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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 Moore-Penrose pseudoinverse.
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, non-negative singular values $\sigma_i$ $i=1,\ldots,n$ We find
$$ A^+ = V(S^TS)^{-1}S^TU^T $$
If the rank $r$ of $A$ is smaller than $n$ the inverse of $S^TS$ 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.
The term ``pseudoinverse'' is actually used for any operator $\operatorname{pinv}$ satisfying
$$ M \operatorname{pinv}(M) M = M $$
for a $m \times n$ matrix $M$ Beyond this, pseudoinverses can be defined on any reasonable matrix identity.
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