optimally selecting elements in matrix to satisfy the system of equations

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sarath yadav
sarath yadav 2018-7-27
编辑： Aquatris 2018-8-3
Hi I am sarath, I am working on system of equations that have unknown matrices, which needs to be solved using optimization.
I have 3 different equations which are function of matrices (A,B,C)
A=[0 1 0 0; 0 0 1 0; 0 0 0 1; -a1 -a2 -a3 -a4]
B=[b1;b2;b3;b4]
C=[c1 c2 c3 c4]
I know the right hand side scalar values of all the three equations. I need to solve for this A, B, C matrices such that, those matrices should satisfy the right hand side scalar values.
Requesting for suggestions.
  2 个评论
Amal George M
Amal George M 2018-8-2
编辑：Amal George M 2018-8-2
Hi Sarath,
Can you provide details on how these matrices are related or the equation which needs solving?
sarath yadav
sarath yadav 2018-8-2
编辑：per isakson 2018-8-2
Hi amal,
Thanks for your response. Here are the equations below.
equation.1:
2.12=C*(exp(0.9*A)*(exp(0.1*A)*(inv(I-exp(4.14*A)*(inv(A)*((exp(4.14*A)-2*exp(4.04*A)+2*exp(1.97*A)-I)*B))))+inv(A)*(exp(0.1*A)-I)*1*B)-(inv(A)*(exp(0.9*A)-I)*B))
equation.2:
-0.1=C*(inv(A)*(inv(I-exp(4.14*A))*(exp(4.14*A)*(I-exp(0.1*A)))-(exp(0.1*A)-I))*1*B)
equation.3:
0.1=C*(inv(A)*(inv(I-exp(4.14*A))*(exp(4.24*A)-2*exp(4.14*A)+2*exp(2.07*A)-exp(0.1*A))+(exp(0.1*A)-I))*1*B)
Above are the three equations. All the three equations are the functions of A, B and C.
I need to optimally get A, B and C such that, it satisfies all the three equations.
Structure of A, B and C are as follows:
A=[0 1 0 0; 0 0 1 0; 0 0 0 1; -a1 -a2 -a3 -a4]
B=[b1;b2;b3;b4]
C=[c1 c2 c3 c4]

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回答（1 个）

Aquatris
Aquatris 2018-8-3
编辑：Aquatris 2018-8-3
First of all, in Matlab you need to use expm() function for matrix exponentials.
Secondly, your 1st equation is wrong. the right hand side gives 4x1 row vector instead of scalar.
Other than that you can create a function like (i commented out 1st equation since it has errors, but you can look into what the error is in your equation derivation and fix it);
function err = deltah(x)
a1 =x(1);
a2 =x(2);
a3 =x(3);
a4 =x(4);
b1 =x(5);
b2 =x(6);
b3 =x(7);
b4 =x(8);
c1 =x(9);
c2 =x(10);
c3 =x(11);
c4=x(12);
A=[0 1 0 0; 0 0 1 0; 0 0 0 1; -a1 -a2 -a3 -a4];
B=[b1;b2;b3;b4];
C=[c1 c2 c3 c4];
I = eye(4);
% eq(1) = 2.12-C*(expm(0.9*A)*(expm(0.1*A)*(inv(I-expm(4.14*A)*(inv(A)* ...
% ((expm(4.14*A)-2*expm(4.04*A)+2*expm(1.97*A)-I)*B))))+inv(A)*(expm(0.1*A)-I)*1*B)- ...
% (inv(A)*(expm(0.9*A)-I)*B));
eq(2) = -0.1-C*(inv(A)*(inv(I-expm(4.14*A))*(expm(4.14*A)*(I-expm(0.1*A)))- ...
(expm(0.1*A)-I))*1*B);
eq(3) = 0.1-C*(inv(A)*(inv(I-expm(4.14*A))*(expm(4.24*A)-2*expm(4.14*A)+2*expm(2.07*A)- ...
expm(0.1*A))+(expm(0.1*A)-I))*1*B);
err = norm(eq);
end
Than, in the command window, you can use fminunc or any other optimization functions to have the solution and assign them to the correct variables;
x = fminunc(@deltah,rand(12,1));
a1 =x(1);
a2 =x(2);
a3 =x(3);
a4 =x(4);
b1 =x(5);
b2 =x(6);
b3 =x(7);
b4 =x(8);
c1 =x(9);
c2 =x(10);
c3 =x(11);
c4=x(12);
With the 1st equation omitted, the value of x I got is;
x =
0.3449
-0.0326
0.3539
1.0273
0.2967
0.7542
-0.5809
-0.4496
-1.0049
0.2925
-0.9052
0.8451
and using these, the values of right hand side of the 2nd equation was exactly -0.1000 and the 3rd equation was exactly 0.1000.

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