solve a linear system of equations with several unknown parameters

hello ,

I have a system V*F=u with

V= [2 1 1; 4 1 1; 4 2 1; 2 2 1]

F=[f11, f21;f12, f22;g1, g2]

and u= [10,10;10,5 ;10,4; 40 ,40]

and i want to sove it in order to find f11,f12,f21,f22,g1 ang g2. I tried with this code but i get warning msg :Solution does not exist because the system is inconsistent.

I am not sure this is the right way to solve such a system

syms f11 f12 f21 f22 g1 g2

V=[2 1 1; 4 1 1; 4 2 1; 2 2 1] ;

F=[f11 f21;f12 f22;g1 g2];

U=[10,10;10,5 ;10,4; 40 ,40] ;

L=V*F;

eqn1 = L(1,1) == U(1,1);

eqn2 = L(1,2) == U(1,2);

eqn3 =L(2,1) == U(2,1);

eqn4 =L(2,2) == U(2,2);

eqn5 =L(3,1) == U(3,1);

eqn6 =L(3,2) == U(3,2);

eqn7 =L(4,1) == U(4,1);

eqn8 =L(4,2) == U(4,2);

[D,E] = equationsToMatrix([eqn1, eqn2, eqn3, eqn4, eqn5,eqn6,eqn7,eqn8], [f11, f12,f21,f22 g1,g2]);

Ug = linsolve(D,E);

thank you in advance.

Answer 1

[Steven Lord]

syms f11 f12 f21 f22 g1 g2
V=[2    1    1;   4     1     1;  4     2     1;  2     2     1] ;
F=[f11 f21;f12 f22;g1 g2];
U=[10,10;10,5 ;10,4;  40 ,40] ;
L=V*F == U

L = 

Subtract the second equation in the first column from the third.

L(3, 1)-L(2, 1)

ans = 

This tells us that f12 must be 0. Now subtract the fourth equation in the first column from the first.

L(4, 1)-L(1, 1)

ans = 

This tells us that f12 must be 30. Is there a number that is simultaneously equal to 0 and equal to 30?

Now if you wanted a least-squares solution use the mldivide function or the \ operator:

FM = V\U

FM = 3×2

-7.5000 -10.2500 15.0000 14.5000 17.5000 23.7500

check = V*FM-U

check = 4×2

7.5000 7.7500 -7.5000 -7.7500 7.5000 7.7500 -7.5000 -7.7500

[tty]

thanks for your answer. I have a qst though : is the system solvable although f12 has two different values which is impossible ?

[Torsten]

This is what Steven Lord tried to explain: your equations are contradictory and thus do not have a "solution" in the usual sense.

But as always, you can find FM that "best" solves your equations in the sense that norm(V*FM-U) is minimized. For a "solution" in the usual sense, norm(V*FM-U) would be 0.

[tty]

ah ok. Thank you both for the replies.

Answer 2

[Torsten]

You have 8 equations for 6 unknowns. The consequence in general is that there is no solution that satisfies the 8 equations exactly, but only in the least-squares sense. This is also the case here:

syms f11 f12 f21 f22 g1 g2
V=[2    1    1;   4     1     1;  4     2     1;  2     2     1] ;
F=[f11 f21;f12 f22;g1 g2];
U=[10,10;10,5 ;10,4;  40 ,40] ;
L=V*F;
eqn1 = L(1,1) == U(1,1);
eqn2 = L(1,2) == U(1,2);
eqn3 =L(2,1) == U(2,1);
eqn4 =L(2,2) == U(2,2);
eqn5 =L(3,1) == U(3,1);
eqn6 =L(3,2) == U(3,2);
eqn7 =L(4,1) == U(4,1);
eqn8 =L(4,2) == U(4,2);
[D,E] = equationsToMatrix([eqn1, eqn2, eqn3, eqn4, eqn5,eqn6,eqn7,eqn8], [f11, f12,f21,f22 g1,g2])

D = 

E = 

%Ug = linsolve(D,E);
Ug = double(D)\double(E)

Ug = 6×1

-7.5000 15.0000 -10.2500 14.5000 17.5000 23.7500

double(D)*Ug-double(E)

ans = 8×1

7.5000 7.7500 -7.5000 -7.7500 7.5000 7.7500 -7.5000 -7.7500