# Proving that the product of two invertible matrices is also invertible, without determinants

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?

• What you quote as "Ans 2" is saying exactly what you want. First, only square matrices have inverses. Then "Ans 2" is showing that for $C=AB$, there is a matrix $D$ such that $CD=DC=I$. That matrix $D$ is $B^{-1}A^{-1}$. The computation is showing that that choice works. It requires using that matrix multiplication is associative. – user566930 Jul 7 '18 at 14:18
• I agree that it is written in the wrong order of the inference. You can order it as $CD=(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AIA^{-1}=AA^{-1}=I$. And then you can also compute $DC$ if you want and again get $I$. – user566930 Jul 7 '18 at 14:22
• @minghan Ohh, okay. Makes a lot more sense after seeing it reordered. Thanks. – user2719875 Jul 7 '18 at 14:26

Let $C$ be $n\times n$ and $y=n\times 1$. Then if for every $y$ there is $x$ such that $Cx=y$ we are saying that the columns of $C$ span $\mathbb{R}^n$.
In particular, for each elementary vector $e_i$ ($1$ in ith place, zero elsewhere) there is $x_i$ such that $Cx_i=e_i$. The inverse of $C$ then contains these $x_i$s as its columns.