Sure, assuming that A is full-rank and small, then its perfectly reasonable to invert A with adjoint(A)&#x2F;discriminant(A).  But this is very expensive, and sensitive to roundoff for larger A.<p>Lets say you are using QR factorization.  Then R\b is easy to compute by back-substitution, because R is triangular.  Q\b is also easy, because Q is orthogonal, and Q&#x27;s inverse is its transpose, so you just calculate transpose(Q)*b.<p>Same thing for LDL^T factorization, LU factorization, etc.  Its easy because back-substitution with a triangular matrix is easy.  In the case of LDL^T factorization, D is diagonal, and again D\b is easy (its a bunch of rank-1 division operations).