解:∵a1,a2,a3线性相关∴a1,a2,a3行列式为0故: a 1 1A=1 -2 1 1 1 -2 ,该矩阵的行列式为0,则|A|=(4-1)a-(-2-1)x1+(1+2)x1=0 3a+3+3=0 3a+6=0 3a=-6 a=-2