getting a wrong solution to a simple linear system using "solve"

I am trying to solve the following simple symbolic singular 6X6 linear system using "solve":

[xx,yy,zz,yz,xz,xy]=solve(' 0*xx -f1*yy + f1*zz + (b1-c1)*yz + d1*xz -e1*xy=0' , 'e1*xx -e1*zz -d1*yz + (c1-a1)*xz + f1*xy=0' , '-d1*xx + d1*yy + e1*yz -f1*xz + (a1-b1)*xy=0'... ,' -f2*yy + f2*zz + (b2-c2)*yz + d2*xz -e2*xy=0' , 'e2*xx -e2*zz -d2*yz + (c2-a2)*xz + f2*xy=0' , '-d2*xx + d2*yy + e2*yz -f2*xz + (a2-b2)*xy=0' )

The correct answer should be xy=yz=xz=0 and xx=yy=zz=z (z free parameter). Instead I am getting xx=yz=xz=z and xy=yy=zz=0.

Why is that?

Also, is there a way to solve symbolicallty using matrices? Something like linsolve with symbolic arguments?

Thanks Uri

Answer 1

[Friedrich]

Hi Uri,

this is because you are using a wrong syntax for your problem. With the following syntax everything works fine

sol =solve(' 0*xx -f1*yy + f1*zz + (b1-c1)*yz + d1*xz -e1*xy=0' , 'e1*xx -e1*zz -d1*yz + (c1-a1)*xz + f1*xy=0' , '-d1*xx + d1*yy + e1*yz -f1*xz + (a1-b1)*xy=0'... 
    ,' -f2*yy + f2*zz + (b2-c2)*yz + d2*xz -e2*xy=0' , 'e2*xx -e2*zz -d2*yz + (c2-a2)*xz + f2*xy=0' , '-d2*xx + d2*yy + e2*yz -f2*xz + (a2-b2)*xy=0' );

xx = sol.xx
yy = sol.yy
zz = sol.zz
xy = sol.xy
yz = sol.yz
xz = sol.xz

This behavior is also stated in the documentation:

For a system of equations and an equal number of outputs, the results are sorted alphabetically and assigned to the outputs.

http://www.mathworks.com/access/helpdesk/help/toolbox/symbolic/solve.html

I hope I could help,

Friedrich

[uri albocher]

Thank you Fredrich! This does help.

In this syntax, where are the unknowns declared? How does matlab know, for example, to solve for xx and not for a1?

Can I give "solve" input in matrix form? for example provide a 6X6 matrix with the coefficients a1..f2 ?

Thanks again

Uri

Answer 2

[Friedrich]

I think that MATLAB gues the right values is a little bit luck. Normally you would specify what you want to solve:

sol =solve(' 0*xx -f1*yy + f1*zz + (b1-c1)*yz + d1*xz -e1*xy=0' , 'e1*xx -e1*zz -d1*yz + (c1-a1)*xz + f1*xy=0' , '-d1*xx + d1*yy + e1*yz -f1*xz + (a1-b1)*xy=0'... 
    ,' -f2*yy + f2*zz + (b2-c2)*yz + d2*xz -e2*xy=0' , 'e2*xx -e2*zz -d2*yz + (c2-a2)*xz + f2*xy=0' , '-d2*xx + d2*yy + e2*yz -f2*xz + (a2-b2)*xy=0','xx','yy','zz','xy','yz','xz');

xx = sol.xx
yy = sol.yy
zz = sol.zz
xy = sol.xy
yz = sol.yz
xz = sol.xz

I think you can pass a matrix to solve. This works for me:

>> syms a b c d
>> A = [a b ; c d],
>> sol = solve(A)

Friedrich

[uri albocher]

I see. Thanks.

I tied to copy-paste the matrix form into the command window, but I am getting the following error:

??? Error using ==> sym.maple

at offset 26, `;` unexpected

Error in ==> sym.transpose at 18

B = maple('transpose',A);

Error in ==> sym.findsym at 26

sc = char(S(:).');

Error in ==> solve at 99

vars = ['[' findsym(sym(eqns),neqns) ']'];

Error in ==> sym.solve at 49

[varargout{1:max(1,nargout)}] = solve(S{:});

Did you actually try to run this and got it to work?

Thanks

Uri

[Friedrich]

Yes i tried it and it works fine:

syms a b c d

A = [a b ; c d],

sol = solve(A)

A =

[ a, b]

[ c, d]

sol =

a: [1x1 sym]

b: [1x1 sym]

c: [1x1 sym]

d: [1x1 sym]

I am using Matlab R2010b and Win7 64bit.