Let $A$ and $B$ be $n\times n$ matrices such that $AB$ is invertible.

  1. Prove that $A$ and $B$ are invertible.

  2. Give an example to show a product of nonsquare matrices can be invertible even though the factors are not.

  1. If $AB$ is invertible, then there exists $(AB)^{-1}$ such that $(AB)(AB)^{-1}=I_n$. So we find inverses for both $A$ and $B$, hence $A$ and $B$ are invertible.

  2. Let $$ A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{pmatrix} $$ and $$ B=\begin{pmatrix} 1 & 1 \\ 0 & 1 \\ 1 & 0 \\ \end{pmatrix} $$ Then $AB$ is invertible while $A$ and $B$ are not.

Is it fine?

  • 2
    $\begingroup$ In (a) you are not showing anything. How do you find the inverses for $A$ and $B$ ? Also, (b) is fine! $\endgroup$ – azif00 Feb 13 '20 at 4:14
  • $\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. See also the section on titles in How to ask a good question. (The part entitled "Make your title your question" is especially relevant to this.) $\endgroup$ – Brian Feb 13 '20 at 4:16
  • $\begingroup$ In (b), what is a non-invertible nonsquare matrix? If it means "not full rank" then $A$ and $B$ are not valid. $\endgroup$ – Riley Feb 13 '20 at 4:20
  • $\begingroup$ Are you saying that $A^{-1}$ exists because $(AB)^{-1}$ exists? That's exactly the problem. $\endgroup$ – WishofStar Feb 13 '20 at 4:21

Hints. Let $E$ be a square matrix. Then, the following are equivalent:

$\rm (a)$ $E$ is invertible.

$\rm (b)$ The only solution for $E\vec{x} = \vec0$ is $\vec{x} = \vec0$.

$\rm (c)$ For any vector $\vec y$, there exists a vector $\vec x$ such that $\vec{y} = E\vec{x}$.

Now, use $\rm (b)$ to show that $B$ is invertible, and use $\rm (c)$ to show that $A$ is invertible.


If you know some things about determinants: if $AB = I$ then


So, $\det(A),\det(B)\neq 0$ and so both are invertible.


For Part 1, think about how you might use the fact that $\det(AB) = \det(A)\det(B)$ to prove that statement in a more rigorous fashion.

Part 2 looks good.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.