线性代数问题求解,需要过程!
1个回答
展开全部
(A, b) =
[5 10 -8 11 12]
[3 6 -5 7 a]
[1 2 -1 1 3]
初等行变换为
[1 2 -1 1 3]
[0 0 -3 6 -3]
[0 0 -2 4 a-9]
初等行变换为
[1 2 0 -1 4]
[0 0 1 -2 1]
[0 0 0 0 a-7]
a = 7 时, r(A,b) = r(A) = 2 < 4, 方程组有无穷多解,
此时方程组化为
x1 = 4 - 2x2 + x4
x3 = 1 - 2x4
取 x2 = x4 = 0, 得特解 (4, 0, 1, 0)^T.
导出组为
x1 = - 2x2 + x4
x3 = - 2x4
取 x2 = -1, x4 = 0, 得基础解系 (2, -1, 2, 0)^T,
取 x2 = 0, x4 = 1, 得基础解系 (1, 0, -2, 1)^T.
方程组通解是
x = (4, 0, 1, 0)^T + k (2, -1, 2, 0)^T + c(1, 0, -2, 1)^T。
[5 10 -8 11 12]
[3 6 -5 7 a]
[1 2 -1 1 3]
初等行变换为
[1 2 -1 1 3]
[0 0 -3 6 -3]
[0 0 -2 4 a-9]
初等行变换为
[1 2 0 -1 4]
[0 0 1 -2 1]
[0 0 0 0 a-7]
a = 7 时, r(A,b) = r(A) = 2 < 4, 方程组有无穷多解,
此时方程组化为
x1 = 4 - 2x2 + x4
x3 = 1 - 2x4
取 x2 = x4 = 0, 得特解 (4, 0, 1, 0)^T.
导出组为
x1 = - 2x2 + x4
x3 = - 2x4
取 x2 = -1, x4 = 0, 得基础解系 (2, -1, 2, 0)^T,
取 x2 = 0, x4 = 1, 得基础解系 (1, 0, -2, 1)^T.
方程组通解是
x = (4, 0, 1, 0)^T + k (2, -1, 2, 0)^T + c(1, 0, -2, 1)^T。
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