求矩阵X (1 2 3)
求矩阵X (1 2 3)
第4题怎么算呀
矩阵的四则运算。
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第1个回答 2018-04-02
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求矩阵(1 2 3)(2 1 3)(3 3 6)的特征值和特征向量。答:= (-1-λ)[(3-λ)(6-λ)-18]= (-1-λ)[λ^2-9λ]= λ(9-λ)(1+λ)所以A的特征值为 0, 9, -1 AX = 0 的基础解系为: a1 = (1,1,-1)'所以,A的属于特征值0的全部特征向量为: c1(1,1,-1)', c1为非零常数.(A-9E)X = 0 的基础解系为: a2 = (1,1,2)...
如图,求矩阵(x1, x2, x3)=?答:③-①*3,14x2-14x3=28,x2=x3+2,④-①*4,7x2-7x3=14,x2=x3+2.⑤ 把⑤代入①,x1-2(x3+2)+4x3=-5,x1=-2x3-1,∴(x1,x2,x3)=(-2t-1,t+2,t)=t(-2,1,1)+(-1,2,0).还可以用增广矩阵解。