试确定a为何值时,使得向量组a1(3,a,0),a2(a,1,2),a3(1,-2,1),a4(2,-4,2)的秩为3
设A=(a1,a2,a3,a4),ai(i=1,2,3,4)为5维向量,若a2,a3,a4线性无关,且a4=a1+2a2
设A=(a1,a2,a3,a4),ai(i=1,2,3,4)为5维列向量,若a2,a3,a4线性无关,且a4=a1+2a
已知向量组a1,a2,a3,a4,A=(a1,a2,a3),B=(a2,a3,a4,R(A)=2,R(B)=3,证明a1
设矩阵A=[a1.a2.a3.a4],其中a2.a3.a4线性无关,a1=2a3-3a4.向量b=a1+2a2+3a3+
设a1,a2,a3均为3维列向量,矩阵A=(a1,a2,a3)并且|A|=1,B=(a1+a2+a3,a1+2a2+4a
a1a2a3a4为n元向量且r(a1,a2,a3)=2r(a2,a3,a4)=3证明 a1能由[a2,a3]线性表出 a
已知四阶方阵A=(a1,a2,a3,a4),a1,a2,a3,a4均为四维列向量,其中a2,a3,a4线性无关,a1=2
求向量组a1=(1,-1,2,4),a2=(0,3,1,2),a3=(1,8,5,10),a4=(2,-5,3,6),a
已知一个向量组a1=(1.-1,2,4),a2=(0.3.1.2)a3=(3,0,7,14)a4=(1.-1,2,0)a
高代题,设四阶方阵A=(2A1,3A2,4A3,A4),B=(A1,A2,A3,A5)其中Ai均为4×1矩阵,且detA
设矩阵A=(a1,a2,a3,a4),其中a2,a3,a4线性无关,a1=2a2-a3,向量b=a1+a2+a3+a4,
设矩阵A=(a1,a2,a3,a4)其中a2,a3,a4线性无关,a1=2a2-a3,向量b=a1+a2+a3+a4,求