线性方程组问题,若方程组x+y+z=a;x+y-z=b;x-y+z=c;有唯一解,且x=1,那么abc1-1111-1该行列式的值是多少?

Question

线性方程组问题,
若方程组
x+y+z=a;
x+y-z=b;
x-y+z=c;
有唯一解,且x=1,那么
a b c
1 -1 1
1 1 -1
该行列式的值是多少?

Answer 1

由Crammer 法则
x = D1/D = 1
即 D1 = D =
1 1 1
1 1 -1
1 -1 1
= -4.
即有 D1 =
a 1 1
b 1 -1
c -1 1
=
a b c
1 1 -1
1 -1 1
= -4.
故所求行列式等于 4.