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pseudoinverse
The inverse $A^{{1}}$ of a matrix $A$ exists only if $A$ is square and has full rank. In this case, $Ax=b$ has the solution $x=A^{{1}}b$.
The pseudoinverse $A^{+}$ (beware, it is often denoted otherwise) is a generalization of the inverse, and exists for any $m\times n$ matrix. We assume $m>n$. If $A$ has full rank ($n$) we define:
$A^{+}=(A^{T}A)^{{1}}A^{T}$ 
and the solution of $Ax=b$ is $x=A^{+}b$.
More accurately, the above is called the MoorePenrose pseudoinverse.
1 Calculation
The best way to compute $A^{+}$ is to use singular value decomposition. With $A=USV^{T}$ , where $U$ and $V$ (both $n\times n$) orthogonal and $S$ ($m\times n$) is diagonal with real, nonnegative singular values $\sigma_{i}$, $i=1,\ldots,n$. We find
$A^{+}=V(S^{T}S)^{{1}}S^{T}U^{T}$ 
If the rank $r$ of $A$ is smaller than $n$, the inverse of $S^{T}S$ does not exist, and one uses only the first $r$ singular values; $S$ then becomes an $r\times r$ matrix and $U$,$V$ shrink accordingly. see also Linear Equations.
2 Generalization
$M\operatorname{pinv}(M)M=M$ 
for a $m\times n$ matrix $M$. Beyond this, pseudoinverses can be defined on any reasonable matrix identity.
References

Originally from The Data Analysis Briefbook (http://rkb.home.cern.ch/rkb/titleA.html)
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Comments
Generalized PseudoInverse
As far as I know a matrix B is a (generalized) pseudoinverse of A if
(1) ABA=A
(2) BAB=B
I am missing (2) in the article.
Re: Generalized PseudoInverse
There are different notions of pseudoinverse (or generalized inverse). At the moment I'm using a book called "Generalized Inverses" by BenIsrael and Greville in which its first chapter you can find all these different notions. They give four conditions,
(1) ABA=A
(2) BAB=B
(3) (AB)*=AB
(4) (BA)*=BA
There's a unique B satisfying the four conditions (a {1,2,3,4}inverse in BenIsrael and Greville's notation), which is the MoorePenrose pseudoinverse. What TN talks about is a {1,2}inverse and so on.