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Let $A,B\in\mathbb{F}^{n\times n}$ be two $n\times n$ matrices over the underlying field $\mathbb{F}$. In addition, $A$ is guaranteed to be a symmetric matrix, i.e, $A=A^{T}$. We assume complexity measure to be the number of field operations over $\mathbb{F}$. Now consider at the following problem:-

Given $A,B$ and $AB$ as input, can we compute $AB^{T}$ in $o(n^{\omega})$ time?

Here $\omega$ is the exponent of matrix multiplication. Note that $AB^{T}=A^{T}B^{T}=(BA)^{T}$. Thus computing $BA$ in $o(n^{\omega})$ time is also enough.

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No. Consider block matrices $A = \left(\begin{matrix} 0 & 0 \\ 0 & X \end{matrix}\right)$ (with symmetric $X$) and $B = \left(\begin{matrix} 0 & Y \\ 0 & 0 \end{matrix} \right)$. Computing $AB^T$ from $A$, $B$ and $AB$ means computing $XY^T$ from $X$ and $Y$, since $AB = 0$

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