Properties of orthogonal, singular and antisymmetric matrices. 
$A$ and $B$ are real non-zero $3 \times 3$ matrices and satisfy the
  equation
$$\begin{align}   (AB)^T + B^{−1}A = 0. \end{align}$$
(a) Prove that if $B$ is orthogonal then $A$ is antisymmetric.
(b) Without assuming that $B$ is orthogonal, prove that $A$ is singular.

My proof:
(a) Using property of orthogonal matrix:
\begin{align}
B^T = B^{−1}
\end{align}
We have:
\begin{align}
(AB)^T + B^{−1}A = B^TA^T + B^{−1}A = B^TA^T + B^TA = B^T(A^T + A) = 0.
\end{align}
Since B is non-zero, product is zero when:
\begin{align}
A^T + A = 0. \\
A = -A^T  \blacksquare
\end{align}
I can't prove (b). Could you please help?

$$B^{-1}A=-(AB)^T$$
$$A=-BB^TA^T$$
$$\det(A)=(-1)^3\det(A)\det(B)^2=-\det(A)\det(B)^2$$
$$(1+\det(B)^2)\det(A)=0$$
Can you conclude anything about $A$?
