If $\bfA$ is an invertible matrix, show that $\bigl(\bfA^t \bigr)^{-1} = \bigl( \bfA^{-1} \bigr) ^t$. That is, show that the inverse of the transpose is the transpose of the inverse.
Because we are trying to show that $\bigl( \bfA^{-1} \bigr)^t$ is the inverse of $\bfA^t$, we need to establish that $$\bfA^t \bigl( \bfA^{-1} \bigr)^t = \bfI \quad \text{ and } \quad \bigl( \bfA^{-1} \bigr)^t \bfA^t = \bfI$$
In order to capitalize on $\bfA^{-1}$ being the matrix inverse of $\bfA$, we would like to manipulate the terms to use the fact that $\bfA \bfA^{-1} = \bfI$ or $\bfA^{-1} \bfA = \bfI$.