
线性代数,计算行列式
2个回答
展开全部
不是范德蒙行列式。后 3 列分别减去第 1 列, 得 D =
|1 0 0 0|
|a1 a2-a1 a3-a1 a4-a1|
|(a1)^2 (a2)^2-(a1)^2 (a3)^2-(a1)^2 (a4)^2-(a1)^2|
|(a1)^4 (a2)^4-(a1)^4 (a3)^4-(a1)^4 (a4)^4-(a1)^4|
D =
|a2-a1 a3-a1 a4-a1|
|(a2)^2-(a1)^2 (a3)^2-(a1)^2 (a4)^2-(a1)^2|
|(a2)^4-(a1)^4 (a3)^4-(a1)^4 (a4)^4-(a1)^4|
D = (a2-a1)(a3-a1)(a4-a1)*
|1 1 1|
|a2+a1 a3+a1 a4+a1|
|[(a2)^2+(a1)^2](a2+a1) [(a3)^2+(a1)^2](a3+a1) [(a4)^2+(a1)^2](a4+a1)|
D = (a2-a1)(a3-a1)(a4-a1)*
|1 0 0|
|a2+a1 a3-a2 a4-a2|
|[(a2)^2+(a1)^2](a2+a1) (a3-a2)x (a4-a2)y|
其中:x = (a1)^2+(a2)^2+(a3)^2+a1a2+a1a3+a2a3
y = (a1)^2+(a2)^2+(a4)^2+a1a2+a1a4+a2a4
则 D = (a2-a1)(a3-a1)(a4-a1)(a3-a2)(a4-a2)(y-x)
= (a2-a1)(a3-a1)(a4-a1)(a3-a2)(a4-a2)(a4-a3)(a1+a2+a3+a4)
|1 0 0 0|
|a1 a2-a1 a3-a1 a4-a1|
|(a1)^2 (a2)^2-(a1)^2 (a3)^2-(a1)^2 (a4)^2-(a1)^2|
|(a1)^4 (a2)^4-(a1)^4 (a3)^4-(a1)^4 (a4)^4-(a1)^4|
D =
|a2-a1 a3-a1 a4-a1|
|(a2)^2-(a1)^2 (a3)^2-(a1)^2 (a4)^2-(a1)^2|
|(a2)^4-(a1)^4 (a3)^4-(a1)^4 (a4)^4-(a1)^4|
D = (a2-a1)(a3-a1)(a4-a1)*
|1 1 1|
|a2+a1 a3+a1 a4+a1|
|[(a2)^2+(a1)^2](a2+a1) [(a3)^2+(a1)^2](a3+a1) [(a4)^2+(a1)^2](a4+a1)|
D = (a2-a1)(a3-a1)(a4-a1)*
|1 0 0|
|a2+a1 a3-a2 a4-a2|
|[(a2)^2+(a1)^2](a2+a1) (a3-a2)x (a4-a2)y|
其中:x = (a1)^2+(a2)^2+(a3)^2+a1a2+a1a3+a2a3
y = (a1)^2+(a2)^2+(a4)^2+a1a2+a1a4+a2a4
则 D = (a2-a1)(a3-a1)(a4-a1)(a3-a2)(a4-a2)(y-x)
= (a2-a1)(a3-a1)(a4-a1)(a3-a2)(a4-a2)(a4-a3)(a1+a2+a3+a4)

2024-10-28 广告
作为上海华然企业咨询有限公司的一员,我们深知大模型测试对于企业数字化转型与智能决策的重要性。在应对此类测试时,我们注重数据的精准性、算法的先进性及模型的适用性,确保大模型能够精准捕捉市场动态,高效分析企业数据,为管理层提供科学、前瞻的决策支...
点击进入详情页
本回答由上海华然企业咨询提供
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询
广告 您可能关注的内容 |