Question

2.求下列齐次线性方程组的基础解系: x1+x2+x3+4x4-3x5=0

2.求下列齐次线性方程组的基础解系:
x1+x2+x3+4x4-3x5=0
x1-x2+3x-2x4-x5=0
2x1+x2+3x3+5x4-5x5=0,
3x1+x2+5x3+6x4-7x5=0

Answer 1

x1+x2+x3+4x4-3x5=0
x1-x2+3x-2x4-x5=0
2x1+x2+3x3+5x4-5x5=0
3x1+x2+5x3+6x4-7x5=0
系数矩阵 A =
[1 1 1 4 -3]
[1 -1 3 -2 -1]
[2 1 3 5 -5]
[3 1 5 6 -7]
初等行变换为
[1 1 1 4 -3]
[0 -2 2 -6 2]
[0 -1 1 -3 1]
[0 -2 2 -6 2]
初等行变换为
[1 0 2 1 -2]
[0 1 -1 3 -1]
[0 0 0 0 0]
[0 0 0 0 0]
方程组化为
x1 = -2x3 - x4 + 2x5
x2 = x3 - 3x4 +x5
取 x3 = 1, x4 = x5 = 0, 得基础解系 (-2, 1, 1, 0, 0)^T;
取 x4 = -1, x3 = x5 = 0, 得基础解系 (1, 3, 0, -1, 0)^T;
取 x5 = 1, x2 = x4 = 0, 得基础解系 (2, 1, 0, 0, 1)^T;
方程组通解是
x = k1(-2, 1, 1, 0, 0)^T + k2(1, 3, 0, -1, 0)^T + k3(2, 1, 0, 0, 1)^T