Question

求矩阵A的秩 1 2 3 4 -1 -1 -4 -2 3 4 11 8

Answer 1

A =

1 2 3 4

-1 -1 -4 -2

3 4 11 8

--> r1+r2,r3+3r2

0 1 -1 2

-1 -1 -4 -2

0 1 -1 2

r3-r1

0 1 -1 2

-1 -1 -4 -2

0 0 0 0

r(A)=2.