设向量a1,a2,a3,a4线性无关,证明向量组a1-a2,a1-a3,a1-a4线性无关?答:方法1:方法2:由定义:因为a1,a2,a3,a4线性无关,所以k1a1+k2a2+k3a3+k4a4=0,当且仅当k1,k2,k3,k4全为0.又因为:c1(a1-a2)+c2(a1-a3)+c3(a1-a4)=(c1+c2+c3)a1-c1a2-c2a3-c3a4 令k1=(c1+c2+c3),k2=c1,k3=c2,k4=c3,则当且仅当(c1+c2+c3),c1,c2,c3全为0时c1(...
已知向量组a1,a2,a3线性无关,证明向量组a1+a2,3a2+2a3,a1-2a2+a3线性...答:重新分组:a1(k1+k3) + a2(k1+3k2-2k3) + a3(2k2+k3)=0 因为a1,a2,a3线性无关,所以有方程组:k1+k3=0; k1+3k2-2k3=0; 2k2+k3=0 ...行列式:1 0 1 1 3 -2 0 2 1 不等于0,所以方程只有零解,即k1,k2,k3都等于0,所以向量组a1+a2,3a2+2a3,a1-...
已知向量组a1,a2,a3,a4线性无关,证明:a1+a2,a2+a3,a3+a4,a4-a1线性无...答:设k1(a1+a2)+k2(a2+a3)+k3(a3+a4)+k4(a4-a1)=0整理后 得到(k1-k4)a1+(k1+k2)a2+(k2+k3)a3+(k3+k4)a4=0由于a1,a2,a3,a4 线性无关,则k1-k4=k1+k2=k2+k3=k3+k4=0解得k1=k2=k3=k4=0所以a1+a2,a2+a3,a3+a4,a4-a1线性无关...
向量组a1,a2,a3线性无关,β=k1a1+k2a2+k3a3,证明若k1不等于0,β,a2,a...答:(β,a2,a3) = (a1,a2,a3)K K= k1 0 0 k2 1 0 k3 0 1 因为 a1,a2,a3 线性无关 所以 r(β,a2,a3) = r(K)所以 β,a2,a3 线性无关 <=> r(K)=3 <=> |K|≠0 <=> k1≠0.