大一线性代数题

A = (a1, a2, a3, a4, b) =
[1 1 1 1 1]
[3 2 1 -3 0]
[0 1 2 6 k]
[5 4 3 -1 2]
初等行变换为
[1 1 1 1 1]
[0 -1 -2 -6 -3]
[0 1 2 6 k]
[0 -1 -2 -6 -3]
初等行变换为
[1 0 -1 -5 -2]
[0 1 2 6 3]
[0 0 0 0 k-3]
[0 0 0 0 0]
得 k - 3 = 0， k = 3
此时 b = -2a1 + 3a2 ( 即 b = -2a1 + 3a2 + 0a3 + 0a4 )