λ取何值,如图非齐次线性方程组(1)有唯一解?(2)无解?(3)有无穷多解?
2个回答
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系数矩阵行列式 |A| =
|3-2λ 2-λ 1|
|2-λ 2-λ 1|
|1 1 2-λ|
|A| =
|1-λ 0 0|
|2-λ 2-λ 1|
|1 1 2-λ|
|A| = (1-λ)(3-4λ+λ^2) = (3-λ) (1-λ)^2
当 λ ≠ 3 且 λ ≠ 1 时, |A| ≠ 0, 方程组有唯一解;
当 λ = 3 时,增广矩阵 (A, b) =
[-3 -1 1 3]
[-1 -1 1 1]
[ 1 1 -1 1]
初等行变换为
[ 1 1 -1 1]
[0 2 -2 6]
[0 0 0 2]
初等行变换为
[ 1 0 0 -2]
[0 1 -1 3]
[0 0 0 1]
r(A, b) = 3, r(A) = 2 , 方程组无解;
当 λ = 1 时,增广矩阵 (A, b) =
[ 1 1 1 1]
[1 1 1 1]
[ 1 1 1 1]
初等行变换为
[ 1 1 1 1]
[ 0 0 0 0]
[ 0 0 0 0]
r(A, b) = r(A) =1 < 3, 方程组有无穷多解, 方程组已化为
x1 = 1-x2-x3
取 x2 = x3 = 0 得特解 (1,0,0)^T
导出组 是
x1 = x2-x3
取 x2 = -1, x3 = 0 得基础解系 (1,-1,0)^T
取 x2 = 0, x3 = -1 得基础解系 (1,0,-1)^T
此时通解是 x = k (1,-1,0)^T + c (1,0,-1)^T + (1,0,0)^T。
|3-2λ 2-λ 1|
|2-λ 2-λ 1|
|1 1 2-λ|
|A| =
|1-λ 0 0|
|2-λ 2-λ 1|
|1 1 2-λ|
|A| = (1-λ)(3-4λ+λ^2) = (3-λ) (1-λ)^2
当 λ ≠ 3 且 λ ≠ 1 时, |A| ≠ 0, 方程组有唯一解;
当 λ = 3 时,增广矩阵 (A, b) =
[-3 -1 1 3]
[-1 -1 1 1]
[ 1 1 -1 1]
初等行变换为
[ 1 1 -1 1]
[0 2 -2 6]
[0 0 0 2]
初等行变换为
[ 1 0 0 -2]
[0 1 -1 3]
[0 0 0 1]
r(A, b) = 3, r(A) = 2 , 方程组无解;
当 λ = 1 时,增广矩阵 (A, b) =
[ 1 1 1 1]
[1 1 1 1]
[ 1 1 1 1]
初等行变换为
[ 1 1 1 1]
[ 0 0 0 0]
[ 0 0 0 0]
r(A, b) = r(A) =1 < 3, 方程组有无穷多解, 方程组已化为
x1 = 1-x2-x3
取 x2 = x3 = 0 得特解 (1,0,0)^T
导出组 是
x1 = x2-x3
取 x2 = -1, x3 = 0 得基础解系 (1,-1,0)^T
取 x2 = 0, x3 = -1 得基础解系 (1,0,-1)^T
此时通解是 x = k (1,-1,0)^T + c (1,0,-1)^T + (1,0,0)^T。
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