Solve system of linear equations ...matrix output is not as expected

Hi All,

I am trying to solve system of equations, as attached,

I have written a code, but my final matrix is 9*9 however, it should be 3*3....I understand i have 9 unknowns with 9 equations, but i wish to understand how could i solve it..

clearvars;
clc; 

syms A11 A12 A13 A21 A22 A23 A31 A32 A33 real; 

eq1 = A11*-0.001179+A12*-6.581+A13*0.008718 == -0.5;
eq2 = A12*-0.001179+A22*-6.581+A23*0.008718 == 0;
eq3 = A31*-0.001179+A32*-6.581+A33*0.008718 == 0;

eq4 = A11*-0.00251+A12*0.05848+A13*-0.039518 == 0;
eq5 = A21*-0.00251+A22*0.05848+A23*-0.039518 == -0.5;
eq6 = A31*-0.00251+A32*0.05848+A33*-0.039518 == 0;

eq7 = A11*-0.0051200+A12*0.2779500+A13*0.0001469 == 0;
eq8 = A21*-0.0051200+A22*0.2779500+A23*0.0001469 == 0;
eq9 = A31*-0.0051200+A32*0.2779500+A33*0.0001469 == -0.5;

[A] = equationsToMatrix([eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8, eq9], [A11, A12, A13, A21, A22,A23, A31, A32, A33]);

Answer 1

[Steven Lord]

A being a 9-by-9 matrix is correct.

syms A11 A12 A13 A21 A22 A23 A31 A32 A33 real; 

eq1 = A11*-0.001179+A12*-6.581+A13*0.008718 == -0.5;
eq2 = A12*-0.001179+A22*-6.581+A23*0.008718 == 0;
eq3 = A31*-0.001179+A32*-6.581+A33*0.008718 == 0;

eq4 = A11*-0.00251+A12*0.05848+A13*-0.039518 == 0;
eq5 = A21*-0.00251+A22*0.05848+A23*-0.039518 == -0.5;
eq6 = A31*-0.00251+A32*0.05848+A33*-0.039518 == 0;

eq7 = A11*-0.0051200+A12*0.2779500+A13*0.0001469 == 0;
eq8 = A21*-0.0051200+A22*0.2779500+A23*0.0001469 == 0;
eq9 = A31*-0.0051200+A32*0.2779500+A33*0.0001469 == -0.5;

equations = [eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8, eq9];
v = [A11, A12, A13, A21, A22,A23, A31, A32, A33];
[A, b] = equationsToMatrix(equations, v);

Let's attempt to recreate the original equations using A and b.

A*v.' == b

ans = 

equations.'

ans = 

Those look like they match to me. Now to solve for v:

v2 = A\b

v2 = 

vpa(A*v2-b, 5)

ans = 

[Mark Sc]

@Steven Lord

Thanks for your prompt reply..

I think, I meant another thing, as I will write below:

sol = solve([eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8, eq9], [A11, A12, A13, A21, A22, A23, A31, A32, A33]);
A11Sol = double(sol.A11);
A12Sol = double(sol.A12);
A13Sol = double(sol.A13);
A21Sol = double(sol.A21);
A22Sol = double(sol.A22);
A23Sol = double(sol.A23);
A31Sol = double(sol.A31);
A32Sol = double(sol.A32);
A33Sol = double(sol.A33);
C=[A11Sol A12Sol A13Sol; A21Sol A22Sol A23Sol;A31Sol A32Sol A33Sol];

May I just ask if my new modifications do what is correct ?

[Steven Lord]

syms A11 A12 A13 A21 A22 A23 A31 A32 A33 real; 

eq1 = A11*-0.001179+A12*-6.581+A13*0.008718 == -0.5;
eq2 = A12*-0.001179+A22*-6.581+A23*0.008718 == 0;
eq3 = A31*-0.001179+A32*-6.581+A33*0.008718 == 0;

eq4 = A11*-0.00251+A12*0.05848+A13*-0.039518 == 0;
eq5 = A21*-0.00251+A22*0.05848+A23*-0.039518 == -0.5;
eq6 = A31*-0.00251+A32*0.05848+A33*-0.039518 == 0;

eq7 = A11*-0.0051200+A12*0.2779500+A13*0.0001469 == 0;
eq8 = A21*-0.0051200+A22*0.2779500+A23*0.0001469 == 0;
eq9 = A31*-0.0051200+A32*0.2779500+A33*0.0001469 == -0.5;

equations = [eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8, eq9];
v = [A11, A12, A13, A21, A22,A23, A31, A32, A33];
sol = solve(equations, v);

Let's check by substituting back into the original equations.

check = subs(equations, sol)

check = 

Looks good to me. 0 is equal to 0 and -1/2 is equal to -1/2. In addition:

all(isAlways(check))

ans = logical
   1

isAlways says that all the elements in check are always true.

[Mark Sc]

@Steven Lord

Thanks I see, but I am just afraid as the answer should has some zeros elements,

but thanks again