I think that the idea is not that you don&#x27;t take steps that, taken together, amount to inverting `A`, but rather that you don&#x27;t <i>just</i> try to invert `A`.  Thus, I think that the idea is to do something like &quot;Compute `A = LU`, and then invert `L` and `U`&quot;—which of course can be <i>used</i> to invert `A`, but which isn&#x27;t <i>exactly</i> the same as &quot;Compute `A^{-1}`&quot; (using Cramer&#x27;s rule, or some other absurdity).<p>Incidentally, while solving `Ax = b` <i>in general</i> is definitely equivalent to inverting `A` (simply by taking `b` to be basis vectors, so that the solutions `x` are the columns of the inverse matrix in that basis), it is certainly possible to solve it <i>in particular cases</i> without inverting `A`; for example, the solution to `Ax = b`, where `b` is the first column of `A` in a particular basis, is the first vector `x` of that basis—no inversion required (nor even computation of `b`)!