已知 $abcd=1$，求证：\[\begin{vmatrix}a^2+\dfrac 1{a^2}&a&\dfrac 1a&1\\ b^2+\dfrac1{b^2}&b&\dfrac 1b&1\\ c^2+\dfrac1{c^2}&c&\dfrac 1c&1\\ d^2+\dfrac1{d^2}&d&\dfrac 1d&1\end{vmatrix}=0.\]
解析 根据行列式运算的性质，有\[\begin{split} LHS&=\begin{vmatrix}a^4+1&a^3&a&a^2\\ b^4+1&b^3&b&b^2\\ c^4+1&c^3&c&c^2\\ d^4+1&d^3&d&d^2\end{vmatrix}\\ &=\begin{vmatrix}a^4&a^3&a&a^2\\ b^4&b^3&b&b^2\\ c^4&c^3&c&c^2\\ d^4&d^3&d&d^2\end{vmatrix}+\begin{vmatrix}1&a^3&a&a^2\\ 1&b^3&b&b^2\\ 1&c^3&c&c^2\\ 1&d^3&d&d^2\end{vmatrix}\\ &=\begin{vmatrix}a^3&a^2&1&a\\ b^3&b^2&1&b\\ c^3&c^2&1&c\\ d^3&d^2&1&d\end{vmatrix}+\begin{vmatrix}1&a^3&a&a^2\\ 1&b^3&b&b^2\\ 1&c^3&c&c^2\\ 1&d^3&d&d^2\end{vmatrix}\\ &=0. \end{split}\]
备注 事实上，原行列式展开后为\[\dfrac{(a-b)(b-c)(c-d)(d-a)(a-c)(b-d)(abcd-1)}{a^2b^2c^2d^2}.\]