how to resolve linear equations of system

[enjoywm at cs.wm.edu]

Thank you very much!
>> 1.
>
> In general the answer is no, as there is no solution.  But, there are
> some situations where an acceptable solution is found using
> appropriate pre-conditioning.  It all depends on the exact
> circumstance you are in. (For example, if the singularity is due to a
> zero row in K, then using the Jacobi pre-conditioner will make it an
> identity row.  This may or may not be acceptable depending on your
> application.)
>
>> 2.
>
> Solving the linear system is almost always preferred as finding the
> inverse to K explicitly boils down to repeatedly solving Kx=u where u
> are individual columns of I.  Even if you are interested in solving
> for many u's it is preferable to save K's factorization rather than
> K's inverse.
>
> Read Matrix Computations by Gene Golub, or an introductory Scientific
> Computing book.  Many of these questions are addressed there.
>
> -Andrew
>
> On Wed, Jun 25, 2008 at 4:29 PM,  <enjoywm at cs.wm.edu> wrote:
>> Hi,
>> I have a linear equations of system such as Kx=u.
>> 1. If K is singular, can I use KSPSolve to solve it?
>> 2. If K is non-singular I also can use x= inverse(K)u to do it.
>> Which one is better?
>>
>> Best,
>> Yixun
>>
>>
>
>