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As I mentioned before the question is If matrix $A$ is not symmetric matrix then $A^{-1}$ is not symmetric.

I have already known that if matrix is symmetric then $A^{-1}$ is symmetric. But for this question I wrote the followings: If $A$ is not symmetric then $ A \neq A^T$. Now, what about $A^{-1}$. Consider the following $(A^{-1})^T = (A^T)^{-1}$ and thats it.

Actually, intuitively I guess this is true even though I can not complete the proof.

Fuat Ray
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This is equivalent to showing: if $A^{-1}$ is symmetric, then $A$ is symmetric. The assumption is that $(A^{-1})^T=A^{-1}$, and we know that $(A^{-1})^T=(A^T)^{-1}$. Now use uniqueness of inverse.

Dave
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  • YES! How couldn't I see this. We have this $\neg p\rightarrow \neg q \cong q\to p$ – Fuat Ray Feb 08 '22 at 18:26
  • That's right. I think this contraposition makes it clearer, but you could still do it directly as you were; the key is still using the identity $(A^{-1})^T=(A^T)^{-1}$ (as you do) together with uniqueness of inverses. – Dave Feb 08 '22 at 18:44