1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Significance of a singular matrix

  1. Nov 7, 2007 #1
    So I have a system of equations (composed of force and moment eqns) and I can split them up into matrices which will then look like this:
    Ax = B
    I know the matrices A and B are correct, because when I plug in known values for x from a working prog, I get the correct values for B. So that must mean A and B are correct, right?
    HOWEVER, I am not able to invert A so that I may solve x = inv(A) * B.
    The determinant of A is 0, so it's a singular matrix.
    So what is the significance of an singular matrix?
    Maybe there are redundant eqns involved? Am I looking at a statically indeterminant problem?
    How does one fix this?
  2. jcsd
  3. Nov 7, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    A singular matrix doesn't have an inverse! And yes, that means that the rows making up the matrix A are not independent which, in turn, means that the equations you are using are not independent- at least one is just a combination of the others. You don't have enough data to solve those equations.
  4. Nov 29, 2009 #3
    Singularity of a matrix means that it becomes discontinuous at that point: take for example a black hole: it has a radius zero, which means that its gravitational force becomes undefined then; the same case here is when the adjoint of the co-efficient matrix we got is to be divided by the inverse, which is zero. Thus, the matrix is undefined there.

    This is not a dead end :smile:. We have numerical methods like Gauss-Siedel and Jacobi iteration to find out the solution, that is, X.

    let the eqn be of 3 variables, i.e., f(x1, x2, x3) = 0.

    then we can write it as:
    a11x1 + a12x2 + a13x3 = b1.......1
    a21x1 + a22x2 + a23x3 = b2.......2
    a31x1 + a32x2 + a33x3 = b3.......3

    then take initial gues values of x2(0) & x3(0). Put in eqn 1 n get x1(1).
    Put x1(1), x3(0) n get x2(1) from eqn 2.
    Put x1(1), x2(1) n get x3(1) from eqn 3.

    Now we have the second guess values as: x1(1), x2(1) & x3(1).
    Relaxation: x1(after iteration) = (some weightage)*x1(after iteration) + (1-weightage)*x1(before iteration).

    same for x2 and x3.
    Repeat the iterations till you get:
    abs{x1(k+1),x2(k+1), x3(k+1) - x1(k),x2(k),x3(k)} < tolerance.

    this will be your final set of x1, x2, x3 values.
    Hope this helps.
    The jacobi iteration is simpler than this: Google it yourself.
    here's a reference: http://www.math-linux.com/spip.php?article48
  5. Nov 29, 2009 #4
  6. Nov 30, 2009 #5
    I don't know what is your application, so my hint maybe not be useful.
    However, if you cannot invert that matrix you can still find the pseudoinverse of A (a.k.a. Moore-Penrose inverse, or generalized inverse).
    This will give you the vector x which "gets closest" to the solution in terms of L2-norm. Namely, it is a closed form solution for the following:


    See: http://en.wikipedia.org/wiki/Moore–Penrose_pseudoinverse
    Last edited: Nov 30, 2009
  7. Nov 30, 2009 #6
    First Gauss-Seidel, then Jacobi.
    Now Moore-Penrose.
    I just remembered one more: Gauss-Jordan elimination method of finding a MATRIX INVERSE

  8. Nov 30, 2009 #7
    try using a software named MATLAB: it has readymade methods of finding matrix inverse if you don't need to show to your professor/teacher how you got the answer matrix: X
  9. Nov 30, 2009 #8
    Are you sure the methods you mentioned still work when the matrix A is singular?

    I thought the Moore-Penrose pseudoinverse was the only one that provides the "best solution" in terms of least squares.
    Keep also in mind that the Moore-Penrose pseudoinverse works also when A is not a square matrix!
  10. Nov 30, 2009 #9
    cent percent sure! we have used it in our FORTRAN classes to solve sums.... fresh out of grad...can't be mistaken
  11. Dec 1, 2009 #10
    That means your truss topology or whatever you are working on is overconstrained. Similar to defining a plane with 4 points.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Significance of a singular matrix
  1. Singular matrix theory (Replies: 3)