秋风无泪1
春芽
共回答了24个问题采纳率:91.7% 举报
x1 x2 x3
x3 x1 x2
x2 x3 x1
c1+c2+c3
x1+x2+x3 x2 x3
x1+x2+x3 x1 x2
x1+x2+x3 x3 x1
r2-r1,r3-r1
x1+x2+x3 x2 x3
0 x1-x2 x2-x3
0 x3-x2 x1-x3
行列式= (x1+x2+x3)[(x1-x2)(x1-x3)-(x2-x3)(x2-x3)]
= (x1+x2+x3)(x1^2 - x1x2 - x1x3 + x2^2 - x2x3 + x3^2)
由已知x1,x2,x3是x*3+qx+p=0的解
由要与系数的关系得 x1+x2+x3=0
所以行列式 = 0.
1年前
9