设x1b1+x2b2+x3b3=0,则有[(m-1)x1+x2-x3]a1+[3x1+(m+1)x2-(m+1)x3]a2+[x1+x2+(m-1)x3]a3=0由于向量组a1,a2,a3线性无关所以有关于x1,x2,x3的方程组(1)(m-1)x1+x2-x3=03x1+(m+1)x2-(m+1)x3=0x1+x2+(m-1)x3=0若要向量组...