Question

线性代数：计算行列式Dn=[1 2 3 …n-1 n][2 1 2 … n-2 n-1][3 2 1 …n-3 n-2]

Answer 1

Dn=|1 2 3 ... n| 【由rn到r1进行变换：ri-r(i-1)

1 -1 -1 ... -1

1 1 -1 ... -1

..............

1 1 1 ... -1

=|1-n 2-n 3-n .... n|

0 -2 -2 .... -1 【ck-cn

0 0 -2 ..... -1

..............................

0 0 0 ... -1 【是个《上三角》

=(1-n)*(-2)*(-2)*...*(-2)*(-1)

=[(-1)^n]*(n-1)*2^(n-2)