MATLAB程序问题解一个方程组如下syms x y z t a b c d m s; f = x+z-a;g = (2^(1/2)*m*x)/2 + (2^(1/2)*m*y)/2+(2^(1/2)*m*t)/2 - (2^(1/2)*m*z)/2-b;h = m^2*(y- t)-m^2*c;k = (2^(1/2)*m^3*(y-x+z+t))/2-m^2*d;[x,y,z,t] = solve(f,g,h,k); x = simplify(

MATLAB程序问题解一个方程组如下syms x y z t a b c d m s; f = x+z-a;g = (2^(1/2)*m*x)/2 + (2^(1/2)*m*y)/2+(2^(1/2)*m*t)/2 - (2^(1/2)*m*z)/2-b;h = m^2*(y- t)-m^2*c;k = (2^(1/2)*m^3*(y-x+z+t))/2-m^2*d;[x,y,z,t] = solve(f,g,h,k); x = simplify(
MATLAB程序问题
解一个方程组如下
syms x y z t a b c d m s;
f = x+z-a;
g = (2^(1/2)*m*x)/2 + (2^(1/2)*m*y)/2+(2^(1/2)*m*t)/2 - (2^(1/2)*m*z)/2-b;
h = m^2*(y- t)-m^2*c;
k = (2^(1/2)*m^3*(y-x+z+t))/2-m^2*d;
[x,y,z,t] = solve(f,g,h,k);
x = simplify(x),
y = simplify(y),
z = simplify(z),
t = simplify(t)
x =
(2^(1/2)*(b + d))/(4*m) - c/2
y =
a/2 + (2^(1/2)*(b - d))/(4*m)
z =
c/2 + (2^(1/2)*(b + d))/(4*m)
t =
a/2 - (2^(1/2)*(b - d))/(4*m)
将x,z带入f方程,明显不成立,运行给的这个答案把x,y 的值和z,t的对调了.求教原因?

MATLAB程序问题解一个方程组如下syms x y z t a b c d m s; f = x+z-a;g = (2^(1/2)*m*x)/2 + (2^(1/2)*m*y)/2+(2^(1/2)*m*t)/2 - (2^(1/2)*m*z)/2-b;h = m^2*(y- t)-m^2*c;k = (2^(1/2)*m^3*(y-x+z+t))/2-m^2*d;[x,y,z,t] = solve(f,g,h,k); x = simplify(
楼上的回答存在问题.
 
诚然,Mathematica在符号运算方面总体上优于MATLAB,推荐使用Mathematica没问题；但楼上关于MATLAB符号运算的说法却纯粹是想当然,像这样误导人的做法以后还是应该少一些.
 
其实,当solve返回多个输出参数的时候,其顺序是按照字母表顺序,而不是你通过输入参数指定的变量顺序,也不是楼上所说的【按照变量出现的先后顺序】.所以不小心很容易搞错（对于当前这个例子没问题）.
 
从solve函数的说明中摘录一段：
>> help solve
 SOLVE  Symbolic solution of algebraic equations.
    . 
    Three different types of output are possible.  For one equation and one
    output,the resulting solution is returned,with multiple solutions to
    a nonlinear equation in a symbolic vector.  For several equations and
    an equal number of outputs,the results are sorted in lexicographic
    order and assigned to the outputs.  For several equations and a single
    output,a structure containing the solutions is returned.

现在应该明白怎么做了吧?调用时应该是
[t,x,y,z] = solve(f,g,h,k);
再检验一下结果看看：
f=simple(subs(f))
g=simple(subs(g))
h=simple(subs(h))
k=simple(subs(k)) 

f = 
0 
g =
0 
h =
0 
k = 
0
 
比 较好的做法是返回一个输出参数,该参数为结构体,然后再获得结构体的域即可：
s=solve(.);
fns = fieldnames(s);
for i=1:length(fns)
    eval([fns{i} '=s.' fns{i}]);
end
在MATLAB中同样也可以用一个命令解决：
>> s=solve('x+z-a', '(2^(1/2)*m*x)/2 + (2^(1/2)*m*y)/2+(2^(1/2)*m*t)/2 - (2^(1/2)*m*z)/2-b', 'm^2*(y- t)-m^2*c', '(2^(1/2)*m^3*(y-x+z+t))/2-m^2*d')
s =
    t: [1x1 sym]
    x: [1x1 sym]
    y: [1x1 sym]
    z: [1x1 sym]
后面用s.t、s.x之类的符号就可以引用求得的结果了.