Just to elaborate on this, the matrix B such that BA = I_6 can be found (iff A has rank 6) by noting that (A^t * A) will also be of rank 6, and be a 6x6 matrix, and thus have some inverse B'. Then B' * A^t * A = I_6, so by setting B = B' * A^t we have B * A = I_6. (Here, ^t is used to denote the matrix transpose.)
On 12月3日, 午前6:09, John Cremona <[email protected]> wrote: > On Dec 2, 7:38 am, Santanu Sarkar <[email protected]> > wrote: > > > Let A be a matrix of order (24,6). How one can find a matrix B in Sage of > > order (6,24) such that > > AB=Identity matrix? > > There is no such matrix, since AB cannot have rank more than 6. > > Did you mean to solve BA = 6x6 identity? That would have a solution > iff A has rank 6. > > John Cremona -- To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
