Error solving three equations in three unknowns

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

댓글 수: 5

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
The online version at https://matlab.mathworks.com/
Even in MATLAB online, I am getting the correct result. Can you copy and paste the code in my answer and run it?
The only diff I used
Tan=simplify((solve(S==(T^f),f)))==0
Man=simplify((solve(S1==(M^f),f)))==0
Lan=simplify((solve(S2==(L^f),f)))==0
eqns = [Tan Man Lan];
vars = [a b c];
L = solve(eqns,vars,'ReturnConditions',true);
Instead of writing the equations
The answer =
a: [0×1 sym]
b: [0×1 sym]
c: [0×1 sym]
parameters: [1×0 sym]
conditions: [0×1 sym]
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.

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답변 (1개)

John D'Errico
John D'Errico 2020년 10월 25일
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)
av = 
0
bv = 
2
cv = 
1
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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질문:

2020년 10월 25일

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2020년 10월 25일

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