Question

线性代数，解方程组，有没有会的大佬！

Answer 1

增广矩阵 (A, b) =
[2 1 0 6 6 7]
[1 0 -1 4 2 6]
[3 0 -3 12 2 10]
初等行变换为
[1 0 -1 4 2 6]
[0 1 2 -2 2 -5]
[0 0 0 0 -4 -8]
初等行变换为
[1 0 -1 4 0 2]
[0 1 2 -2 0 -9]
[0 0 0 0 1 2]
r(A, b) = r(A) = 3 < 5, 方程组有无穷多解。

方程组化为
x1 = 2 + x3 - 4x4
x2 = -9 - 2x3 + 2x4
x5 = 2
取 x3 = x4 = 0， 得特解 (2, -9, 0, 0, 2)^T,
导出组是
x1 = x3 - 4x4
x2 = - 2x3 + 2x4
x5 = 0
取 x3 = 1，x4 = 0 得基础解系 (1, -2, 1, 0, 0)^T,
取 x3 = 0，x4 = 1 得基础解系 (-4, 2, 0, 1, 0)^T,
方程组通解是
x = (2, -9, 0, 0, 2)^T + k(1, -2, 1, 0, 0)^T + c (-4, 2, 0, 1, 0)^T。
k， c 为任意常数。