Question

线性代数题目，急，在线等

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Answer 1

4 |λE-A| =
|λ-3 1 2|
|-2 λ 2|
|-2 1 λ+1|
|λE-A| =
|λ-3 1 2|
|1-λ λ-1 0|
|-2 1 λ+1|
|λE-A| =
|λ-2 1 2|
| 0 λ-1 0|
|-1 1 λ+1|
|λE-A| = (λ-1)(λ^2-λ) = λ(λ-1)^2
特征值 λ = 0, 1， 1。
对于重特征值 λ = 1， （λE-A）=
[-2 1 2]
[-2 1 2]
[-2 1 0]
初等行变换为
[-2 1 2]
[ 0 0 -2]
[ 0 0 0]
初等行变换为
[-2 1 0]
[ 0 0 1]
[ 0 0 0]
对应的特征向量 只有一个， 是 (1, 2, 0)^T，
故该矩阵不能相似于对角矩阵。
5. (A, b) =
[1 2 3 1 -3 5]
[2 1 0 2 -6 1]
[3 4 5 6 -3 12]
[1 1 1 3 1 4]
初等行变换为
[1 2 3 1 -3 5]
[0 -3 -6 0 0 -9]
[0 -2 -1 3 6 -3]
[0 -1 -2 2 4 -1]
初等行变换为
[1 0 -1 1 -3 -1]
[0 1 2 0 0 3]
[0 0 3 3 6 3]
[0 0 0 2 4 2]
初等行变换为
[1 0 0 2 -1 0]
[0 1 0 -2 -4 1]
[0 0 1 1 2 1]
[0 0 0 1 2 1]
初等行变换为
[1 0 0 0 -5 -2]
[0 1 0 0 0 3]
[0 0 1 0 0 0]
[0 0 0 1 2 1]
r(A, b)) = r(A) = 4 < 5, 方程组有无穷多解。
同解方程组是
x1 = -2+5x5
x2 = 3
x3 = 0
x4 = 1-2x5,
得特解 (-2, 3, 0, 1, 0)^T,
导出组是
x1 = 5x5
x2 = 0
x3 = 0
x4 = -2x5,
得基础解系 (5, 0, 0, -2, 1)^T,
则方程组通解是 x = k (5, 0, 0, -2, 1)^T+ (-2, 3, 0, 1, 0)^T

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