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, 2020 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, 2020 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, 2020 at 4:20
  • $\begingroup$ Are you saying that $A^{-1}$ exists because $(AB)^{-1}$ exists? That's exactly the problem. $\endgroup$
    – WishofStar
    Feb 13, 2020 at 4:21

3 Answers 3


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.


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