If a matrix has an inverse, then the transpose also has an inverse I just have simple one question.

"If a matrix $A$ has an inverse, then the transpose $A^T$also has an inverse."

I think the statement is 'true'.

Because

$AA^{-1} = I \\
(AA^{-1})^{T} = I^T \\
(AA^{-1})^T = I \\
(A^{-1})^TA^T = I \\$

since $(A^{-1})^TA^T=\operatorname{Id}$, $A^T$ has an inverse (which is $(A^{-1})^T$).

Is it right?
Yes, it is right. The statement is true (not just 'true' as you wrote). At the end, instead of writing &ldquo;since $A^{-1}$ exists, $(A^{-1})^T$also exists&rdquo;, I would have written that, since $(A^{-1})^TA^T=\operatorname{Id}$, $A^T$ has an inverse (which is $(A^{-1})^T$).
Suppose $A \in M_n(R)$. If $R/J(R)$ is commutative and $A$ has an inverse, then so does its transpose. Otherwise the transpose of $A$ may not have an inverse. See Journal of Algebra 322 (2009) 1627–1636.