$A$ 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 $A$?