第1个回答 2014-11-26
(A, b) =
[1 3 5 -4 0 1]
[1 3 2 -2 1 -1]
[1 -2 1 -1 -1 3]
[1 -4 1 1 -1 3]
[1 2 1 -1 1 -1]
行初等变换为
[1 3 5 -4 0 1]
[0 0 -3 2 1 -2]
[0 -5 -4 3 -1 2]
[0 -7 -4 5 -1 2]
[0 -1 -4 3 1 -2]
行初等变换为
[1 0 -7 5 3 -5]
[0 1 4 -3 -1 2]
[0 0 -3 2 1 -2]
[0 0 16 -12 -6 12]
[0 0 24 -16 -8 16]
行初等变换为
[1 0 -7 5 3 -5]
[0 1 4 -3 -1 2]
[0 0 -3 2 1 -2]
[0 0 24 -18 -9 18]
[0 0 3 -2 -1 2]
行初等变换为
[1 0 -7 5 3 -5]
[0 1 4 -3 -1 2]
[0 0 3 -2 -1 2]
[0 0 0 -2 -1 2]
[0 0 0 0 0 0]
行初等变换为
[1 0 -7 5 3 -5]
[0 1 4 -3 -1 2]
[0 0 3 0 0 0]
[0 0 0 2 1 -2]
[0 0 0 0 0 0]
行初等变换为
[2 0 -14 10 6 -10]
[0 2 8 -6 -2 4]
[0 0 1 0 0 0]
[0 0 0 2 1 -2]
[0 0 0 0 0 0]
行初等变换为
[2 0 -14 0 1 0]
[0 2 8 0 1 -2]
[0 0 1 0 0 0]
[0 0 0 2 1 -2]
[0 0 0 0 0 0]
行初等变换为
[2 0 0 0 1 0]
[0 2 0 0 1 -2]
[0 0 1 0 0 0]
[0 0 0 2 1 -2]
[0 0 0 0 0 0]
r(A) = r(A<b)=4<5, 方程组有无穷多解。
方程组同解变形为
2x1=-x5
2x2=-2-x5
x3=0
2x4=-2-x5
令 x5=0 , 得特解 (0, -1, 0, -1, 0)^T,
导出组即对应的齐次方程是
2x1=-x5
2x2=-x5
x3=0
2x4=-x5
令 x5=-2 , 得基础解系 (1, 1, 0, 1, -2)^T,
方程组的通解是
x=(0, -1, 0, -1, 0)^T+k(1, 1, 0, 1, -2)^T,
其中 k 为任意常数。本回答被提问者采纳