If $A$ and $B$ are $n×n$ matrices, $AB = -BA$ , and $n$ is odd, show that either $A$ or $B$ has no inverse. If $A$ and $B$ are $n×n$ matrices, $AB = -BA$ , and $n$ is odd, show that either $A$ or $B$ has no inverse. 
I have no clue how to do this and any help/guidance would be appreciated!
Thanks in advance!

$$det(AB) = det(-BA)$$ 
$$det(AB)= det(-B)det(A)$$ 
$$det(AB) = (-1)^ndet(BA)$$ since n is odd 
$$det(AB) = -det(BA)$$
$$det(A)det(B) = -det(B)det(A)$$
$$2det(A)det(B) = 0$$
Therefore $det(A)=0$ or $det(B)=0$
 A: Take determinants of both sides. $$\begin{align}\det(AB) &= \det(-BA)\\&=(-1)^n\det(BA)\\&=-\det(BA)\end{align}$$ but $\det(AB) = \det(A)\det(B)$. So, we have $\det(A) \det(B) = -\det(A) \det(B)$ and hence $\det(A) \det(B) = 0$. Can you finish it off from here?
A: $$det(AB) = det(-BA)$$ 
$$det(A)det(B) = det(-B)det(A)$$ 
$$det(A)det(B) = (-1)^ndet(B)det(A)$$
You are done right up to here.
add $det(A)det(B)$ on both sides, 
$\implies$
$$2det(A)det(B)=0$$ ($\because$ $n$ is odd)
$\implies$
$\det A=0$ or $\det B=0$.
A: I'll assume you're working in a field of characteristic zero (like reals, complex, etc.), but this is true more generally in any ring that is a commutative integral domain of characteristic not equal to $2$:
$$x = \det(AB) = \det(A)\det(B) = \det(B)\det(A) = \det(BA) = -\det(AB) = -x$$
The steps are justified as: determinant is multiplicative, scalars commute, determinant is multiplicative (again), and finally using the equation you derived in your post.  
Thus $x = -x \implies 2x=0$.  We can say $x=0$ as long as the field (or whatever ring the matrices take values in) has characteristic not equal to $2$.  Then $x = \det(AB) = 0$ and $\det(A)\det(B) = 0$, and it must be the case that $\det(A) = 0$ or $\det(B) = 0$.  
NB that it's crucial that we're working in an algebraic structure with characteristic not equal to $2$ and without zero divisors (an integral domain) so that we can reason $2x = 0$ implies $x = 0$ and $ab=0$ implies $a=0$ or $b=0$.  If we were working in characteristic $2$, then $x = -x$ for all $x$, and in the integers modulo $6$, for example, $2x = 0$ would be true for $x=3$ just the same.  
