第5节行列式按行列展开

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1、2011-09-21251215223714例2计算4阶行列式D0113592711(3)390030461200031522152217340216解：D2957295716421642152215221522021601130113011302160030012001200033带有一定技巧的例子：3111111111111311rr0200rr310200rr例1计算4阶行列式D216641113111

2、310020111311131113解：行列式的特点：各列之和是相等的；11113111666611110200661222481311rrrr131113110020D123461131113111310002111311131113加边法111x1说明：11x11例2计算4阶行列式D1)对于各行(列)之和相等的行列式可采用加1x111x1111边法求其值.解：行列式的特点：各行之和是相等的；2)加边法的步骤：xx111(1)先把其余各行(列)的

3、元素加到某一行(列)；ccccxx1111432(2)对该行(列)提取公因子；Dxx111x111(3)再化简；12011-09-21111x11111x例3计算n1阶行列式11x1100xxa0111xx1a00箭形行列式1x1110x0x11111000xD10a0n11111x100anrr2300xxx1xx(x)4111xxa11100xx01a12aancc000x11iai000a1解：

4、Din1,2,,000an1000ann1n1例4计算行列式()a0aa12anaa12aan()0i1aii1ai1a0001110aa0箭形行列式12011a002形如0001aann100011a的行列式常称为箭形行列式；n求箭形行列式的方法：利用对角元素或次对角元素将一条边消为零.1a00011a000101a00201a002011a002rr00100解：D2110001aann10001an00011an000011a000101

5、a002rr32001000001aann100011an22011-09-21例5解方程解：aaaaa1231nnrraaaaa210000ax123nn11a1a1a2xa3ann1a0000ax2rraaaaxaa左边n11223nn100000axn2a1a2a3an2an1xan0000axn1aaaaaax123n1n1naa(xa)(x)()(axa)x011221nn解得x1ax1,2a2,,,x

6、n2an2xn1an1是方程的n1个根.§1.4行列式按行(列)展开§1.4行列式按余子式和代数余子式行(列)展开按行(列)展开计算行列式小结、作业、思考题aaa()aaa()aaaa11223323321223312133aaa()aaaaaij11121321322231ij(1)二阶行列式aa1122a1221aaa2122aa2223aaaaa2123212211aaa12a13三阶行列式3233aa3133aa3132aaa111213aaaaa

7、aaaa112233122331132132a21a22a23a11a12a13aaa132231aaa122133aaa112332降低按行a31a32a331阶a21a22a23展开aaa31323332011-09-21三阶行列式按第2行展开三阶行列式按第1列展开aaaaaa111213111213aa21a2223aaa112233aaa122331aaa132132aa21a2223aaa112233aaa122331aaa132132aa31a3233aaa13

8、2231aaa122133aaa112332aa31a3233aaa132231aaa122133aaa112332a()aaaaa()aaaaaaa()aaa()aaaa2113321233221133133111223323322113321233a()aaaaa()aaaa23123111323112231322aa1213aa1113aa1112aaa2223aa1213aaa1213a21a22a2311aaa2131aaaa