r4?r3r3?r2r2?r11a?(?1)3[a?b?c?d]2aa31bb2b31cc2c31d d2d3??(a?b?c?d)(d?c)(d?b)(d?a)(c?b)(c?a)(b?a).
1a例13 计算行列式D?2aa41bb2b41cc2c41d. d2d41a2解 构造范德蒙德行列式 f(x)?aa3a4则由范德蒙德行列式知
1bb2b3b41cc2c3c41dd2d3d41xx2 x3x4f(x)?(x?d)(x?c)(x?b)(x?a)?(d?c)(d?b)(d?a)(c?b)(c?a)(b?a)其中x的系数为
3，
?(a?b?c?d)(d?c)(d?b)(d?a)(c?b)(c?a)(b?a)
将f(x)按第五列展开可得到f(x)的另一种表示形式，由待定系数法知x的系数为(?1)34?5D在两个不同的展式中相等，从而
(?1)4?5D??(a?b?c?d)(d?c)(d?b)(d?a)(c?b)(c?a)(b?a)故 D?(a?b?c?d)(d?c)(d?b)(d?a)(c?b)(c?a)(b?a).
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