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Is there any method for finding inverse of a matrix other than Gauss-Jordan and (1/detA)(adjA methods?

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The inverse of a matrix is a critically important conceptual tool. And there are even good algorithms, of many kinds, including ones for very special but important matrices in which you just have a narrow band of non-zero entries close to the main diagonal. (One needs these for numerical solution of PDE.) But by and large, one tries to bypass calculation of the inverse. –  André Nicolas Sep 14 '13 at 7:02
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You should be clear about your context. Are you a student needing to invert $3\times3$ matrices with integer coefficient, or an engineer needing to solve a $2000\times2000$ sparse linear system with floating point coefficients? The most relevant answers will vary accordingly. –  Marc van Leeuwen Sep 14 '13 at 7:27
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3 Answers

If you have a decomposition of a matrix then it will usually help you find the inverse - e.g. after computing the Singular Value Decomposition $M = U D V^*$, we simply have $M^{-1} = VD^{-1}U^*$ where the inverse of the diagonal matrix is very easy to compute.

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I don't think finding the SVD is any easier than doing Gauss-Jordan. In addition it involves the inner product structure that is irrelevant for defining inverses; this probably only involves additional complications –  Marc van Leeuwen Sep 14 '13 at 7:25
    
@MarcvanLeeuwen: Sure, computing a decomposition like SVD is certainly not an efficient way to invert an arbitrary matrix. Depending on how your matrix arises, however, there may be an easy decomposition available that simplifies things. –  Anthony Carapetis Sep 14 '13 at 7:27
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Numerical linear algebra has much to say about how to solve $Ax = b$ -- QR factorization, SVD, iterative methods -- and any method for solving $Ax = b$ can be used to compute $A^{-1}$. (Though you rarely want to form $A^{-1}$ explicitly in practice.)

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One possibility is to use Hamilton-Cayley theorem. See this Wikipedia article. Another possibility would be to consider a series of powers of $I-A$ that can be proven convergent in some cases. See this and this Wikipedia articles.

Edit: As the Wikipedia article does not supply an example for the second method here is one. Take a square matrix $A$ of the form $I - D$ where $D$ is a diagonal matrix such that all the elements of its diagonal have modulus strictly smaller than $1$. Consider these equalities:

$A^{-1} = (I-I+A)^{-1}=(I-D)^{-1}=\sum_{n=0}^{\infty} D^{n}$

As $D$ is a diagonal matrix it's easy to see that the series converges to a diagonal matrix whit a convergent geometric series in each element of its diagonal. Write down the matrix and verify that it is the inverse of $A$. In order to apply the method it is not essential for D to be diagonal but it must have all of it eigenvalues of modulus smaller than $1$ and, unless you have a good canonical form for it, forget about calculating the sum of the series.

A note on the convergence of the series: any "decent" notion of convergence will be equivalent as you are in a finite dimensional space. You can just think that the matrix is a vector of $n^2$ components written in the form of an square array and consider componentwise convergence.

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