层次分析法中的成对比较矩阵是自己定的值吗?权值能用哪种软件计算吗?

clcclear %修改对比矩阵、一致性检验就可以a=[1,1,1,4,1,1/21,1,2,4,1,1/21,1/2,1,5,3,1/21/4,1/4,1/5,1,1/3,1/31,1,1/3,3,1,12,2,2,3,3,1];[x,y]=eig(a);eigenvalue=diag(y);lamda=eigenvalue(1);ci1=(lamda-6)/5;cr1=ci1/1。24w1=x(:,1)/sum(x(:,1))b1=[1,1/4,1/2;4,1,3;2,1/3,1];[x,y]=eig(b1);eigenvalue=diag(y);lamda=eigenvalue(1);ci21=(lamda-3)/2;cr21=ci21/0。58w21=x(:,1)/sum(x(:,1))b2=[1 1/4 1/5;4 1 1/2;5 2 1];[x,y]=eig(b2);eigenvalue=diag(y);lamda=eigenvalue(1);ci22=(lamda-3)/2;cr22=ci22/0。58w22=x(:,1)/sum(x(:,1))b3=[1 3 1/3;1/3 1 1/7;3 7 1];[x,y]=eig(b3);eigenvalue=diag(y);lamda=eigenvalue(1);ci23=(lamda-3)/2;cr23=ci23/0。58w23=x(:,1)/sum(x(:,1))b4=[1 1/3 5;3 1 7;1/5 1/7 1];[x,y]=eig(b4);eigenvalue=diag(y);lamda=eigenvalue(1);ci24=(lamda-3)/2;cr24=ci24/0。58w24=x(:,1)/sum(x(:,1))b5=[1 1 7;1 1 7;1/7 1/7 1];[x,y]=eig(b5);eigenvalue=diag(y);lamda=eigenvalue(2);ci25=(lamda-3)/2;cr25=ci25/0。58w25=x(:,2)/sum(x(:,2))b6=[1 7 9;1/7 1 1 ;1/9 1 1];[x,y]=eig(b6);eigenvalue=diag(y);lamda=eigenvalue(1);ci26=(lamda-3)/2;cr26=ci26/0。58w26=x(:,1)/sum(x(:,1))w_sum=[w21,w22,w23,w24,w25,w26]*w1ci=[ci21,ci22,ci23,ci24,ci25,ci26];cr=ci*w1/sum(0。58*w1)这个我以前写的,你拿去改改吧就可以用了,matla