Finding inv(A) for Ax=b system

I tried 3 methods to solve the system for inv(A), where A is highly sparse matrix spy(A) is given below:

1) pinv(A)......... Matlab solve it in 190 sec without any warnings.

2) A\b............. Matlab took only 10 sec but gives warning "Matrix is singular to working precision"

3) [L,U]=lu(A); inv(L)*inv(U) .......... Matlab took 50 sec but give the same warning as in 2nd method.

Is there anyway, I can get inv(A) without warnings.

[Stephen23]

Did you read the inv documentation? It states: "It is seldom necessary to form the explicit inverse of a matrix. A frequent misuse of inv arises when solving the system of linear equations Ax = b. One way to solve the equation is with x = inv(A)*b. A better way, from the standpoint of both execution time and numerical accuracy, is to use the matrix backslash operator x = A\b. This produces the solution using Gaussian elimination, without explicitly forming the inverse. See mldivide for further information."

So really the best answer to your question is to stop using inv. Using inv is almost always the wrong way to solve a problem.

[Jan]

@Asif Arshid: What is your question? You observed that the slash operator has a different sensitivity to detect near to singular matrices than lu and inv(L)*inv(U). What is the condition number of the matrix? Do you think that slash is to pessimistic or the lu method too sloppy? Or are you surprised by the speed of the slash operator?

[Asif Arshid]

@Stephen: I tried "mldivide", but its gives warning of "Matrix is singular to working precision".

@Jan: my question is, I need to solve my system for "x" in less time than "pinv" (190 sec). The condition number is 4.7167e+17.

Answer 1

[Star Strider]

Use the lsqr (link) or similar function to solve sparse matrix problems.

[Asif Arshid]

Thanks @Star for your suggestions, but its also not working for me.

It says, "lsqr stopped at iteration 20 without converging to the desired tolerance 1e-06 because the maximum number of iterations was reached. The iterate returned (number 20) has relative residual 1."

[Walter Roberson]

The 4th parameter to lsqr() allows you to specify a higher number of iterations.

[Star Strider]

@Walter — Thank you.

[M.Shaarawy]

Is there regularized parameterized trust region sub problem (RPTRS) in MATLAB to solve this kind of problem?