层次分析法实验

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1、题日：用方根法解矩阵A的最人特征值及对应的特征向最，并将特征向最归一化,対A进行一致性检验。四、1%%%%实验平台：MATLAB7.4问题描述：已知矩阵A,它进行归一化，最后进行一致性检验实验步骤：已知矩阵A,采用方根法求解分以下的步骤:法一（方根法）：用方根法来求它的最人特征值及对应的特征向址，还要对61.计算A每一行元索的乘积，耐，=匸［勺1i=12345,6W：Mj=i=12345,6122512252/251/14701/2452.计算的6次方根:i=12345,6W]=3.2711vv2

2、=3.2711yv>=1.2009vv4=0.6564w5=0.2966vv6=0.3998“3.2711、3.27111.20090.65640.2966,0.3998丿进行归一化，即6_/=!心12345,66其中工石=3.2711+3.2711+1.2009+0.6564+0.2966+0.3988=9.09491=1则W二[W],w2-«-w6]=(0.3597,0.3597,0.1320,0.0722,0.0326,0.0440)即为所求的特征向量。3.最人特征值入“狀的计算：其中，（A

3、W表示向量AW的第i个元索。AW=(2.2766,2.2766,0.8755,0.4459,0.2078,0.2745)'2.27662.27660.87550.44590.2078++++6*0.35976*0.35976*0.13206*0.07226*0.0326=6.57346*0.0440»(2.2766/0.3597+2.766/0.3597+0.8755/0.1320+0.4459/0.0722+0.2078/0.0326+0.2745/0.0440)/6ans=6.57342-6

4、Ci二——=(6.5734-6)/(6-1)=0.11476-1由于这个矩阵是6阶矩阵，所以取修正值RI二1.24d检验一致性:CR=—=1.1147/1.24=0.0925<0.1RI法二(MATLAB直接求解).运用MATLAB7.4求得矩阵A的最大特征值如图所示>>A二[115577;115577;1/51/51355;1/51/51/3132;1/71/71/51/311/2;1/71/71/51/221]A=115r*077115L□771/51/513551/51/51/31321/7

5、1/71/51/311/21/71/71/51/221求得其特征值为：V=3994/629000000-127/3025+2407/1630i000000-127/3025-2407/1630i000000*000000-643/3763000000-299/3150W*^XMEQZT3>■

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