),a3=(-1,-2,1,3),a4=(3,0,1,-4)的秩

(a1,a2,a3,a4)=
1 0 -1 3
2 -2 -2 0
1 1 1 1
1 1 3 -4

r2-2r1,r3-r1,r4-r1
1 0 -1 3
0 -2 0 -6
0 1 2 -2
0 1 4 -7

r2*(-1/2),r3-r2,r4-r2
1 0 -1 3
0 1 0 3
0 0 2 -5
0 0 4 -10

r4-2r3
1 0 -1 3
0 1 0 3
0 0 2 -5
0 0 0 0

r3*(1/2),r1+r3
1 0 0 1/2
0 1 0 3
0 0 1 -5/2
0 0 0 0

所以向量组的秩为3, a1,a2,a3 是一个极大无关组
a4 = (1/2)a1+3a2-(5/2)a3.

向量组a1=(1,2,1,1),a2=(0,-2,1,1),a3=(-1,-2,1,3),a4=(3,0,1,-4)的秩是4
b1=(1,0,0,0),b2=(0,1,0,0),b3=(0,0,1,0),b4=(0,0,0,1)是其一个极大线性无关组
a1=b1+2b2+b3+b4
a2=-2b2+b3+b4
a3=-b1-2b2+b3+3b4
a4=3b1...