若3维向量a1,a2,a3有行列式∣（a3+a2,2a2,10a1-a3）∣=1000,则行列式∣（a2+ea1+πa3,10a2-9a1,πa1）∣=

Question

若3维向量a1,a2,a3有行列式∣（a3+a2,2a2,10a1-a3）∣=1000,则行列式∣（a2+ea1+πa3,10a2-9a1,πa1）∣=

Answer 1

用这个方法试试:
(a3+a2,2a2,10a1-a3) = (a1,a2,a3)K1
K1 =
0 0 10
1 2 0
1 0 -1
|K1|=-20
两边取行列式得 |a3+a2,2a2,10a1-a3| = |a1,a2,a3||K1|
所以 1000 = |a1,a2,a3|*(-20)
所以 |a1,a2,a3|=-50.
同样有 (a2+ea1+πa3,10a2-9a1,πa1)=(a1,a2,a3)K2
K2=
e -9 π
1 10 0
π 0 0
|K2|=-10π^2
所以有 |a2+ea1+πa3,10a2-9a1,πa1|=|a1,a2,a3||K2|
= -10π^2 * (-50)
= 500π^2