线代第2章习题答案(1).pdfVIP免费

第2章1.1321A，3012B，计算得762318AB，3303ABBA，22160511AB.2.222200442ABBA，668()211202TAB，444013004TTAB.3.(1)326510121101224012421010211;(2)1212323210113710324815;(3)1231121;(4)123112312311231;(5)111121311111211322122232212222233313233331332333aaaaaaaaaaaaaaaaaa;(6)133131112111212123323212212122223333331321312323433434142141242aaaaaaaaaaaaaaaaaaaaaaaa;(7)1112131123212223211112123133132333()aaaxxxxaaaxxaxaxaxaaax2121222323()xaxaxax3131232333()xaxaxax.4.解：因为A与B可交换，所以ABBA，又因为A是对角矩阵，所以可得111112111112121221222221212222121122nnnnnnnnnnnnnnnnnnbbbbbbbbbbbbbbbbbb，其中主对角线元素都相等，对于非主对角元，应有()0,ijijbij又因为ij，所以只能有0ijb，当ij时。即B也是对角矩阵。5.(1)1516()823fA;(2)140()6110201fA;(3)004()251034fA.6.(1)*131/411211241/3AAA;(2)516A,*5111351111A,所以*15/61/61/613/65/61/61/61/61/6AAA.7.(1)21223100110010|110010223100121001121001AErr2132311100101100102043120011011011011043120rrrrrr...