solve a system of equations symbolically

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Mahsa Babaee
Mahsa Babaee am 20 Okt. 2021
Kommentiert: Mahsa Babaee am 28 Okt. 2021
Hello
I want to solve a system of equations symbolically. But I faced some problems. I was wondering if you could guide me on why MATLAB is incapable of the determination of the Z as a function of a1 and a2? And how can I find Z?
Thanks for your attention in advance.
Here is my code:
clc
clear
syms a3 a2 a1
eqn=[a3-3*a2+10*a1==68,5*a3+6*a2+2*a1==31];
S= solve(eqn);
X=S.a1
X = 
Y=S.a2
Y = 
Z=S.a3
Unrecognized field name "a3".

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Paul
Paul am 20 Okt. 2021
It looks like there are two equations with three unknowns for which there are many solutions. So solve() gives the results for two of the variables in terms of the third, which you can set to an arbitray value. In this case, it chose to solve for a1 and a2 in terms of a3. But you can make it solve for any two in terms of the third, for example a2 and a3 in terms of a1
syms a3 a2 a1
eqn = [a3-3*a2+10*a1==68,5*a3+6*a2+2*a1==31];
S = solve(eqn,[a2 a3])
S = struct with fields:
a2: (16*a1)/7 - 103/7 a3: 167/7 - (22*a1)/7
Now pick an arbitrary value for a1, say sqrt(3), and show that the solution satisfies the eqn
subs(eqn,[a1 a2 a3],[sqrt(3) subs([S.a2 S.a3],a1,sqrt(3))])
ans = 
  3 Kommentare
Paul
Paul am 28 Okt. 2021
It seems like you did everything correct. Are you getting results different than shown as follows?
syms S1 S2 S3 A1 A2 A3 B1 B2 B3 M21 M32
eqn=[A1*sinh(S1)+B1*sinh(S1)-B2==0, A2*sinh(S2)+B2*cosh(S2)-B3==0, A3*sinh(S3)+B3*cosh(S3)==0, A1*cosh(S1)+B1*sinh(S1)-A2*M21==0, A2*cosh(S2)+B2*sinh(S2)-A3*M32==0];
S= solve(eqn, [A2 A3 B1 B2 B3]);
S.A2
ans = 
S.A3
ans = 
S.B1
ans = 
S.B2
ans = 
S.B3
ans = 
Mahsa Babaee
Mahsa Babaee am 28 Okt. 2021
Thanks a million.
It works!

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