I want to prove that if there are two matices A and B and A and B are both invertible, then the product A * B is also invertible.
This question is similar: Prove that the product of two invertible matrices also invertible but the answer uses determinants, which I have not learned about yet.
There is an answer provided in the thread above which does not use determinants.
Ans 2 is:
It is that $(AB)^{-1}=B^{-1}A^{-1}$, because $AB(AB)^{-1}=ABB^{-1}A^{-1}=1\!\!1$, but only for $n\times n$ matrices.
But I can't see how this shows that AB = matrix C such that C is invertible (such that there exists a matrix D where CD = DC = I, which is the definition of invertible given to me).
The other answer starts off by saying:
C is invertible iff for all y there is some x such that Cx = y.
I thought C is invertible iff there exists a matrix D such that CD = DC = I. How is it the case that for every y, there is some x such that Cx = y? Is y a matrix in this case? and is x a scalar?