求矩阵A的秩 1 2 3 4 -1 -1 -4 -2 3 4 11 8_数学解答

求矩阵A的秩 1 2 3 4 -1 -1 -4 -2 3 4 11 8

人气：494 ℃　时间：2020-06-06 09:43:15

解答

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.