求矩阵的秩A=（11-120,2-2-200,-1-1110,101-12)

Question

求矩阵的秩A=（1 1-1 2 0,2 -2 -2 0 0,-1 -1 1 1 0,1 0 1 -1 2)

Answer 1

r2-2r1,r3+r1,r4-r1
1 1 -1 2 0
0 -4 0 -4 0
0 0 0 3 0
0 -1 2 -3 2
r2*(-1/4),r4+r2
1 1 -1 2 0
0 1 0 1 0
0 0 0 3 0
0 0 2 -2 2
r3r4
1 1 -1 2 0
0 1 0 1 0
0 0 2 -2 2
0 0 0 3 0
所以 r(A)=4.