Question

用初等变换求矩阵的秩 1 0 1 0 0 1 1 0 0 0 A= 0 1 1 0 0 0 0 1 1 0 0 1 0 1 1

Answer 1

1 0 1 0 0
1 1 0 0 0
A= 0 1 1 0 0
0 0 1 1 0
0 1 0 1 1
r2-r1得:
1 0 1 0 0
0 1 -1 0 0
0 1 1 0 0
0 0 1 1 0
0 1 0 1 1
拭r3-r2,r5-r2 得:
1 0 1 0 0
0 1 -1 0 0
0 0 2 0 0
0 0 1 1 0
0 0 1 1 1
r3 /2得:
1 0 1 0 0
0 1 -1 0 0
0 0 1 0 0
0 0 1 1 0
0 0 1 1 1
r4-r3,r5-r3
1 0 1 0 0
0 1 -1 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 1 1
r5-r4
1 0 1 0 0
0 1 -1 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
已经化为阶梯形,非零行有5个,故秩这5.