计算行列式3个行列式 求详解 1 2 3...n2 3 4...n+13 4 5...n+2.n n+1 n+2...2n-10 1 2 3...n-11 0 1 2...n-22 1 0 1...n-33 2 1 0...n-4.n-1 n-2 n-3 n-4...0b1 b2 b3 ...bn-1 bn-a1 a2 0...0 00 -a2 a3...0 0.0 0 0 -an-1 an (所有的 1 2 3 n-1 n 都
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![计算行列式3个行列式 求详解 1 2 3...n2 3 4...n+13 4 5...n+2.n n+1 n+2...2n-10 1 2 3...n-11 0 1 2...n-22 1 0 1...n-33 2 1 0...n-4.n-1 n-2 n-3 n-4...0b1 b2 b3 ...bn-1 bn-a1 a2 0...0 00 -a2 a3...0 0.0 0 0 -an-1 an (所有的 1 2 3 n-1 n 都](/uploads/image/z/3157588-28-8.jpg?t=%E8%AE%A1%E7%AE%97%E8%A1%8C%E5%88%97%E5%BC%8F3%E4%B8%AA%E8%A1%8C%E5%88%97%E5%BC%8F+%E6%B1%82%E8%AF%A6%E8%A7%A3+1+2+3...n2+3+4...n%2B13+4+5...n%2B2.n+n%2B1+n%2B2...2n-10+1+2+3...n-11+0+1+2...n-22+1+0+1...n-33+2+1+0...n-4.n-1+n-2+n-3+n-4...0b1+b2+b3+...bn-1+bn-a1+a2+0...0+00+-a2+a3...0+0.0+0+0+-an-1+an+%28%E6%89%80%E6%9C%89%E7%9A%84+1+2+3+n-1+n+%E9%83%BD)
计算行列式3个行列式 求详解 1 2 3...n2 3 4...n+13 4 5...n+2.n n+1 n+2...2n-10 1 2 3...n-11 0 1 2...n-22 1 0 1...n-33 2 1 0...n-4.n-1 n-2 n-3 n-4...0b1 b2 b3 ...bn-1 bn-a1 a2 0...0 00 -a2 a3...0 0.0 0 0 -an-1 an (所有的 1 2 3 n-1 n 都
计算行列式3个行列式 求详解
1 2 3...n
2 3 4...n+1
3 4 5...n+2
.
n n+1 n+2...2n-1
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
b1 b2 b3 ...bn-1 bn
-a1 a2 0...0 0
0 -a2 a3...0 0
.
0 0 0 -an-1 an (所有的 1 2 3 n-1 n 都是字母的下标)
没人?第一道自己算出来了。
计算行列式3个行列式 求详解 1 2 3...n2 3 4...n+13 4 5...n+2.n n+1 n+2...2n-10 1 2 3...n-11 0 1 2...n-22 1 0 1...n-33 2 1 0...n-4.n-1 n-2 n-3 n-4...0b1 b2 b3 ...bn-1 bn-a1 a2 0...0 00 -a2 a3...0 0.0 0 0 -an-1 an (所有的 1 2 3 n-1 n 都
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.