Question

已知矩阵A=[1 1 1,1 3 -1,2 -x 6,-2 -2x 0]的秩为2,求X的值

Answer 1

对A矩阵做行的初等变换,解出x=2,r(A)=2

[1 1 1 ] [1 1 1 ] [1 1 1] [ 1 1 1] [1 1 1]

[1 3 -1 ] [0 2 -2 ] [0 1 -1 ] [ 0 1 -1] [ 0 1 -1]

[2 -x 6 ] [0 -x-2 4 ] [0 -x-2 4] [ 0 0 2-x] [ 0 0 0 ]

[-2 -2x 0 ] [0 2-2x 2 ] [ 0 1-x 1] [ 0 0 2-x] [ 0 0 0]

可见只有：x = 2 时,矩阵：A 的秩为2.