线性代数,向量,以知n维向量a1,a2,a3线性无关,证明3a1+2a2,a2-a3,4a3...答:由已知 (3a1+2a2,a2-a3,4a3-5a1)=(a1,a2,a3)K K= 3 0 -5 2 1 0 0 -1 4 = 22 ≠0 所以K是可逆矩阵 所以 r(3a1+2a2,a2-a3,4a3-5a1)=r((a1,a2,a3)K)=r(a1,a2,a3)=3 所以 3a1+2a2,a2-a3,4a3-5a1线性无关....
设向量组a1,a2,a3线性无关, 则下列向量组线性相关的是答:因为向量组a1,a2,a3线性无关,所以可以知道det(a1,a2,a3)≠0,所以可以知道矩阵(a1,a2,a3)为非奇异矩阵,即矩阵(a1,a2,a3)为可逆矩阵。因为矩阵(a1,a2,a3)为可逆矩阵,所以会存在(a1,a2,a3)逆,可以令A=(a1,a2,a3)逆,所以有A(b1,b2,b3)=K, K为一3阶方阵 。令X=(b1,...
已知n维向量a1,a2,a3,a4,a5线性无关,A是n阶可逆矩阵,证明Aa1,Aa2,Aa3...答:因为 (Aa1,Aa2,Aa3,Aa4,Aa5) = A(a1,a2,a3,a4,a5)且A可逆 所以 r(Aa1,Aa2,Aa3,Aa4,Aa5) =r[ A(a1,a2,a3,a4,a5)] = r(a1,a2,a3,a4,a5) = 5 所以 Aa1,Aa2,Aa3,Aa4,Aa5 线性无关.
设n维向量组a1,a2,a3线性无关,证明向量组a1+2a2, a2+2a3, a3+2a4线性...答:设存在一组数,k1,k2,k3使得 k1(a1+2a2)+k2( a2+2a3)+k3( a3+2a1)=0 整理得:(k1+2k3)a1+(2k1+k2)a2+(2k2+k3)a3=0 因为a1,a2,a3线性无关 所以 k1 + 2k3=0 2k1+k2 =0 2k2+k3=0 解得:k1=k2=k3=0 所以向量组a1+2a2, a2+2a3, a3+2a1线性无关 ...
线性代数:设a1,a2,a3,a4是n维向量,已知a1,a2线性无关,a3,a4线性无...答:线性代数:设a1,a2,a3,a4是n维向量,已知a1,a2线性无关,a3,a4线性无关,且(a1,a3)=(a1,a4)=(a2,a3)=(a2,a4)=0,证明a1,a2,a3,a4线性无关。... 线性代数:设a1,a2,a3,a4是n维向量,已知a1,a2线性无关,a3,a4线性无关,且(a1,a3)=(a1,a4)=(a2,a3)=(a2,a4)=0,证明a1,a2,a3,a4线性无关。