I get that inverting a matrix can be computationally intensive, but what&#x27;s the alternative?<p>&quot;What if you have to solve Ax = b for a lot of different b‘s? Surely then it’s worthwhile to find A-1. No. The first time you solve Ax = b, you factor A and save that factorization. Then when you solve for the next b, the answer comes much faster.&quot;<p>I don&#x27;t get it, although I&#x27;m not a numerics expert so maybe I&#x27;m missing something. He says, &quot;the first time you solve Ax = b, you factor A and save that factorization.&quot;<p>Great. But don&#x27;t you then use the factorization to determine the inverse of A (LAPACK example: call dgetrf to get the factorization, then call dgetri to compute the inverse using the factorization)? Maybe what he&#x27;s talking about is going over my head. How else would you solve Ax = b for x, if not by computing the inverse of A (assuming A is a 3x3 or 4x4)?