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证:由克莱姆法则, 上式右边第$\,i\,$行8"t,= ©Elinkage数学论坛 -- Elinkage极酷超级论坛 "*Y $\qquad\lambda_i=\frac{\begin{vmatrix}\omega_1^{-1}&\cdots&\omega_{i-1}^{-1}&1&\omega_{i+1}^{-1}&\cdots&\omega_n^{-1}\\\vdots& &\vdots&\vdots&\vdots& &\vdots\\\omega_1^{-n}&\cdots&\omega_{i-1}^{-n}&1&\omega_{i+1}^{-n}&\cdots&\omega_n^{-n} \end{vmatrix}}{\begin{vmatrix}\omega_1^{-1}&\cdots&\omega_{n}^{-1}\\\vdots& \ddots&\vdots\\\omega_1^{-n}&\cdots&\omega_{n}^{-n}\end{vmatrix}}\quad$消去行列式列的公因子3:z.` $\qquad\quad=\frac{\begin{vmatrix}1&\cdots&1&1&1&\cdots&1\\\omega_1^{-1}&\cdots&\omega_{i-1}^{-1}&1&\omega_{i+1}^{-1}&\cdots&\omega_n^{-1}\\\vdots& &\vdots&\vdots&\vdots& &\vdots\\\omega_1^{-(n-1)}&\cdots&\omega_{i-1}^{-(n-1)}&1&\omega_{i+1}^{-(n-1)}&\cdots&\omega_n^{-(n-1)} \end{vmatrix}}{\omega_i^{-1}\begin{vmatrix}1&\cdots&1\\\omega_1^{-1}&\cdots&\omega_{n}^{-1}\\\vdots& \ddots&\vdots\\\omega_1^{-(n-1)}&\cdots&\omega_{n}^{-(n-1)}\end{vmatrix}}\;$去连乘式公因子d@):q $\qquad\quad=\displaystyle\frac{\omega_i\prod_{j\ne i}(w_j^{-1}-1)}{\prod_{j\ne i}(w_j^{-1}-w_i^{-1})}=\frac{\omega_i^n\prod_{j\ne i}(\omega_j-1)}{\prod_{j\ne i}(\omega_j-\omega_i)}.\quad\square$?YkBD
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