How to solve multiple equation and get all value of the variable?

Question

DoinK 2022-11-1

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1840623-how-to-solve-multiple-equation-and-get-all-value-of-the-variable

评论： Torsten 2022-11-1

采纳的回答： Torsten

This is the code

syms p1 p2 p3 p4 p5 a

R =[1 2 3 1 2;4 5 1 2 6;7 1 2 1 2;1 4 3 2 2;1 5 4 3 2];

I =[1 0 0 0 0;0 1 0 0 0;0 0 1 0 0;0 0 0 1 0;0 0 0 0 1];

pi =[p1 p2 p3 p4 p5];

e =[1;1;1;1;1];

a=[p1 p2 p3 p4 p5]*(I-R)^(-1)*e

t=a - 1

z=solve(t,p1) %get p1

A_0=[2 1 1 2 4;1 2 2 1 1;2 2 1 1 3;1 3 6 2 2;1 3 4 1 2];

B =[1 3 2 1 2;2 1 2 7 1;1 2 1 2 3;1 4 2 1 2;2 3 1 2 2];

y=pi*(A_0+R*B)

j =subs(y,p1,z) %substitude p1 to 5 equations

[A,Z]=equationsToMatrix(j,pi)

sol=double(A)\double(Z)

After run, i get this warning and solutions

Warning: Matrix is singular to working precision.

> In tes_matlab_baru (line 30)

sol =

-Inf

1.4983

-0.6005

3.7351

-5.4810

What i want to get is all value of pi, how to get it?

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Answer 1

Torsten 2022-11-1

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1840623-how-to-solve-multiple-equation-and-get-all-value-of-the-variable#answer_1088558

编辑：Torsten 2022-11-1

在 MATLAB Online 中打开

p1 can be chosen arbitrarily since the corresponding column in the matrix A is zero.

(I set p1 = 1).

Note that sol does not satisfy A*sol = Z exactly since you have 5 equations for 4 unknowns (p1 does not count since it cannot influence the system).

syms p1 p2 p3 p4 p5 a

R =[1 2 3 1 2;4 5 1 2 6;7 1 2 1 2;1 4 3 2 2;1 5 4 3 2];

I =[1 0 0 0 0;0 1 0 0 0;0 0 1 0 0;0 0 0 1 0;0 0 0 0 1];

pi =[p1 p2 p3 p4 p5];

e =[1;1;1;1;1];

a=pi*(I-R)^(-1)*e

a =

t=a - 1

t =

z=solve(t,p1) %get p1

z =

A_0=[2 1 1 2 4;1 2 2 1 1;2 2 1 1 3;1 3 6 2 2;1 3 4 1 2];

B =[1 3 2 1 2;2 1 2 7 1;1 2 1 2 3;1 4 2 1 2;2 3 1 2 2];

y=pi*(A_0+R*B)

y =

j =subs(y,p1,z) %substitude p1 to 5 equations

j =

[A,Z]=equationsToMatrix(j,pi)

A =

Z =

sol = lsqlin(double(A),double(Z),[],[],[1 0 0 0 0],1)

Minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance.

sol = 5×1

1.0000 0.8534 -0.4176 4.5550 -5.3446

double(A)*sol - double(Z)

ans = 5×1

4.8257 -2.9487 2.6561 -1.1261 -0.7202

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DoinK 2022-11-1

i'm sorry sir, but the sum of pi must equal to 1

so i think we cannot assume p1=1

can we use z (p1 value) to get all value of pi?

Torsten 2022-11-1

在 MATLAB Online 中打开

syms p1 p2 p3 p4 p5 a

R =[1 2 3 1 2;4 5 1 2 6;7 1 2 1 2;1 4 3 2 2;1 5 4 3 2];

I =[1 0 0 0 0;0 1 0 0 0;0 0 1 0 0;0 0 0 1 0;0 0 0 0 1];

pi =[p1 p2 p3 p4 p5];

e =[1;1;1;1;1];

a=pi*(I-R)^(-1)*e

a =

t=a - 1

t =

z=solve(t,p1) %get p1

z =

A_0=[2 1 1 2 4;1 2 2 1 1;2 2 1 1 3;1 3 6 2 2;1 3 4 1 2];

B =[1 3 2 1 2;2 1 2 7 1;1 2 1 2 3;1 4 2 1 2;2 3 1 2 2];

y=pi*(A_0+R*B)

y =

j =subs(y,p1,z) %substitude p1 to 5 equations

j =

[A,Z]=equationsToMatrix(j,pi)

A =

Z =

sol = lsqlin(double(A),double(Z),[],[],[1 1 1 1 1],1)

Minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance.

sol = 5×1

1.3539 0.8534 -0.4176 4.5550 -5.3446

double(A)*sol - double(Z)

ans = 5×1

4.8257 -2.9487 2.6561 -1.1261 -0.7202

sum(sol)

ans = 1.0000

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