2)无解?(3)有无穷多解?并求出其通解. 讨论a取何值时,下列非齐次线性方程组(1)无解（2）有唯一解...
\u8ba8\u8bbaa\u53d6\u4f55\u503c\u65f6\uff0c\u7ebf\u6027\u65b9\u7a0b\u7ec4\uff081\uff09\u6709\u60df\u4e00\u89e3\uff082\uff09\u65e0\u89e3\uff083\uff09\u6709\u65e0\u7a77\u591a\u89e3\uff0c\u5e76\u6c42\u901a\u89e3\u3002\u9700\u8981\u8be6\u7ec6\u8fc7\u7a0b\u3002\u7cfb\u6570\u884c\u5217\u5f0f |A|=a(1-a)
\u6240\u4ee5 a\u22600 \u4e14 a\u22601 \u65f6\u65b9\u7a0b\u7ec4\u6709\u552f\u4e00\u89e3
\u5f53a=0\u65f6, \u589e\u5e7f\u77e9\u9635=
0 1 1 2
1 0 1 2
1 0 1 1
-->
0 1 1 2
1 0 1 2
0 0 0 -1
\u65b9\u7a0b\u7ec4\u65e0\u89e3
\u5f53a=1\u65f6, \u589e\u5e7f\u77e9\u9635=
1 1 1 2
1 1 1 2
1 2 1 1
-->
1 1 1 2
0 0 0 0
0 1 0 -1
-->
1 0 1 3
0 1 0 -1
0 0 0 0
\u65b9\u7a0b\u7ec4\u6709\u65e0\u7a77\u591a\u89e3 (3,-1,0)^T+k(-1,0,1)^T
\u539f\u9898
x1 + x2 - x3 = 1 x1 = 1 + x3 - x2
2x1 +3x2 + ax3 = 3
x1 + ax2 + 3x3 = 2
\u6d88\u53bb x1 \u5f97\u5230
x2 + \uff08a + 2 )x3 = 1 x2 = 1 - \uff08a + 2 )x3
( a -1 )x2 + 4x3 = 1
\u6d88\u53bb x2 \u5f97\u5230
( 6-a -a^2)x3 = 2 - a
( a-2 )(a + 3) x3 = a - 2
\u5f53 a = -3 \u65f6\uff0c \u5de6\u4fa7\u59cb\u7ec8\u4e3a0\uff0c \u56e0\u6b64\u65e0\u89e3
a \u2260 2 \u4e14 a \u2260 -3 \u65f6 \u6709 \u552f\u4e00\u89e3
a = 2 \u65f6\uff0c \u6709 \u65e0\u7a77\u591a\u89e3
x2 = 1 - \uff08a + 2 )x3
x1 = 1 + x3 - x2 = ( a + 3 )x3
\u5373
x1 = ( a + 3 )x3
x2 = 1 - \uff08a + 2 )x3
x3 = x3
计算过程见上图，结果你自己分析一下，希望能帮助到你！
系数矩阵行列式 |A| =
| 2 2 4-2λ|
|2-λ 2-λ 1|
|3-2λ 2-λ 1|
第 3 行 减去 第 2 行， 得
| 2 2 4-2λ|
|2-λ 2-λ 1|
|1-λ 0 0|
|A| = (1-λ)*
| 2 4-2λ|
|2-λ 1|
|A| = (1-λ)(2-8+8λ-2λ^2)
= -2(1-λ)(λ^2-4λ+3) = 2(λ-3)(λ-1)^2.
(1) λ ≠ 1 且 λ ≠ 3 时， |A| ≠ 0， 方程组有唯一解。
(2) λ = 3 时，增广矩阵 (A, b) =
[ 2 2 -2 2]
[-1 -1 1 1]
[-3 -1 1 3]
初等行变换为
[ 1 1 -1 1]
[ 0 0 0 2]
[ 0 2 -2 6]
初等行变换为
[ 1 1 -1 1]
[ 0 1 -1 3]
[ 0 0 0 1]
r(A, b) = 3， r(A) = 2， 方程组无解。
(3) λ = 1 时，增广矩阵 (A, b) =
[ 2 2 2 2]
[ 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。
此时方程组化为 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 ;
λ = 1 时，方程组的通解是
x = (1, 0, 0)^T + k(1, -1, 0)^T + c(1, 0, -1)^T
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