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matlaba=-5.0000-2.00001.00004.00001.5000-4.80000.20001.30001.00000.6000-2.5000-0.50000.9000-3.0000-1.5000-7.0000symsxyz;equation1=abs(x*a(1,3))+abs(a(2,3))+a(3,3)+abs(a(4,3))-(a(1,1)+abs(1/x*a(2,1))+abs(1/x*a(3,1))+abs(1/x*a(4,1))
题目详情
matlab
a =-5.0000 -2.0000 1.0000 4.0000
1.5000 -4.8000 0.2000 1.3000
1.0000 0.6000 -2.5000 -0.5000
0.9000 -3.0000 -1.5000 -7.0000
syms x y z ;
equation1=abs(x*a(1,3))+abs(a(2,3))+a(3,3)+abs(a(4,3))-(a(1,1)+abs(1/x*a(2,1))+abs(1/x*a(3,1))+abs(1/x*a(4,1)));
equation2=abs(a(1,2)*y)+a(2,2)+abs(a(3,2))+abs(a(4,2))- (a(1,1)+abs(1/y*a(2,1))+abs(1/y*a(3,1))+abs(1/y*a(4,1)));
equation3=abs(z*a(1,4))+abs(a(2,4))+abs(a(3,4))+a(4,4)-(a(1,1)+abs(1/z*a(2,1))+abs(1/z*a(3,1))+abs(1/z*a(4,1)));
x=solve(eval(equation1));
y=solve(eval(equation2));
z=solve(eval(equation3));
和用
[x y z]=solve(eval(equation1),eval(equation2),eval(equation3))
后一个解法中含有新引入的符号变量.不知为何?非诚勿扰~
a =-5.0000 -2.0000 1.0000 4.0000
1.5000 -4.8000 0.2000 1.3000
1.0000 0.6000 -2.5000 -0.5000
0.9000 -3.0000 -1.5000 -7.0000
syms x y z ;
equation1=abs(x*a(1,3))+abs(a(2,3))+a(3,3)+abs(a(4,3))-(a(1,1)+abs(1/x*a(2,1))+abs(1/x*a(3,1))+abs(1/x*a(4,1)));
equation2=abs(a(1,2)*y)+a(2,2)+abs(a(3,2))+abs(a(4,2))- (a(1,1)+abs(1/y*a(2,1))+abs(1/y*a(3,1))+abs(1/y*a(4,1)));
equation3=abs(z*a(1,4))+abs(a(2,4))+abs(a(3,4))+a(4,4)-(a(1,1)+abs(1/z*a(2,1))+abs(1/z*a(3,1))+abs(1/z*a(4,1)));
x=solve(eval(equation1));
y=solve(eval(equation2));
z=solve(eval(equation3));
和用
[x y z]=solve(eval(equation1),eval(equation2),eval(equation3))
后一个解法中含有新引入的符号变量.不知为何?非诚勿扰~
▼优质解答
答案和解析
clearclcclose alla =[-5.0000 -2.0000 1.0000 4.00001.5000 -4.8000 0.2000 1.30001.0000 0.6000 -2.5000 -0.50000.9000 -3.0000 -1.5000 -7.0000];syms x y zequation1=abs(x*a(1,3))+abs(a(2,3))+a(3,3)+abs(a(4,...
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