已知 abcd=1,求证:|a2+1a2a1a1b2+1b2b1b1c2+1c2c1c1d2+1d2d1d1|=0.
解析 根据行列式运算的性质,有LHS=|a4+1a3aa2b4+1b3bb2c4+1c3cc2d4+1d3dd2|=|a4a3aa2b4b3bb2c4c3cc2d4d3dd2|+|1a3aa21b3bb21c3cc21d3dd2|=|a3a21ab3b21bc3c21cd3d21d|+|1a3aa21b3bb21c3cc21d3dd2|=0.
备注 事实上,原行列式展开后为(a−b)(b−c)(c−d)(d−a)(a−c)(b−d)(abcd−1)a2b2c2d2.