Prove $(A^T)^{-1}$ = $(A^{-1})^T$ whenever $A$ is invertible. Prove $(A^T)^{-1}$ = $(A^{-1})^T$ for any invertible matrix $A.$
I actually don't know where to start. I do not think I can just apply index laws.
Any help is cool! Thanks.
 A: As we know $AA^{-1}=I$.
Now taking transpose both sides, we get
                   $$(AA^{-1})^T = I^T$$ which implies
               $$[ (A^{-1})^T ](A^T) = I$$
Now multiply both sides with $[(A^T)^{-1}]$ at right side,
      $$[(A^{-1})^T]{(A^T)(A^T)^{-1}} = (A^T)^{-1}$$
Here  $(A^T)(A^T)^{-1}$  will form identity $I$,
Since we know $AA^{-1} = I$,
Therefore     $$(A^{-1})^T = (A^T)^{-1}$$
Hence Proved!
A: first we define a "B" matrix like below :
$$B = (A^{-1})$$
so :
$$B^T=(A^{-1})^T    *$$
and we can write :
$$AB = I$$ 
then 
$$(AB)^T=I^T \textrm{ and } 
B^T A^T =I$$
From this equation we can say that : 
$$B^T=(A^T)^{-1}$$ 
and finally from * we can write :
$$B^T=(A^{-1})^T=(A^T)^{-1}$$
A: Try taking the transpose of the equation
$$AA^{-1}=I.$$
A: Alternatively:
$$A^{-1}=\frac{\text{adj} (A)}{|A|}$$
Transpose:
$$\begin{align}(A^{-1})^T&=\left(\frac{\text{adj}(A)}{|A|}\right)^T=\\
&=\frac{(\text{adj} (A))^T}{|A|}=\\
&=\frac{\text{adj} (A^T)}{|A|}=\\
&=\frac{\text{adj} (A^T)}{|A^T|}=\\
&=(A^T)^{-1}.\end{align}$$
The following properties of adjoint and transpose were used:
$$\begin{align}(\text{adj} (A))^T&=\text{adj} (A^T);\\
(cA)^T&=c(A)^T;\\
|A|&=|A^T|.\end{align}$$
A: $A^T(A^T)^{-1} = I$   ;                        $ eq^n 1$
taking   transpose  on  both   side of $ eq^n 1 $
$[(A^T)^{-1}]^T A = I $;    $eq^n 1.1$
[Since $(AB)^T= B^TA^T $ & $I^T=I$ , $(A^T)^T=A$]
Now  to hold above $eq^n 1.1$ it must satisfy the relation below
$[(A^T)^{-1}]^T  =  A^{-1} $   ;           $eq^n 2 $
taking transpose on $eq^n 2 , $ we get
$ (A^T)^{-1} = (A^{-1})^T $
