2.2-矩阵的运算

第2.2节矩阵的运算一、矩阵的加减法二、数与矩阵相乘三、矩阵与矩阵相乘四、矩阵的转置五、方阵的行列式主要内容:六、思考与练习mnmnmmmmnnnnbababababababababaBA221122222221211112121111一、矩阵的加减法设有两个矩阵那末矩阵与的和记作，规定为nm,,ijijbBaAABBA加法:注意：只有当两个矩阵是同型矩阵时，才能进行加法运算.例如：12345698186309153121826334059619583112.98644741113减法:负矩阵：)(BABAmnmmnnaaaaaaaaaA112222111211ija。的称为矩阵负矩阵A矩阵加法满足的运算规律:.1ABBA交换律：.2CBACBA结合律：.4OAA.,03是同型矩阵与其中OAAA二、数与矩阵相乘数乘:.112222111211mnmmnnaaaaaaaaaAA规定为或的乘积记作与矩阵数,AAA注意：矩阵数乘与行列式数乘的区别.25023124100462522125422;1AA;2AAA.3BABA数乘矩阵满足的运算规律：矩阵相加与数乘矩阵合起来,统称为矩阵的线性运算.（设为矩阵，为数）,nmBA、例1:设,502134A301021B求A－2B解：6020422B6020425021342BA104172定义：skkjiksjisjijiijbabababac12211,,,2,1;,2,1njmi并把此乘积记作.ABC三、矩阵与矩阵相乘ijcC设是一个矩阵，是一个矩阵，那末规定矩阵与矩阵的乘积是一个矩阵，其中ijaAsmijbBnsnmAB例2：222263422142C221632816415003112101A121113121430B例3：?求AB故121113121430415003112101ABC.解:,43ijaA,34ijbB.33ijcC567102621710注意:只有当第一个矩阵的列数等于第二个矩阵的行数时，两个矩阵才能相乘.106861985123321例如:123321132231.10不存在.例4:计算下列矩阵的乘积.213221解:21322122212221323.634242332222112bababa321bbb.222322331132112233322222111bbabbabbabababa331221111bababa=(333223113bababa）3213332312322211312113212bbbaaaaaaaaabbb例5：计算下列矩阵的乘积.43337311020854121111444311117311020854121437311020854121437311020854121nnnnnnbbbaaa2121nnnnbababa2211矩阵乘法满足的运算规律：;:1BCACAB结合律,:2ACABCBA分配律;CABAACBBABAAB3（其中为数）;AEA;EAA;4矩阵乘法不满足交换律,BAAB即：例:设1111A1111B则,0000AB,2222BA.BAAB注意：矩阵乘法不满足消去律CBAACAB0,不能推出1111A1111B,0000AB,2222C例如：,0000AC有CB但是ACAB000BAAB或不能推出若A是n阶方阵，则为A的次幂，即kAk个kkAAAA,kmkmAAA.mkkmAA为正整数km,方阵的幂：并且,时当BAAB.BAABkkk方阵的多项式：0111)(axaxaxaxfkkkk0111)(aAaAaAaAfkkkkE四、矩阵的转置定义:把矩阵的行换成同序数的列得到的新矩阵，叫做的转置矩阵，记作.AAA例:,854221A;825241TA,618B.618TB转置矩阵满足的运算规律：;1AATT;2TTTBABA;3TTAA.4TTTABAB例6：已知,102324171,231102BA.TAB求解法1：102324171231102AB,1013173140.1031314170TAB解法2：TTTABAB213012131027241.1031314170.称为反对称的则矩阵如果AATA对称阵的元素以主对角线为对称轴对应相等.说明：AAAATTAA是反对称矩阵是对称矩阵例7:.,都是对称矩阵和则设TTnmBBBBB设为阶方阵，如果满足，即那末称为对称阵.Annjiaajiij,,2,1,A.6010861612为对称阵例如A对称阵:TAA例9：.2.,1:.对称矩阵之和可表示为对称矩阵和反是反对称矩阵是对称矩阵证明阶矩阵对于任意的AAAAAAnTT注：对称矩阵的乘积不一定是对称矩阵311121111100001010例311111121例8：.n,,,阶反对称矩阵是则BAABBBBAAATnnTnn五、方阵的行列式定义：由阶方阵的元素所构成的行列式，叫做方阵的行列式，记作或nAAA.detA8632:A例8632A则.2运算规律：:;1AAT;2AAnBAAB3.ABBA注：虽然,ABBA但定义：行列式的各个元素的代数余子式所构成的如下矩阵AijAnnnnnnAAAAAAAAAA212221212111称为矩阵的伴随矩阵.A故ijAAAijA.EA同理可得nkkjkiaAAA1ijAijA.EA性质：.EAAAAA证明：,ijaA设,ijbAA记则jninjijiijAaAaAab2211,ijA例10:设解111020011A求52AB,110051142,111020011BA,2110051142B052051142,552AB532AB58AB58BA55108528四、思考与练习设其中是数,,,21nnEkBEkA21,kk求及BABA解：nnEkEkBA21nnnnEkEk21nnkk21nnEkEkBA21nEkk)(21nnEkk21nkk21注意:一般地BABA