Question

线性数学 求通解 求大神详细步骤

Answer 1

8. 增广矩阵 (A, b) =
[1 1 -1 1 2]
[2 1 3 -2 4]
[1 -a -1 1 0]
初等行变换为
[1 1 -1 1 2]
[0 -1 5 -4 0]
[0 -a-1 0 0 -2]
初等行变换为
[1 0 4 -3 2]
[0 1 -5 4 0]
[0 -a-1 0 0 -2]
将 解 (1, 1, 1, 1)^T 代入得 -a-1 = -2, a = 1.

此时，进一步初等行变换为
[1 0 4 -3 2]
[0 1 -5 4 0]
[0 1 0 0 1]
初等行变换为
[1 0 4 -3 2]
[0 1 0 0 1]
[0 0 -5 4 -1]
初等行变换为
[1 0 0 1/5 6/5]
[0 1 0 0 1]
[0 0 1 -4/5 1/5]
方程组化为

x1 = 6/5 - (1/5)x4
x2 = 1
x3 = 1/5 + (4/5)x4
特解是 (1, 1, 1, 1)^T;
导出组 是
x1 = - (1/5)x4
x2 = 0
x3 = (4/5)x4
取 x4 = 5， 得 Ax = 0 的基础解系 (-1, 0, 4, 5)^T
Ax = b 的通解是 x = k(-1, 0, 4, 5)^T + (1, 1, 1, 1)^T

Answer 2

8. 增广矩阵 (A, b) =
[1 1 -1 1 2]
[2 1 3 -2 4]
[1 -a -1 1 0]
初等行变换为
[1 1 -1 1 2]
[0 -1 5 -4 0]
[0 -a-1 0 0 -2]
初等行变换为
[1 0 4 -3 2]
[0 1 -5 4 0]
[0 -a-1 0 0 -2]
将 解 (1, 1, 1, 1)^T 代入得 -a-1 = -2, a = 1.

此时，进一步初等行变换为
[1 0 4 -3 2]
[0 1 -5 4 0]
[0 1 0 0 1]
初等行变换为
[1 0 4 -3 2]
[0 1 0 0 1]
[0 0 -5 4 -1]
初等行变换为
[1 0 0 1/5 6/5]
[0 1 0 0 1]
[0 0 1 -4/5 1/5]
方程组化为

x1 = 6/5 - (1/5)x4
x2 = 1
x3 = 1/5 + (4/5)x4
特解是 (1, 1, 1, 1)^T;
导出组 是
x1 = - (1/5)x4
x2 = 0
x3 = (4/5)x4
取 x4 = 5， 得 Ax = 0 的基础解系 (-1, 0, 4, 5)^T
Ax = b 的通解是 x = k(-1, 0, 4, 5)^T + (1, 1, 1, 1)^T