用行列式解线性方程组
1个回答
展开全部
2. 增广矩阵 (A, b) =
[2 1 -5 1 8]
[1 -3 0 -6 9]
[0 2 -1 2 -5]
[1 4 -7 6 0]
初等行变换为
[1 -3 0 -6 9]
[0 7 -5 13 -10]
[0 2 -1 2 -5]
[0 7 -7 12 -9]
初等行变换为
[1 -3 0 -6 9]
[0 7 -5 13 -10]
[0 2 -1 2 -5]
[0 0 -2 -1 1]
初等行变换为
[1 0 -3/2 -3 3/2]
[0 1 -1/2 1 -5/2]
[0 0 -3/2 6 15/2]
[0 0 -2 -1 1]
初等行变换为
[1 0 0 -9/4 3/4]
[0 1 0 5/4 -11/4]
[0 0 1 1/2 -1/2]
[0 0 0 27/4 27/4]
初等行变换为
[1 0 0 0 3]
[0 1 0 0 -4]
[0 0 1 0 -1]
[0 0 0 1 1]
得 x1 = 3, x2 = -4, x3 = -1, x4 = 1
[2 1 -5 1 8]
[1 -3 0 -6 9]
[0 2 -1 2 -5]
[1 4 -7 6 0]
初等行变换为
[1 -3 0 -6 9]
[0 7 -5 13 -10]
[0 2 -1 2 -5]
[0 7 -7 12 -9]
初等行变换为
[1 -3 0 -6 9]
[0 7 -5 13 -10]
[0 2 -1 2 -5]
[0 0 -2 -1 1]
初等行变换为
[1 0 -3/2 -3 3/2]
[0 1 -1/2 1 -5/2]
[0 0 -3/2 6 15/2]
[0 0 -2 -1 1]
初等行变换为
[1 0 0 -9/4 3/4]
[0 1 0 5/4 -11/4]
[0 0 1 1/2 -1/2]
[0 0 0 27/4 27/4]
初等行变换为
[1 0 0 0 3]
[0 1 0 0 -4]
[0 0 1 0 -1]
[0 0 0 1 1]
得 x1 = 3, x2 = -4, x3 = -1, x4 = 1
追问
用克拉默法则如何解答?
追答
用克拉默法代公式 xi = Di/D (D ≠ 0, i = 1, 2, ...... , n)
其中 D = |A| (系数方阵的行列式)
Di 是将 A 中第 i 列换为常数向量后的方阵的行列式。
计算 n+1 个 n 阶行列式,比较麻烦。
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询