Answer
See the proof below.
Work Step by Step
Consider the following statement:
If $A$ and $B$ are square matrices of order $n$ such that $\det\left(AB\right)=1$ then both $A$ and $B$ are nonsingular.
We will prove by contradiction that the statement is true.
$\textbf{Proof: }$ Suppose there exist square matrices $A$ and $B$ of order $n$ with at least one of them singular such that $\det\left(AB\right)=1$.
First, please remember that the determinant of a singular square matrix is $0$.
Thus, since at least one of the matrices $A$ and $B$ is singular it follows that at least one of the numbers $\det\left(A\right)$ or $\det\left(B\right)$ is $0$ which gives us that
$$\det\left(A\right)\det\left(B\right)=0.$$
Now, recall that
$$\det\left(AB\right)=\det\left(A\right)\det\left(B\right)$$
and therefore, we obtain that $\det\left(AB\right)=0$ which contradicts the fact that $\det\left(AB\right)=1$.
Hence, $A$ and $B$ are both nonsingular.
