Basic matrix equation Ax=B with restrictions

Hi! Im modeling a truss using FEM and im trying to solve a very basic equation A*x=b where a is the stiffness matrix , x and b are the displacements and forces vectors. So A and b are known and i want to find x, the displacements. I know i could type x=b/A in matlab, the thing is i know some of the entries of the x vector due to boundary conditions, and typing a=b/A it works as if all x entries where unknowns. Is there a function to do this in MATLAB ?

Thanks a lot!!

Pd: Sorry for my poor english!

Answer 1

[Greg Heath]

x = [ x1 ; x2 ]% x2 known

A = [ A1 A2 ]% A1 & A2 known

A*x = A1*x1 + A2*x2 % last term known

A1*x1 = b - A2*x2

Voila!

Hope this helps.

Greg

P.S. Ain't Linear Systems grand?

[bym]

+1 Much better answer than my feeble attempt

Answer 2

[bym]

you need to reduce the size of your stiffness and force matrix to eliminate those entries where you know the value of x. I call these Ared and bred. Then perform

xred = Ared\bred

then substitute the known values of x into xred and multiply by the stiffness matrix to get the full force matrix.

It is difficult to explain in words, if you can post your A b x then an example would be better

Answer 3

[Dr. Seis]

If "x" and "b" are column vectors, then your system of equations is like:

A(1,1) * x(1,1) + A(1,2) * x(2,1) + ... + A(1,N) * x(N,1) = b(1,1)
A(2,1) * x(1,1) + A(2,2) * x(2,1) + ... + A(2,N) * x(N,1) = b(2,1)
...
A(M,1) * x(1,1) + A(M,2) * x(2,1) + ... + A(M,N) * x(N,1) = b(M,1)

If you already know some values of x, then move them to the right-hand side of your system of linear equations. For example, if you know the value of x(2,1) then your system of linear equations would be:

A(1,1) * x(1,1) + A(1,3) * x(3,1) + ... + A(1,N) * x(N,1) = b(1,1) - A(1,2) * x(2,1)
A(2,1) * x(1,1) + A(2,3) * x(3,1) + ... + A(2,N) * x(N,1) = b(2,1) - A(2,2) * x(2,1)
...
A(M,1) * x(1,1) + A(M,3) * x(3,1) + ... + A(M,N) * x(N,1) = b(M,1) - A(M,2) * x(2,1)

In this example (and if x(2,1) = 5), you would then redefine you matrices as follows:

b = b - A(:,2)*5;
A = A(:,1:N~=2);

You will find your new "x" as follows:

x = A\b;

[Walter Roberson]

If you are using inv() you are probably doing something wrong!

The mldivide ('\') operator has no problem with non-square matrices.

[Dr. Seis]

The results *should* be identical in theory. However, in practice, using the x=A\b will offer more accuracy (as documented in the Help Navigation under "inv").

[Dr. Seis]

Updated x = A\b;

from x = inv(transpose(A)*A)*transpose(A)*b;