文档介绍：习题九行列式（续）
a
b
b
...b
1
1
1
••- 1
b
a
b
...b
b
a
b
...b
⑴Dn =
b
b
a
...b
=(Q + (〃一 1)Z?)
b
b
a
...b
b
b
b
••- a
b
b
b
••- a
一、计算下列行列式:
1
1
1 ••-
1
0
a-b
0 ••-
0
=(Q +(M - 1)/?)
0
0
a-b ••-
0
= (a + (”_l)b)(a_b)"T
0
0
0 ••-
a-b
an-l
bn
Q] b、
=a
ax b{ cl d{
dn
d”—i
0
an-l
ax
-b”
c
Ci
d{
C”_i
c”
an-l
=(a”d” -b”c”)
ax b、
q £
Fl
d”-i
C”—i
0
d”-i
0
St
(a”d”—b”cJDT = Y{(aidi -Z?,c,) =
Z=1
a + b
a
0 ••-
0
0
b
a + b
a ••-
0
0
0
b
a + b ••-
0
0
0
0
0 ••-
a + b
a
0
0
0 ••-
b
a + b
（3）Dn
解：D” = (a + b)D”_[ 一 abD”_2,因此,
D”
D„
-aDn_\ = Z?(Dn_! -aDQ =…=bn~{(D2 -aD{) = bn+[ -bD心=a(D”r -bD”』) = ■■■ = a(D? - dDJ = an+x
••• Dn
an+i-bn+1 1
,a 丰 b
a-b
(ji + !)an ,a = b
X
-1
0 ••-
0
0
0
X
—1 …
0
0
0
0
0 ••-
X
-1
a”
a”-i
a”-2 •••
a2
可得:
n-2
⑷Dn
解：按最后一行展开,
Dn = axxn~x +a2x
cos 3
1
1
2 cos 3
0
1
••- 0
••- 0
0
0
Dn =
=cos
0
0
0
••- 2 cos 3
1
0
0
0
••- 1
2cos&
证明：
〃 =1时，左边二
0 =
COS0=右边;
假设"
"时
cos 6*
1
0
0
0
1
2 cos 6*
0
0
0
2+1 ■
_ 0
0
2 cos 6*
1
0
0
0
1 2cos0
1
0
0
0
1 2cos&
二、用数学归纳法证明:
nO
,Dk = cos kO, n = k +1 时,
cos 6*
1
1
2 cos 6*
0
0
0
0
=2 cos 0Dk +
=2cos&cosk& — q_i
2 cos 6*
1
= 2cos&cosk0-cos伙 一 1)0 = cos&cos —sin Osin 肋=sin伙 +1)0。
二、计算"阶行列式:
1
1
1 ••-
1
1
1
1 ••-
1
3
2 •
• 2
0
3
2 ••-
2
-2
1
0 ••-
0
2
4 •
• 2
D =
n
=
0
2
4 ••-
2
—
-2
0
2 ••-
0
2
2 •
• n + 2
0
2
2 ••-
n + 2
-2
0
0 ••-
n
n
1
1
1
i=l
1 + 2亡
n 1
(1 + 2.)"!。
i=l
解方程：
1
1
1
1
1
1 - X
1
1
(