1.5行列式按行(列)展开一、引言aaaaaaaaa111213212223313233112233122331132132112332122133132231aaaaaaaaaaaaaaaaaa122331111221221333332132132231112332aaaaaaaaaaaaaaaaaa2223113233aaaaaaaaaa2123123133.2122133132aaaaa1122332332()aaaaa1223312133()aaaaa1321322231()aaaaa结论三阶行列式可以用二阶行列式表示.能否推广?二、余子式和代数余子式例如11121314212223243132333441424344aaaaaaaaDaaaaaaaa定义1在n阶行列式中,划掉元素aij所在的第i行和第j列,留下来的元素按原来的排法构成一个n-1阶行列式叫做元素aij的余子式,记作Mij.称为元素aij的代数余子式.1ijijijAM11131432212324414344,aaaMaaaaaa32323232(1).AMM二、余子式和代数余子式引入元素的余子式和代数余子式后,三阶行列式可表示为:111211112133111213(1)(1)(1)aaaMMM212122331112132333aaaaaaaaa222321232122323331333111213132aaaaaaaaaaaaaaa111213111213.AAaaaA三、行列式按一行(列)展开法则定理1n阶行列式D=det(aij)等于它的任一行(列)的各元素与其对应的代数余子式乘积之和,即,,,iiiiiinnnjiijjDaAaAaAaAin1122112,,,nnnjjjjjjjjiijDaAaAaAaAjn1122112或注:行列式按一行(列)展开法则也称“降阶法”,特别适用于计算某一行(列)含有较多零的行列式.三、行列式按一行(列)展开法则推论一个n阶行列式D=det(aij),如果其中第i行所有元素除aij外都为零,那么这个行列式等于aij与它的代数余子式的乘积,即.ijijDaA1112132331323300,aaaDaaaa.2311122331321aaaaa例如2323DaA1112133132323300,aaaaDaaa三、行列式按一行(列)展开法则5312017252023100414002350D例1计算行列式解5312017252023100414002350D25531202312(1)041402352311007206625531202312(1)04140235三、行列式按一行(列)展开法则231254142352131(2)rrrr7210(2)6620(4212)1080.三、行列式按一行(列)展开法则例2用“降阶法”计算行列式.D311251342011153333511111111550解.3112513420115313D51111113100105530D51162055033511111111550三、行列式按一行(列)展开法则136215540.33151111111055三、行列式按一行(列)展开法则证明对阶数n用数学归纳法.例3证明n(n≥2)阶范德蒙德(Vandermonde)行列式所以n=2时,(1)式成立..12322221231111112311111nnnijjinnnnnnaaaaDaaaaaaaaaaDaaaa2211211.ijjiaa12当n=2时,三、行列式按一行(列)展开法则假设(1)对于n-1阶范德蒙行列式成立,按照第一列展开,并提出每列的公因子,iaa1.21311222212313112121221231311111000nnnnnnnnnnnnaaaaaaDaaaaaaaaaaaaaaaaaa则有始,后一行减去前一行的a1倍,得从第n行开三、行列式按一行(列)展开法则.232131122223111nnnnnnnaaaDaaaaaaaaan−1阶范德蒙德行列式由于包含a1的差全在前面出现了,根据归纳假设,结论对于n阶范德蒙行列式也成立.nnijjinDaaaaaaaa213112.ijjinaa1三、行列式按一行(列)展开法则结论:n阶范德蒙德行列式等于零12,,,naaa这n个元素中至少有两个相等.推论行列式任一行(列)的元素与另一行(列)的对应元素的代数余子式乘积之和等于零,即.11220kikinkniaAaAaAki.11220ljllnnjjajAaAaAl三、行列式按一行(列)展开法则分析我们以3阶行列式为例.把第1行的元素换成第2行的对应元素,则111213111112121313212223...