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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – user566930
    Jul 7, 2018 at 14:18
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    $\begingroup$ 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 $I$. $\endgroup$
    – user566930
    Jul 7, 2018 at 14:22
  • $\begingroup$ @minghan Ohh, okay. Makes a lot more sense after seeing it reordered. Thanks. $\endgroup$ Jul 7, 2018 at 14:26

2 Answers 2


I don’t know if I understood your doubt. However, there is a very simple answer to the first question of your post. Suppose $A$ and $B$ are $n\times n $ matrices, both invertible. Then there are matrices $A^{-1}$ and $B^{-1}$ such that $A^{-1}A=AA^{-1}=I$ and $B^{-1}B=BB^{-1}=I$. We want to find a matrix that satisfies the same kind of relations for $AB$. Well, such matrix is $B^{-1}A^{-1}$. We prove this by direct verification; $$B^{-1}A^{-1} AB=B^{-1}IB=B^{-1}B=I, $$ and, analogously, $$ ABB^{-1}A^{-1}=I.$$

This concludes the proof.


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.

  • $\begingroup$ Oh, okay. So isn't "C is invertible iff for all y there is some x such that Cx = y." non-trivial, and thus, have to be proven? It is not in the definition an what an invertible matrix is, right? $\endgroup$ Jul 7, 2018 at 14:29
  • $\begingroup$ Sure, it is not exactly obvious/trivial and some argument/s need to be given to shiw it (as in the answer), starting from the usual definition of invertibility $\endgroup$
    – AnyAD
    Jul 7, 2018 at 14:42

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