matrix inverse in tensor notation Suppose there is a matrix $A$ that transforms vectors,
$$
   Y = A x
$$
Now express this in some other coordinate system, with $x = B z, \,\, y = B w$, so
\begin{align*}
& Bw = A B z
\\
\Rightarrow & w = B^{-1} A B z
\end{align*}
So $A$ expressed in the other system is $B^{-1} A B$.
What would be the equivalent in tensor notation, in particular of the $B^{-1}$?
Here's what I'm trying 
\begin{align*}
& y^i = A^i_j x_j
\\
& \quad\quad x^j = B^j_k z^k
\\
& \quad\quad y^i = B^i_m w^m
\\
\text{so}\quad & B^i_m w^m = A^i_j B^j_k z_k
\end{align*}
Now what is the tensor equivalent of premultiplying by $B^{-1}$ on the left,
in order to find what $A$ looks like in tensor notation in the new coordinate system?
 A: The tensor equivalent is simply:
$$ (B^{-1})^i_j $$
defined to satisfy
$$ (B^{-1})^i_j B^j_k = \delta^i_k \qquad B^i_j (B^{-1})^j_k = \delta^i_k $$. 
So you write
$$ w^m = (B^{-1})^m_i A^i_j B^j_k z^k $$

Though, I would really, really just advice you to use abstract index notation instead of the concrete tensor notation you seem to be using. The advantage is that you treat $z$ now as an element in your vector space $V$ and not as the coordinate representation, and $A$ the linear mapping from $V$ to itself instead of its matrix representation, so you don't have to worry about matrices coming from "changes of variables". 
A: Using a basis, you could use the common method of calculation given by
$$A^{-1} = \frac{\text{adj}(A)}{\det(A)} $$
where $\det$ is the determinant, and $\text{adj}$ is the adjugate, i.e. the transpose of the matrix of cofactors $\text{cof}(A)$.
One can show that
$$ {(\text{adj}(A))^{i}}_{j} = \frac{\partial \det(A)}{\partial {A^{j}}_{i}}. $$
In order to calculate this, we can use the expression for the determinant in terms of the generalized Kronecker delta
$$\det(A) = \frac{1}{n!}\delta^{a_1\dots a_n}_{b_1\dots b_n}{A^{b_1}}_{a_1}\cdots{A^{b_n}}_{a_n}$$
where $n$ is the dimension of the vector space.
Hence the adjugate is
\begin{align*}
{(\text{adj}(A))^{a}}_{b}
&= \frac{\partial}{\partial {A^{b}}_{a}} \left( \frac{1}{n!}\delta^{c_1\dots c_n}_{d_1\dots d_n}{A^{d_1}}_{c_1}\cdots{A^{d_n}}_{c_n} \right)\\
&= \frac{1}{n!} \delta^{c_1\dots c_n}_{d_1\dots d_n} \frac{\partial}{\partial {A^{b}}_{a}} \left({A^{d_1}}_{c_1}\cdots{A^{d_n}}_{c_n} \right)
\end{align*}
If we perform the calculation (for instance, in $n=3$) we get
$$\begin{align}
{(\text{adj}(A))^{a}}_{b}
&= \frac{1}{3!} \delta^{ijk}_{lmn}
\frac{\partial}{\partial {A^{b}}_{a}} \left( {A^{l}}_{i}{A^{m}}_{j}{A^{n}}_{k} \right)\\
&= \frac{1}{3!} \delta^{ijk}_{lmn}
\left(
\delta^{l}_{b}\delta^{a}_{i}{A^{m}}_{j}{A^{n}}_{k} +
\delta^{m}_{b}\delta^{a}_{j}{A^{l}}_{i}{A^{n}}_{k} +
\delta^{n}_{b}\delta^{a}_{k}{A^{l}}_{i}{A^{m}}_{j}
\right)\\
&= \frac{1}{3!}
\left(
\delta^{ijk}_{lmn} \delta^{l}_{b}\delta^{a}_{i}{A^{m}}_{j}{A^{n}}_{k} +
\delta^{ijk}_{lmn} \delta^{m}_{b}\delta^{a}_{j}{A^{l}}_{i}{A^{n}}_{k} +
\delta^{ijk}_{lmn} \delta^{n}_{b}\delta^{a}_{k}{A^{l}}_{i}{A^{m}}_{j}
\right)\\
&= \frac{1}{3!}
\left(
\delta^{ajk}_{bmn} {A^{m}}_{j}{A^{n}}_{k} +
\delta^{iak}_{lbn} {A^{l}}_{i}{A^{n}}_{k} +
\delta^{ija}_{lmb} {A^{l}}_{i}{A^{m}}_{j}
\right)\\
&= \frac{1}{2!}\delta^{aij}_{bmn} {A^{m}}_{i}{A^{n}}_{j}
\end{align}$$
hence the inverse (in the $n=3$ case) is
$${(A^{-1})^{a}}_{b} = \frac{{(\text{adj}(A))^{a}}_{b}}{\det(A)}
= \frac{\frac{1}{2!}\delta^{aij}_{bmn} {A^{m}}_{i}{A^{n}}_{j}}
{\frac{1}{3!} \delta^{cde}_{fgh}{A^{f}}_{c}{A^{g}}_{d}{A^{h}}_{e} }
= 3\frac{\delta^{aij}_{bmn} {A^{m}}_{i}{A^{n}}_{j}}
{\delta^{cde}_{fgh}{A^{f}}_{c}{A^{g}}_{d}{A^{h}}_{e} }$$
You can check that the $n$-dimensional adjugate is
$${(\text{adj}(A))^{a}}_{b} = \frac{1}{(n-1)!}\delta^{ac_1\dots c_{n-1}}_{bd_1\dots d_{n-1}} {A^{d_1}}_{c_1}\cdots{A^{d_{n-1}}}_{c_{n-1}}$$
and finally we get the $n$-dimensional inverse
$${(A^{-1})^{a}}_{b}
= \frac{{(\text{adj}(A))^{a}}_{b}}{\det(A)}
= \frac{\frac{1}{(n-1)!}\delta^{ac_1\dots c_{n-1}}_{bd_1\dots d_{n-1}} {A^{d_1}}_{c_1}\cdots{A^{d_{n-1}}}_{c_{n-1}}}
{\frac{1}{n!}\delta^{e_1\dots e_n}_{f_1\dots f_n}{A^{f_1}}_{e_1}\cdots{A^{f_n}}_{e_n}}
= n\frac{\delta^{ac_1\dots c_{n-1}}_{bd_1\dots d_{n-1}} {A^{d_1}}_{c_1}\cdots{A^{d_{n-1}}}_{c_{n-1}}}
{\delta^{e_1\dots e_n}_{f_1\dots f_n}{A^{f_1}}_{e_1}\cdots{A^{f_n}}_{e_n}}$$
Note that all expressions (apart from those with the partial derivative) contain exclusively components of tensors. In these cases we can safely assume that the whole expression is a tensor as well, and we can interpret the indices not as components but as abstract tensor indices.
