What is the difference between transpose and inverse?

Question

I have 2 tasks: To show that $A^{-1} = A^T$ and that $A^T = A^{-1}$.

So I proved the first case with:

$$A^T A = I$$

and later according to uniqueness of inverse for matrices we can say that they are equal.

But what about the second case $A^T = A^{-1}$? Is not it the same thing, I mean, can I do the same thing for this case too or I should prove that in another way?

UPDATE

Let p0, p1, p2 ∈ R3 be three vectors that form an orthonormal coordinate system, i.e. pT0 p1 = 0, pT1 p2 = 0, pT2 p0 = 0, and ∥p0∥ = ∥p1∥ = ∥p2∥ = 1. Similarly, let q0, q1, q2 ∈ R3 also be three vectors that form an orthonormal coordinate system.

Answer 1

This statement is false in general; a linear operator $A$ is called orthogonal exactly when $A^T = A^{-1}$.

Answer 2

You can proof it's false by giving a matrix $1 \times 1$. For example $A=[2]$, then you'll have $A^{t}=[2]$ $A^{-1}=[\frac{1}{2}]$

Answer 3

The square matrix of full rank, $\mathbf{Q}\in\mathbb{C}^{m\times m}_{m}$, is defined as unitary iff $$ \mathbf{Q}\, \mathbf{Q}^{*} = \mathbf{Q}^{*} \mathbf{Q} = \mathbf{I}_{m}. $$ This implies $$ \mathbf{Q}^{*} = \mathbf{Q}^{-1} $$ The magnitude of the determinant is unity: $$ |\det \mathbf{Q} | = 1 $$ This property defines a special class of matrices. In other words, the matrix transpose is the matrix inverse only for unitary matrices.