[问题2014A02] 解答三（降阶公式法）
将矩阵 (A) 写成如下形式:
[A=egin{pmatrix} -2a_1 & 0 & cdots & 0 & 0 \ 0 & -2a_2 & cdots & 0 & 0 \ vdots & vdots & vdots & vdots & vdots \ 0 & 0 & cdots & -2a_{n-1} & 0 \ 0 & 0 & cdots & 0 & -2a_n end{pmatrix}]
[+egin{pmatrix} a_1 & 1 \ a_2 & 1 \ vdots & vdots \ a_{n-1} & 1 \ a_n & 1 end{pmatrix}cdot I_2^{-1}cdotegin{pmatrix} 1 & 1 & cdots & 1 & 1 \ a_1 & a_2 & cdots & a_{n-1} & a_n end{pmatrix}.]
由降阶公式可得
[|A|=egin{vmatrix} -2a_1 & 0 & cdots & 0 & 0 \ 0 & -2a_2 & cdots & 0 & 0 \ vdots & vdots & vdots & vdots & vdots \ 0 & 0 & cdots & -2a_{n-1} & 0 \ 0 & 0 & cdots & 0 & -2a_n end{vmatrix}cdotBigg|I_2+egin{pmatrix} 1 & 1 & cdots & 1 & 1 \ a_1 & a_2 & cdots & a_{n-1} & a_n end{pmatrix}egin{pmatrix} -2a_1 & 0 & cdots & 0 & 0 \ 0 & -2a_2 & cdots & 0 & 0 \ vdots & vdots & vdots & vdots & vdots \ 0 & 0 & cdots & -2a_{n-1} & 0 \ 0 & 0 & cdots & 0 & -2a_n end{pmatrix}^{-1}egin{pmatrix} a_1 & 1 \ a_2 & 1 \ vdots & vdots \ a_{n-1} & 1 \ a_n & 1 end{pmatrix}Bigg|]
[=(-2)^nprod_{i=1}^na_iegin{vmatrix} 1-frac{n}{2} & -frac{1}{2}sum_{i=1}^nfrac{1}{a_i} \ -frac{1}{2}sum_{i=1}^na_i & 1-frac{n}{2} end{vmatrix}]
[=(-2)^{n-2}prod_{i=1}^na_iigg((n-2)^2-Big(sum_{i=1}^na_iBig)Big(sum_{i=1}^nfrac{1}{a_i}Big)igg). quadBox]