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计算下列n(>=2)阶方阵的行列式

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计算下列n(>=2)阶方阵的行列式
1.
a+1 -1 1 -1
A= 1 a-1 1 -1
1 -1 a+1 -1
1 -1 1 a-1
2.
x a .a
An= a x .a
...
a a .x
3.
an bn
..
..
..
A2n = a1b1
c1d1
..,其中未写出的元素均为零
..
..
cn dn
4.
a1 x1
x2 a2
..
..
Xn-1 an
/>1. a+1 -1 1 -1 a -1 1 -1 1 -1 1 -1
A= 1 a-1 1 -1 = a a-1 1 -1 = a× 1 a-1 1 -1
1 -1 a+1 -1 a -1 a+1 -1 1 -1 a+1 -1
1 -1 1 a-1 a -1 1 a-1 1 -1 1 a+1
1 0 0 0
=a× 1 a 0 0 = a×a³ = a^4
1 0 a 0
1 0 0 a
2. x a . a x+(n-1)a a ... a 1 a ... a
An= a x . a = x+(n-1)a x ... a = [x+(n-1)a]× 1 x ... a
. . .
a a . x x+(n-1)a a ... x 1 a ... x
1 0 ... 0
=[x+(n-1)a]× 1 x-a 0 = [x+(n-1)a]×(x-a)^(n-1)
1
1 0 ... x-a
3.
an bn
. .
. .
. .
A2n = a1b1
c1d1
. . ,其中未写出的元素均为零
. .
. .
cn dn
Ann An2n
= A2nn A2n2n =|Ann|×|A2n2n|-|An2n|×|A2nn|
=a1···an×d1···dn-b1···bn×c1···cn
4.最后一行倒数第二个不应该是xn-1吧,应该是xn
a1 x1
x2 a2
. .
. .
xn an
a2 x2 a2
x3
=a1× +(-1)^(n+1)x1 an-1
xn an xn
=a1a2···an+(-1)^(n+1)x1x2···xn
辛苦做完,希望帮助到你,望采纳,谢谢~