Question

线性代数题第五题

线性代数题第五题第五题怎么做

Answer 1

5. 系数矩阵行列式 |A| =
|λ -1 0 0|
|0 λ -1 0|
|0 0 λ -1|
|-1 0 0 λ|
|A| = λ^4 - 1.
当 λ ≠ ±1 时， |A| ≠ 0，方程组有唯一解。
当 λ = 1 时，（A,b) =
[ 1 -1 0 0 1]
[ 0 1 -1 0 1]
[ 0 0 1 -1 1]
[-1 0 0 1 1]
前 3 行加到第 4 行知，r(A) = 3, r(A<b) = 4,
方程组无解。
当 λ = -1 时，（A,b) =
[-1 -1 0 0 1]
[ 0 -1 -1 0 1]
[ 0 0 -1 -1 1]
[-1 0 0 -1 1]
初等行变换为
[1 1 0 0 -1]
[0 1 1 0 -1]
[0 0 1 1 -1]
[0 -1 0 1 0]
初等行变换为
[1 0 -1 0 0]
[0 1 1 0 -1]
[0 0 1 1 -1]
[0 0 1 1 -1]
初等行变换为
[1 0 0 1 -1]
[0 1 0 -1 0]
[0 0 1 1 -1]
[0 0 0 0 0]
r(A, b) = r(A) = 3 < 4,
方程组有无穷多解
x1 = -1-x4
x2 = x4
x3 = -1-x4
特解 (-1, 0, -1, 0)^T
导出组
x1 = -x4
x2 = x4
x3 = -x4
基础解系 (1, -1, 1, -1)^T
通解 x = (-1, 0, -1, 0)^T + k(1, -1, 1, -1)^T。