Having the LU factorization of the matrix $A$ on hand makes it really easy to solve problems in the form $Ax = b$. Since $A = LU$, we can write $LUx = b$, then solve $Ly = b$ for $y$ followed by $Ux = y$ for $x$.
This makes sense if you happen to be given $A$ along with its decomposition $LU$. But in practice, my textbook (Strang 4e, section 1.5) is giving me only $A$. To solve the matrix equation, I first write out the elimination matrices that correspond to the Gaussian row operations that will turn $A$ into $U$. Multiplying these in the correct order yields $L^{-1}$, which I then use to solve the whole system in one shot by back substitution: $Ux = L^{-1}b$.
This way of doing things gets me the correct answers, but I feel like I'm not "really" doing LU decomposition as Strang intends for me to. For one thing, while the textbook hasn't officially introduced inverse matrices at this point, I don't see how I can factor $A$ into $LU$ without first finding $L$'s inverse.
More to the point, I never actually determine or use $L$ itself, since the inverse is what solves the $Ly = b$ equation anyway. When the textbook asks me to provide L, I have been writing out the individual matrices that will undo the elimination steps and multiplying them together in the reverse order, which seems pointless.
So, I have two questions:
When solving $Ax = b$, why bother taking the inverse of the Gaussian elimination steps (that is, $[L^{-1}]^{-1}$) to find $L$ when $L^{-1}$ (the elimination matrix) is what we actually use in the end? Why do we prefer $A = LU$ to $L^{-1}A = U$?
Does L mean anything other than "the inverse of the elimination steps"? Can it be computed more simply? The advantage of LU decomposition over Gaussian elimination disappears if finding L requires you to figure out all the Gaussian elimination steps in the first place.
A few notes:
I know L and U can be computed more quickly than Gaussian elimination. My question isn't about computing time; it's about whether the fast algorithms that find L can yield any insight into the information present in L.
There was a very similar question asking why LU decomposition is preferred to elimination. The accepted answer argues that the reason for using LU substitution is so that you can adapt quickly to any $b$ vector. But the process I've described above also doesn't deal with $b$ until the last step, so I believe my question is distinct.
