Error solving three equations in three unknowns
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syms Tan Man Lan eqns vars L av bv cv
Tan =- b - 2 == 0;
Man = c + 1 == 0;
Lan = a + b - 3*c - 1 == 0;
eqns = [Tan Man Lan];
vars = [a b c];
L = solve(eqns,vars,'ReturnConditions',true)
av=L.a
bv=L.b
cv=L.c
Error is Empty sym: 0-by-1
5 commentaires
Ameer Hamza
le 25 Oct 2020
Which release are you using. I get following result
syms Tan Man Lan eqns vars L av bv cv b c a
Tan =- b - 2 == 0;
Man = c + 1 == 0;
Lan = a + b - 3*c - 1 == 0;
eqns = [Tan Man Lan];
vars = [a b c];
L = solve(eqns,vars,'ReturnConditions',true)
Result
>>
av=L.a
bv=L.b
cv=L.c
av =
0
bv =
-2
cv =
-1
Mohamed Mahmoud
le 25 Oct 2020
Ameer Hamza
le 25 Oct 2020
Even in MATLAB online, I am getting the correct result. Can you copy and paste the code in my answer and run it?
Mohamed Mahmoud
le 25 Oct 2020
John D'Errico
le 25 Oct 2020
First, I would point out you seem to think you need to define all of your variables.
syms Tan Man Lan eqns vars L av bv cv
Yet ALL of those variables were defined later.
In fact, the only variables in your code that you never defined as symbolic, were the ones you really need to define! That is, a,b,c all need to be created as symbolic variables, you you never did that.
Réponses (1)
syms a b c
Tan = -b - 2 == 0;
Man = c + 1 == 0;
Lan = a + b - 3*c - 1 == 0;
[av,bv,cv] = solve(Tan,Man,Lan,a,b,c)
This is a linear system, so there is no need even to specify the returnconditions.
As you can see, the only thing I needed to define as symbolic were the variables a,b,c. Everything else is then derived from them.
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