第三节 行列式展开定理(1).pptVIP免费

音乐第三节2,312213332112322311322113312312332211aaaaaaaaaaaaaaaaaa333231232221131211aaaaaaaaa)(3223332211aaaaa)(3321312312aaaaa)(3122322113aaaaa333123211333312321123332232211aaaaaaaaaaaaaaa1、余子式与代数余子式3在阶行列式中，把元素所在的第行和第列划去后，留下来的阶行列式叫做元素的余子式，记作nijaij1nija.Mij，记ijjiijMA)1(叫做元素的代数余子式．ija例如44434241343332312423222114131211aaaaaaaaaaaaaaaaD44424134323114121123aaaaaaaaaM233223)1(MA.23M4,44434241343332312423222114131211aaaaaaaaaaaaaaaaD,44434134333124232112aaaaaaaaaM122112)1(MA.12M5,44434241343332312423222114131211aaaaaaaaaaaaaaaaD,33323123222113121144aaaaaaaaaM.)1(44444444MMA.代数余子式个对应着一个余子式和一行列式的每个元素分别6n阶行列式D=|aij|等于它的任意一行(列)的各元素与其对应代数余子式乘积的和,即),,2,1(2211niAaAaAaDininiiii).,,2,1(2211njAaAaAaDnjnjjjjj或按第i行展开按第j列展开证略推论:若行列式某行(列)的元素全为零,则行列式的值为零.(行列式展开定理)7,601504321D例2设543106131061542D列展开得按第2010436154260501D行展开得按第1,58292.582928定理行列式某一行的元素乘另一行对应元素的代数余子式之和等于零,即这是因为第i行第j行022111jninjijinkjkikAaAaAaAa)(jinnnniniiiniinnkjkikaaaaaaaaaaaaAa212121112111.09同样,行列式对列展开,也有).(,02211jiAaAaAanjnijiji;,0,,1jijiDDAaijnkkjki当当;,0,,1jijiDDAaijnkjkik当当.,0,1jijiij当，当引入记号则有102、行列式的计算计算行列式的基本方法：利用性质5将某行(列)化出较多的零,再利用展开定理按该行(列)展开.例13351110243152113D0551111115)1(333131312cc34cc5011501150011110551111115)1(330550261155526)1(31.4012rr12例2计算行列式0532004140013202527102135D解5324141325253204140132021352)1(52D13rr122rr66027013210....