 Solution

The contracted Kronecker delta is ${\delta }_{ll}=3$. The other scalar contraction is $a_i{C}_{ij}{b}_j={\mathbf{a}}^{T}C\mathbf{b}$, while $C_{ik}{d}_i$ is the $k$th component of the matrix transpose applied to $\mathbf{d}$. Hence the resulting vector isBoth terms have the same single free index $k$, as required by the Einstein summation convention.