np.linalg.solve(A, b)
does not compute the inverse of A. Instead it calls one of the gesv
LAPACK routines, which first factorizes A using LU decomposition, then solves for x using forward and backward substitution (see here).
np.linalg.inv
uses the same method to compute the inverse of A by solving for A-1 in A·A-1 = I where I is the identity*. The factorization step is exactly the same as above, but it takes more floating point operations to solve for A-1 (an n×n matrix) than for x (an n-long vector). Additionally, if you then wanted to obtain x via the identity A-1·b = x then the extra matrix multiplication would incur yet more floating point operations, and therefore slower performance and more numerical error.
There’s no need for the intermediate step of computing A-1 – it is faster and more accurate to obtain x directly.
* The relevant bit of source for inv
is here. Unfortunately it’s a bit tricky to understand since it’s templated C. The important thing to note is that an identity matrix is being passed to the LAPACK solver as parameter B
.