Showing \( \left( A B \right) C = A \left( B C \right) \) is relatively a simple matter using the super duper summation convention.
Firstly the ith row, jth column of a matrix \( A \) is designated by \( A_{ij} \), \( A^{i}_{j} \) or \( A^{ij} \).
Matrix multiplication is then defined as
$$
\left[ XY \right]_{ij} = X_{ik}Y_{kj}
$$
where square brackets have been placed around \( X Y \) to designate the \( i j \) element of the matrix \( X \cdot Y \).
\begin{aligned}
\left[ \left( A B \right) C \right]_{ij} & = \left( A B \right)_{ik} C_{kj} \\
& = A_{il} B_{lk} C_{kj} \\
& = A_{il} \left( B_{lk} C_{kj} \right) \\
& = A_{il} \left[ B C \right]_{lj} \\
& = \left[ A \left( B C \right) \right]_{ij} \\
\therefore \left( A B \right) C & = A \left( B C \right)
\end{aligned}
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