If we have an n by n matrix dubbed A. How do we recognize if there is one inverse matrix A^-1 such the the product A * A^-1 is the n by n identity matrix?

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If the determinant the the procession A (detA) is no zero, climate this matrix has an inverse matrix. This residential or commercial property of a matrix deserve to be uncovered in any kind of textbook on greater algebra or in a textbook top top the theory of matrices.

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Yes Jose las vegas is right. Infact, what girlfriend should have actually if det(A) is non-zero A.Inv(A) = Inv(A).A = i the identity matrix. However, for large value of n it is daunting to find det(A). If you apply, Gauss removed method, then during elimintion procedure t some allude your diagonal facet becomes zero can not be made non-zero by elementary row exchange climate the matrix is singular and the station does no exist.
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If A is square you require det(A) nomzero. If A is no square and also you don' t require the other condition A*A^(-1)=I, then you are trying to find "left inverses", and also that is a long story. Ns would examine Wikipedia for help. Cheers!

Yes Jose las vegas is right. Infact, what girlfriend should have actually if det(A) is non-zero A.Inv(A) = Inv(A).A = i the identity matrix. However, for huge value the n that is complicated to uncover det(A). If girlfriend apply, Gauss removed method, then throughout elimintion procedure t some suggest your diagonal facet becomes zero deserve to not be made non-zero by elementary heat exchange climate the procession is singular and the inverse does not exist.
It counts on the matrix. If the is of form integer, climate you have the right to do Gauss-Jordan elimination. If you don't finish up through a zero row, climate your procession is invertible. Of food computation the determinant for tiny n is much more efficient. Other technique is to try to uncover eigenvalues, if zero is not among them, climate again A is invertible. Over there exist nearly ten various equivalent ways for your task.
How you choose to show existence of an station really counts on the matrix. There are instances where finding det(A) is far more an overwhelming than prove .
In other case, the product the the matrices A and, in this situation A^-1, will offer you a procession of a rank same to the minimial location of the matrices A and also A^-1. Therefore, friend cannot achieve the identification matrix i m sorry have complete rank same n.

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I such way, also if the dimension of the matrix is large, we may identify if the procession A is invertible there is no computation that its determinant.
R. Mittal's answer to be the sufficient one in the 19th century (where no one considered big matrices). Now the just reasonable way is to do a singular value decomposition (SVD) and inspect the singular worths (i.e. Let the computer system do that). If at the very least one of lock is zero (I don't comment on the question 'what is zero' for results of an extensive numerical computations) climate the matrix has no classic inverse. It has a pseudo-inverse (Penrose inverse) despite (which is very closely related come the SVD), which have the right to replace the classic inverse in plenty of applications where classically one offered the inverse. Peter's donation to the side that the difficulty is valuable. Of course, all this stuff is in Wikipedia, and everybody who works through matrices in all yet the most trivial applications require to recognize it!