Question

已知线性方程组 X1+X2+2X3-3X4=1 X1+2X2-X3+2X4=3 2X1+3X2+X3-X4=B

已知线性方程组 X1+X2+2X3-3X4=1 X1+2X2-X3+2X4=3 2X1+3X2+X3-X4=B
B取何值，方程组无解
B取何值，方程组有解，并求通解，
请高手帮忙，谢谢

Answer 1

首先是系数矩阵的秩
1 1 2 -3
1 2 -1 2
2 3 1 -1
矩阵初等变换得到
1 1 2 -3
0 1 -3 5
0 0 0 0
秩为2
增广矩阵
1 1 2 -3 1
1 2 -1 2 3
2 3 1 -1 B
初等变换
1 1 2 -3 1
0 1 -3 5 2
0 0 0 0 B-2
使方程组无解 增广矩阵秩和系数矩阵秩不同 当B=2时秩相同 B不=2时秩不同
通解=特解+基础解系
当B=2时 方程组解 {2,0,1,1}+k1{-5,3,1,0}+k2{8,-5,0,1} k1,k2为任意常数 x3和x4为自变量

追问

最后经过初等变换后那个最后一位是不是 B-4啊？
B=4有解，B不等于4，无解，关键是通解如何求啊，请高手给出详细的步骤啊，谢谢

追答

特解是让自变量全为1， 这里的自变量是X3，X4当为1时可以求得一组特解 {2,0,1,1} 基础解析是把非齐次方程变成齐次方程的解
齐次方程是
X1+X2+2X3-3X4=0
X1+2X2-X3+2X4=0
2X1+3X2+X3-X4=0 就是等号后的数都变成0
解齐次方程步骤和上面一样 就是最后变换系数矩阵后取 x3和x4为自变量 令x3=1 x4=0 得到{-5,3,1,0} 令X3=0，X4=1得到 {8,-5,0,1}
k1{-5,3,1,0}+k2{8,-5,0,1} 就是齐次方程组的解也是基础解系

Answer 2

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