Question

计算行列式3个行列式 求详解 在线等

Answer 1

第2题
0
1
2
3...
n-1
1
0
1
2...
n-2
2
1
0
1...
n-3
3
2
1
0...
n-4
........................
n-1
n-2
n-3
n-4...
0
依次作:
c1-c2,c2-c3,...,c(n-1)-cn
得
-1
-1
-1
-1
...
-1
n-1
1
-1
-1
-1
...
-1
n-2
1
1
-1
-1
...
-1
n-3
1
1
1
-1
...
-1
n-4
...
...
...
...
1
1
1
1
...
-1
1
1
1
1
1
...
1
0
第1行乘-1加到其余各行,
得
-1
-1
-1
-1
...
-1
n-1
0
-2
-2
-2
...
-2
2n-3
0
0
-2
-2
...
-2
2n-4
0
0
0
-2
...
-2
2n-5
...
...
...
...
0
0
0
0
...
-2
n
0
0
0
0
...
0
n-1
所以行列式等于
(-1)*(-2)^(n-2)*(n-1)
=
(n-1)2^(n-2)(-1)^(n-1).
第3题
b1
b2
b3
...
bn-1
bn
-a1
a2
0
...
0
0
0
-a2
a3
...
0
0
..........................
0
0
0
an-1
0
0
0
0
-an-1
an
解:
记行列式为Dn,
则
n>=2时有
(n=1平凡)
Dn
=
b1a2a3...an
+
a1b2a3...an
+
a1a2b3...an
+...+
a1a2a3...bn
下面归纳证明.
易证
D2
=
b1a2
+
a1b2.
假设
n-1
成立,
即
D(n-1)
=
b1a2a3...a(n-1)
+
a1b2a3...a(n-1)
+
a1a2b3...a(n-1)
+...+
a1a2a3...b(n-1)
则n时,
按第n列展开得:
Dn
=
anD(n-1)+(-1)^(1+n)bn(-1)^(n-1)a1a2...a(n-1)
=
anD(n-1)+
a1a2...a(n-1)bn
=
an[b1a2a3...a(n-1)
+
a1b2a3...a(n-1)
+
a1a2b3...a(n-1)
+...+
a1a2a3...b(n-1)]
+
a1a2...a(n-1)bn
=
b1a2a3...a(n-1)an
+
a1b2a3...a(n-1)an
+
a1a2b3...a(n-1)an
+...+
a1a2a3...b(n-1)an+
a1a2a3...b(n-1)an.
得证.