Inverse matrix notation question

SamanthaYellow
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I'm hoping that you can help me settle an argument. For a matrix \textbf{M} with elements m_{ij}, is there any sitaution where the notation (M_{ij})^{-1} could be correctly interpreted as a matrix with elements 1/m_{ij}?

Personally I interpret (M_{ij})^{-1} in the usual sense of an inverse matrix, where it would have the property \textbf M \textbf M^{-1} = I, but perhaps there are other interpretations that I don't know about. Thanks!
 
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The only interpretation I can think of is when you define matrix multiplication component-wise. In this case, the invertible matrices would be the ones with non-zero values, and the matrix you describe would be the inverse. Note that the identity matrix would be the one with 1's as values. I don't think this is commonly used.
 
I'm not a fan of the notation ##(M_{ij})^{-1}##, mainly because ##M_{ij}## should refer to the component on row i, column j, not the matrix itself. I'm also not a fan of using a lowercase m for the components, because that prevents us from writing the definition of matrix multiplication in what I consider the obviously best way: ##(AB)_{ij}=\sum_k A_{ik}B_{kj}##. I find it very puzzling that some authors go out of their way to avoid this notation, by writing things like "if ##C=AB##, then ##c_{ij}=\sum_k a_{ik}b_{kj}##".

If M is a diagonal matrix, for example
\begin{pmatrix}2 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 4\end{pmatrix} then its inverse is simply
\begin{pmatrix}\frac 1 2 & 0 & 0\\ 0 & \frac 1 3 & 0\\ 0 & 0 & \frac 1 4\end{pmatrix} But even if M is diagonal, and we use horrible notation, we still don't quite have ##(M_{ij})^{-1}=1/m_{ij}## because the off-diagonal elements of ##M^{-1}## aren't 1/0.
 
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This is helpful. The matrix in question isn't diagonal, and that's a good point about 1/0. Hopefully I can convince this other person to change their notation!
 
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