Question

如何计算下列四阶行列式:|2 1 1 1|.(2)|x a a a||1 2 1 1|.|a x a a||1 1 2

Answer 1

形如

x a a a

a x a a

a a x a

a a a x

行(1)+(2)+(3)+(4)

x+3a x+3a x+3a x+3a

a x a a

a a x a

a a a x

列(4)-(1),(3)-(1),(2)-(1)

x+3a 0 0 0

a x-a 0 0

a 0 x-a 0

a 0 0 x-a

值=(x+3a)*(x-a)^3

所以第一个行列式=(2+3)=5