线性代数,看下对不对,矩阵的-1怎么解

2025-03-14 21:34:20
推荐回答(2个)
回答1:

解题错误。
2 应选 (B)
因初等变换包括倍乘变换和交换变换,二者行列式的值都变化。
det(A) = det(kB) = k^ndet(B), k≠1 时二者不等。
一次交换变换,行列式差一个负号。
5 增广矩阵初等变换正确,但特解和基础解系错误,故通解错误。
根据增广矩阵行初等变换最后结果,方程组同解变形为
x1-x2 -6x4 = 5
x3-3x4 = 3
取 x2, x4 为自由未知量,则
x1 = 5+x2+6x4
x3 = 3 +3x4
取 x2=x4=0, 得特解 (5 0 3 0)^T;
导出组即对应的齐次方程组为
x1 = x2+6x4
x3 = 3x4
取 x2=1, x4=0, 得基础解系 (1 1 0 0)^T
取 x2=0, x4=1, 得基础解系 (6 0 3 1)^T.
原非齐次方程组的通解是
x = c1(6 0 3 1)^T+c2(1 1 0 0)^T+ (5 0 3 0)^T
其中 c1, c2 为任意常数

回答2:

矩阵的逆你们还没学么?

这是矩阵的逆呀

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