For a contravariant basis vector in ${\Bbb R}^n$ that is defined using a covariant basis vector ${\bf Z}_j$ in ${\Bbb R}^n$ in terms of the contravarient metric tensor as:
$${\bf Z}^i = Z^{ij}{\bf Z}_j$$
where the contravariant metric tensor $Z^{ij}$ is defined as the inverse of the covariant metric tensor $Z_{ij} = {\bf Z}_i\cdot{\bf Z}_j$, how does one reach the conclusion that the covariant basis is also the inverse of the contravarient basis, or ${\bf Z}^i\cdot{\bf Z}_j = {\delta}^{i}_{j}$?
As I'm trying to adapt my understanding of this relationship to more traditional matrix operations, I'm getting the following for the covarient and contravariant metric tensor relationship, treating the collection of $n$ covariant basis vectors ${\bf Z}_i$ as the matrix $A$ in ${\Bbb R}^n$, the $n$ contravariant basis vectors as the matrix $A^{-1}$ in ${\Bbb R}^n$, and the covariant metric tensor as the matrix $A^tA$ with it's corresponding inverse as the contravariant metric tensor:
$$(A^tA)^{-1}(A^tA) = I$$
The trouble I run into is that using the initial definition of the contravariant basis as ${\bf Z}^i = Z^{ij}{\bf Z}_j$, where $Z^{ij}$ is the same as $(A^tA)^{-1}$ and the $n$ vectors ${\bf Z}_j$ in ${\Bbb R}^n$ form $A$, I believe we would end up with a matrix operation like:
$$(A^tA)^{-1}A = A^{-1}$$
in order to obtain the contravariant basis $A^{-1}$, but that is not the correct relationship. If I changed ${\bf Z}_j$ to $A^t$ then
$$(A^tA)^{-1}A^t = A^{-1}$$
does work, but I don't see how ${\bf Z}^i = Z^{ij}{\bf Z}_j$ indicates I should transpose the matrix $A$ into $A^t$. How should I think about this relationship if I was to write it out using traditional matrix operations rather than index/tensor notation?