,我不是计算机专业的学生只是选修课要求做请您能不能直接把程序写给我呀,非常感谢啦
追答A = magic(10);
%求所有行元素之和
SumH1 = sum(A(1,:)) ; %第一行
SumH2 = sum(A(2,:)) ;
SumH3 = sum(A(3,:)) ;
SumH4 = sum(A(4,:)) ;
SumH5 = sum(A(5,:)) ;
SumH6 = sum(A(6,:)) ;
SumH7 = sum(A(7,:)) ;
SumH8 = sum(A(8,:)) ;
SumH9 = sum(A(9,:)) ;
SumH10 = sum(A(10,:)) ; %第十行
%求所有列元素之和
SumL1 = sum(A(:,1)) ; %第一列
SumL2 = sum(A(:,2)) ;
SumL3 = sum(A(:,3)) ;
SumL4 = sum(A(:,4)) ;
SumL5 = sum(A(:,5)) ;
SumL6 = sum(A(:,6)) ;
SumL7 = sum(A(:,7)) ;
SumL8 = sum(A(:,8)) ;
SumL9 = sum(A(:,9)) ;
SumL10 = sum(A(:,10)) ; %第十列
%求主对角线元素之和
SumZ = sum(diag(A));
%求副对角线元素之和
SumZ = sum(diag(fliplr(A)));
%将矩阵中大于40,小于60的数用NaN代替
[rows,cols,vals] = find( 40<A & A<60 );
for i=1:length(rows)
A(rows(i),cols(i))=NaN;
end;