请问这道线性代数矩阵题怎么求解,求详细过程,谢谢啦
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增广矩阵 (A, b) =
[1 1 2 a]
[3 -1 -6 a+2]
[1 4 11 a+3]
初等行变换为
[1 1 2 a]
[0 -4 -12 -2a+2]
[0 3 9 3]
初等行变换为
[1 1 2 a]
[0 1 3 1]
[0 -4 -12 -2a+2]
初等行变换为
[1 0 -1 a-1]
[0 1 3 1]
[0 0 0 -2a+6]
当 -2a+6 = 0 即 a = 3 时, r(A, b) = r(A) = 2 < 3,
方程组有无穷多解。 此时方程组化为:
x1 = 2+x3
x2 = 1-3x3
取 x3 = 0 得特解 (2, 1, 0)^T.
导出组即
x1 = x3
x2 = -3x3
取 x3 = 1 得基础解系 (1, -3, 1)^T.
则 a = 3 时方程组通解是 x = k(1, -3, 1)^T + (2, 1, 0)^T.
[1 1 2 a]
[3 -1 -6 a+2]
[1 4 11 a+3]
初等行变换为
[1 1 2 a]
[0 -4 -12 -2a+2]
[0 3 9 3]
初等行变换为
[1 1 2 a]
[0 1 3 1]
[0 -4 -12 -2a+2]
初等行变换为
[1 0 -1 a-1]
[0 1 3 1]
[0 0 0 -2a+6]
当 -2a+6 = 0 即 a = 3 时, r(A, b) = r(A) = 2 < 3,
方程组有无穷多解。 此时方程组化为:
x1 = 2+x3
x2 = 1-3x3
取 x3 = 0 得特解 (2, 1, 0)^T.
导出组即
x1 = x3
x2 = -3x3
取 x3 = 1 得基础解系 (1, -3, 1)^T.
则 a = 3 时方程组通解是 x = k(1, -3, 1)^T + (2, 1, 0)^T.
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