1.11:311211 230202311622211: 34043114331 113323201211(1) ;2401210XXXX 习题、求下列式中的矩阵解、计算:解1212112121 11 212 122212657012:426211(2)(,,,);:.(3)( ,,,);:3,,nnniiinnnnnnnnbba aaba baab bbaa ba ba ba ba ba ba ba ba bA Bn 原式解原式解原式、皆为阶方阵问下322322233; :(2)()(2)0. AA BABBABBAABAABBABBA3列等式成立的条件是什么?(1)(A+B)解当时.解:当时. 24.,,: ,,, ()() ,()(): ()() ()()()()5.() :10(1);1110101010:2, 1111112nABBA ACCAA B CA BCBC A A BCBC AA BCABACBACABC AA BCAB CB ACBC Ann若证明 是同阶矩阵且证明计算为正整数解 用数学归纳法,时122211010 ,,11110101010 11111111010 11110(2) 01;001020:,2, 0102000kknnnkkkknaaaaaanaaaa设当时成立 则当n=k+1时:解 用数学归纳法时121121110(1)102 , 0100 1(1)(1)1010102 010101000000kkkkkkkkkkkkkkkak kakaaankaakaaankk kakaaaaaaaaakaaaaa 设当时当时00(3) 00.00nabc 1110000:000()000001 16(1,2,3),(1,, ),,.2 3:()()()()()() 3111232 33321333127nnnnTnnTTTTTTTnnTnnaabbccAAAA 解原式、已知且计算解、举2222:(1),;00 :,0,10(2),,;10 :,,00,.(3),,.0000000...