Question

设有线性方程组x1+2x2-x3+4x4+3x5=6；x1-x2+2x3+3x4+4x5=9；3x1+3x2 +11x4+10x5=k，k取何值时，上述线性方程组有解，求出方程组的解，并用基础解系表示该方程组的通解

Answer 1

问：设有线性方程组x1+2x2-x3+4x4+3x5=6；x1-x2+2x3+3x4+4x5=9；3x1+3x2 +11x4+10x5=k，k取何值时，上述线性方程组有解，求出方程组的解，并用基础解系表示该方程组的通解

答：x1+x2+x3+4x4-3x5=0x1-x2+3x-2x4-x5=02x1+x2+3x3+5x4-5x5=03x1+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 + 2x5x2 = 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