MATLAB: What is tje problem whit the function “solve”

function solveMATLAB

Hi everyone, I have a problem with de function "solve", I have two programs very similar.

the first works excellent, it's the following:

clear all;
clc;
syms i1 i2 i3 i4 i5 i6 i7 v1 v2 v3 v4 v5 v6 v7 e1 e2 e3 e4 e5 R1 R2 R3 R4 A va vb; 
%KCL

eq0=i1-i2;
eq1=i2+i3;
eq2=i4-i5;
eq3=i5-i6;
eq4=-i3+i7;
%KVL

eq5=  v1-e1;
eq6=  v2+e2-e1;
eq7=  v3+e2-e5;
eq8=  v4-e3;
eq9=  v5+e4-e3;
eq10=  v6-e4;
eq11= v7-e5;
%BR

eq12= v1-va;
eq13= v2-i2*R1;
eq14= v3-i3*R2;
eq15= v4-vb;
eq16= v5-i5*R3;
eq17= v6-i6*R4;
eq18= v7-A*e4+A*e2;
S=solve([eq0,eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14, eq15, eq16, eq17, eq18], [i1,i2,i3,i4,i5,i6,i7,v1,v2,v3,v4,v5,v6,v7,e1,e2,e3,e4,e5]);
sol= S.i2

the second is the following:

clear all;
clc;
syms  i3 i4 i5 i6 i7 i8 i9 i10 i11 i12 v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 v11 v12 e3 e4 e5 e6 e7 e8 e9 R1 R2 R3 Rg A;
%KCL
eq0= i3+i12-i4;
eq1= i4-i5;
eq2= i5+i6;
eq3= i11-i6-i8;
eq4= i7-i3;
eq5= i8-i10;
eq6= i9-i7;
%KVL
eq7=  v3-e7+e3;
eq8=  v4-e3+e4;
eq9=  v5-e4+e5;
eq10= v6-e6+e5;
eq11= v7-e9+e7;
eq12= v8-e6+e8;
eq13= v9-e9;
eq14= v10-e8;
eq15= v11-e6;
eq16= v12-e3;
%BR
eq17= v3-R2*i3;
eq18= v4-R1*i4;
eq19= v5-Rg*i5;
eq20= v6-R1*i6;
eq21= v7-R3*i7;
eq22= v8-R2*i8;
eq23= v9-A*e8+A*e7;
eq24= v10-R3*i10;
eq25= v11-A*v2+A*e5;
eq26= v12-A*v1+A*e3;
S=solve(eq0,eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq17,eq18,eq19,eq20,eq21,eq22,eq23,eq24,eq25,eq26,'i3','i4','i5','i6','i7','i8','i9','i10','i11','i12','v1','v2','v3','v4','v5','v6','v7','v8','v9','v10','v11','v12','e3','e4','e5','e6','e7','e8','e9');
sol=S.e9

I don't know what is the problem because I always get zero.

I hope you can help me, thanks

Best Answer

Is zero a valid solution? It appears to be so. (By the way, you don't put the names of the variables in quotes to call solve. But you knew that from the first time you used solve, in the case that you claim did work.)

Your system appears to be linear. It has no right hand side. Thus it is effectively a linear system with no constant right hand side. zero is avalid solution. Even if I missed a product in there between two of the unknowns, it is still one with a zero right hand side as far as I can see.

A valid solution to any such system is ZERO.

Worse, it looks like you have 27 equations, and 29 unknowns? I'm actually surprised that solve gave you any solution at all, since the system is under-determined. So there will be infinitely many solutions.

So, next, lets see what equationsToMatrix tells us about your linear system.

[M2,rhs2] = equationsToMatrix(eq0,eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq17,eq18,eq19,eq20,eq21,eq22,eq23,eq24,eq25,eq26,[i3,i4,i5,i6,i7,i8,i9,i10,i11,i12,v1,v2,v3,v4,v5,v6,v7,v8,v9,v10,v11,v12,e3,e4,e5,e6,e7,e8,e9]);

As I thought, the system is homogeneous, rhs is identically zero.

rhs2'
ans =
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

We see the system is indeed 27x29, so it is underdetermined.

size(M2)
ans =
    27    29

There are infinitely many solutions, one of which is the all zero solution. Of course this must be true, because it is trivially true that M2*zeros(29,1)=0.

To go back and answer your question, why did the solve work the firt time you used solve?

[M1,rhs1] = equationsToMatrix([eq0,eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14, eq15, eq16, eq17, eq18], [i1,i2,i3,i4,i5,i6,i7,v1,v2,v3,v4,v5,v6,v7,e1,e2,e3,e4,e5]);

It worked because your system is not homogeneous there.

As you can see, the right hand side is NOT identically zero.

So why did you get zero the second time? Linear Algebra 101.

Does a non-zero solution to the second problem exist? Yes, infinitely many of them, and no great way to choose which solution you might like.

So, any linear combination of the columns of

null(M2)
ans =
[                          -1/R3,                                                   (A + 1)/(A*R3)]
[                              0,                             -(R2 + R3 + A*R2)/(A*R3*(2*R1 + Rg))]
[                              0,                             -(R2 + R3 + A*R2)/(A*R3*(2*R1 + Rg))]
[                              0,                              (R2 + R3 + A*R2)/(A*R3*(2*R1 + Rg))]
[                          -1/R3,                                                   (A + 1)/(A*R3)]
[                           1/R3,                                                                0]
[                          -1/R3,                                                   (A + 1)/(A*R3)]
[                           1/R3,                                                                0]
[                           1/R3,                              (R2 + R3 + A*R2)/(A*R3*(2*R1 + Rg))]
[                           1/R3, -(2*R1 + R2 + R3 + Rg + 2*A*R1 + A*R2 + A*Rg)/(A*R3*(2*R1 + Rg))]
[ (R2 + R3 + A*R2 + A*R3)/(A*R3),                             -((A + 1)*(R2 + R3 + A*R2))/(A^2*R3)]
[ (R2 + R3 + A*R2 + A*R3)/(A*R3),                        -(R1*(R2 + R3 + A*R2))/(A*R3*(2*R1 + Rg))]
[                         -R2/R3,                                               (R2 + A*R2)/(A*R3)]
[                              0,                        -(R1*(R2 + R3 + A*R2))/(A*R3*(2*R1 + Rg))]
[                              0,                    -(R2*Rg + R3*Rg + A*R2*Rg)/(A*R3*(2*R1 + Rg))]
[                              0,                         (R1*(R2 + R3 + A*R2))/(A*R3*(2*R1 + Rg))]
[                             -1,                                                        (A + 1)/A]
[                          R2/R3,                                                                0]
[                              0,                                                                1]
[                              1,                                                                0]
[                   (R2 + R3)/R3,                                                                0]
[                   (R2 + R3)/R3,                                         -(R2 + R3 + A*R2)/(A*R3)]
[                   (R2 + R3)/R3,                                         -(R2 + R3 + A*R2)/(A*R3)]
[                   (R2 + R3)/R3,                 -((R1 + Rg)*(R2 + R3 + A*R2))/(A*R3*(2*R1 + Rg))]
[                   (R2 + R3)/R3,                        -(R1*(R2 + R3 + A*R2))/(A*R3*(2*R1 + Rg))]
[                   (R2 + R3)/R3,                                                                0]
[                              1,                                                             -1/A]
[                              1,                                                                0]
[                              0,                                                                1]

will yield a valid solution to your problem.

Again, linear algebra. It is a good class to take, if you will then be using linear algebra.

Related Solutions

MATLAB: SOLVING LARGE SYSTEM OF EQUATIONS

Hi,

try:

a = sym('a',[1 15]);
syms r
p1 = poly2sym(a,r);
diff = sym('diff');
eq = sym('eq');
fprintf('\n polynomial: %s\n', p1);
p2 =poly2sym(a,r);
p2=subs(p2,r,r-1);
%fprintf('\n p2: %s\n', p2);
diff= p1-p2;
%fprintf('\n diff: %s\n', simplify(diff));
diff = coeffs(diff,r);
branch2=prod(r-3:r+5)*(((1/1920)*(r+4)*(r+3)*(r^2+r-10))+((1/5120)*r*(r+1)*(r^2+r-10))+((1/9216)*(r+4)*(r+3)*(r+3)*(r-2)*(r-4)));
%fprintf('\n branch: %s\n', simplify(branch2));
%display(diff);
branch2=coeffs(branch2,r);
for u=1:14
    %fprintf('DIFF: %d = %s\n', u, diff(u));
    %fprintf('BRANCH: %d = %s\n', u, branch2(u));
    eq(u)= diff(u)==branch2(u);
    fprintf('EQ %d : %s\n', u, eq(u));
    %eqcoef(u)=coeffs(diff(u),[a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14]);
    % fprintf('EQCOEF %d : %s\n', u, eqcoef(u));
end
[A,B] = equationsToMatrix([eq(1), eq(2), eq(3),eq(4), eq(5), eq(6),eq(7), eq(8), eq(9),eq(10), eq(11), eq(12),eq(13),eq(14)],a);

this results in:

 polynomial: a15 + a14*r + a1*r^14 + a2*r^13 + a3*r^12 + a4*r^11 + a5*r^10 + a6*r^9 + a7*r^8 + a8*r^7 + a9*r^6 + a10*r^5 + a11*r^4 + a12*r^3 + a13*r^2
EQ 1 : a2 - a1 - a3 + a4 - a5 + a6 - a7 + a8 - a9 + a10 - a11 + a12 - a13 + a14 == 45/2
EQ 2 : 14*a1 - 13*a2 + 12*a3 - 11*a4 + 10*a5 - 9*a6 + 8*a7 - 7*a8 + 6*a9 - 5*a10 + 4*a11 - 3*a12 + 2*a13 == 945/32
EQ 3 : 78*a2 - 91*a1 - 66*a3 + 55*a4 - 45*a5 + 36*a6 - 28*a7 + 21*a8 - 15*a9 + 10*a10 - 6*a11 + 3*a12 == -10541/640
EQ 4 : 364*a1 - 286*a2 + 220*a3 - 165*a4 + 120*a5 - 84*a6 + 56*a7 - 35*a8 + 20*a9 - 10*a10 + 4*a11 == -50033/1280
EQ 5 : 715*a2 - 1001*a1 - 495*a3 + 330*a4 - 210*a5 + 126*a6 - 70*a7 + 35*a8 - 15*a9 + 5*a10 == -275051/23040
EQ 6 : 2002*a1 - 1287*a2 + 792*a3 - 462*a4 + 252*a5 - 126*a6 + 56*a7 - 21*a8 + 6*a9 == 86971/9216
EQ 7 : 1716*a2 - 3003*a1 - 924*a3 + 462*a4 - 210*a5 + 84*a6 - 28*a7 + 7*a8 == 38813/5760
EQ 8 : 3432*a1 - 1716*a2 + 792*a3 - 330*a4 + 120*a5 - 36*a6 + 8*a7 == 7883/23040
EQ 9 : 1287*a2 - 3003*a1 - 495*a3 + 165*a4 - 45*a5 + 9*a6 == -3233/3840
EQ 10 : 2002*a1 - 715*a2 + 220*a3 - 55*a4 + 10*a5 == -181/768
EQ 11 : 286*a2 - 1001*a1 - 66*a3 + 11*a4 == 7/720
EQ 12 : 364*a1 - 78*a2 + 12*a3 == 311/23040
EQ 13 : 13*a2 - 91*a1 == 49/23040
EQ 14 : 14*a1 == 1/9216
>> A
 
A =
 
[    -1,     1,   -1,    1,   -1,    1,  -1,   1,  -1,   1, -1,  1, -1, 1, 0]
[    14,   -13,   12,  -11,   10,   -9,   8,  -7,   6,  -5,  4, -3,  2, 0, 0]
[   -91,    78,  -66,   55,  -45,   36, -28,  21, -15,  10, -6,  3,  0, 0, 0]
[   364,  -286,  220, -165,  120,  -84,  56, -35,  20, -10,  4,  0,  0, 0, 0]
[ -1001,   715, -495,  330, -210,  126, -70,  35, -15,   5,  0,  0,  0, 0, 0]
[  2002, -1287,  792, -462,  252, -126,  56, -21,   6,   0,  0,  0,  0, 0, 0]
[ -3003,  1716, -924,  462, -210,   84, -28,   7,   0,   0,  0,  0,  0, 0, 0]
[  3432, -1716,  792, -330,  120,  -36,   8,   0,   0,   0,  0,  0,  0, 0, 0]
[ -3003,  1287, -495,  165,  -45,    9,   0,   0,   0,   0,  0,  0,  0, 0, 0]
[  2002,  -715,  220,  -55,   10,    0,   0,   0,   0,   0,  0,  0,  0, 0, 0]
[ -1001,   286,  -66,   11,    0,    0,   0,   0,   0,   0,  0,  0,  0, 0, 0]
[   364,   -78,   12,    0,    0,    0,   0,   0,   0,   0,  0,  0,  0, 0, 0]
[   -91,    13,    0,    0,    0,    0,   0,   0,   0,   0,  0,  0,  0, 0, 0]
[    14,     0,    0,    0,    0,    0,   0,   0,   0,   0,  0,  0,  0, 0, 0]
 
>> B
 
B =
 
          45/2
        945/32
    -10541/640
   -50033/1280
 -275051/23040
    86971/9216
    38813/5760
    7883/23040
    -3233/3840
      -181/768
         7/720
     311/23040
      49/23040
        1/9216

Best regards

Stephan

MATLAB: How to find the values of 2 variables that will make the matrix rank 6

for x = -10:1:10
  for y = -10:1:10
    A = [  4, -3,  0,  0,  0, 0,  0
          -2,  0,  1,  0,  0, 0,  0
          -6,  0,  0,  0, 12, 0,  0
          -3,  0,  0,  0,  0, 1,  0
           0,  6,  0, -8,  0, 0,  0
           0,  0,  0,  2,  0, 0, -4
           0,  3,  0,  0, -y, 0,  0
           0,  0, -5,  x,  0, 0,  0] ;
    R = rank(A) ;
    if R == 6
        [x y]
        break
    end
    end
end

Best Answer

Related Solutions

MATLAB: SOLVING LARGE SYSTEM OF EQUATIONS

MATLAB: How to find the values of 2 variables that will make the matrix rank 6

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