Let $A,B \in \text{Mat}_n(\mathbb{R})$ invertible with $n$ impar. Then $AB+BA\not = 0$
My work:
Suppose $AB+BA=0$, then $AB=-BA$. Then, $$det(AB)=det(-BA)\implies det(AB)=-det(BA)\implies det(AB)+det(BA)=0\implies det(A)det(B)+det(B)det(A)=0\implies 2det(A)det(B)=0\implies det(A)det(B)=0$$
If $det(B)\not =0$ then $det(A)=0$ this implies $A$ is not invertible. (Contradiction).
If $det(A)\not =0$ then $det(B)=0$ this implies $B$ is not invertible. (Contradiction).
In consequence,
$$AB+BA\not = 0$$
Please if someone can review my proof i will be very grateful.