Question

设矩阵A=(2 -1 1 0,0 -1 -1 -2,1 0 0 -1,1 1 0-1)的秩

Answer 1

2 -1 1 0

0 -1 -1 -2

1 0 0 -1

1 1 0 -1

r1-2r3,r4-r3

0 -1 1 2

0 -1 -1 -2

1 0 0 -1

0 1 0 0

r1+r4,r2+r4

0 0 1 2

0 0 -1 -2

1 0 0 -1

0 1 0 0

r2+r1

0 0 1 2

0 0 0 0

1 0 0 -1

0 1 0 0

r(A)=3