Continuously differentiable $LU$ factorization matrix

Hello I have the following problem and I hope you can help me:

Suppose the entries of $$A(\epsilon)\in\mathbb R^{n\times n}$$ are continuously differentiable functions of the scalar $$\epsilon$$. Assume that $$A\equiv A(0)$$ and all its principal sub matrices are nonsingular. Show that for sufficiently small $$\epsilon$$ the matrix $$A(\epsilon)$$ has an $$LU$$ factorization $$A(\epsilon)=L(\epsilon)U(\epsilon)$$ and that $$L(\epsilon)$$ and $$U(\epsilon)$$ are both continuously differentiable.

My attempt at reasoning has been as follows:

The fact that the matrix is continuously differentiable implies that each of its inputs are continuously differentiable, the fact that $$A \equiv A(0)$$ I don't see what I can contribute, but the fact that each of its of its matrices is non-singular implies that the matrix has an $$LU$$ factorization and since to obtain the $$LU$$ factorization it is necessary to perform elementary operations on the matrix, the only thing that is done is to combine continuously differentiable functions, which is continuously differentiable.

But the problem is that if my reasoning were true it would work for every $$\epsilon$$, so I think that I am wrong. I would appreciate any help you could give me and in advance, thank you.

This is problem P3.2.2 from Golub's book matrix computations

Your method is correct. The reason for the small $$\epsilon$$ is that the non-singularity assumption will eventually fail. If a matrix is nonsingular, small perturbations of it are too, but once you drop the small, it's the Wild West.
• Thanks for your answer! Is the fact $A\equiv A(0)$ important? – Haus Mar 23 at 17:15