= Solution
The displayed <matrix> is \b[not orthogonal]: its first column has squared <Euclidean norm> three, whereas every column of an <orthogonal matrix> must have squared length one.
For the general assertion, put $C=AB$. In the <Einstein summation convention>, orthogonality gives $A_{ki}A_{kj}=\delta_{ij}$ and $B_{ki}B_{kj}=\delta_{ij}$. The entries of $C$ satisfy
$$
C_{ki}C_{kj}
=A_{k\ell}B_{\ell i}A_{km}B_{mj}
=(A_{k\ell}A_{km})B_{\ell i}B_{mj}
=\delta_{\ell m}B_{\ell i}B_{mj}
=B_{\ell i}B_{\ell j}=\delta_{ij}.
$$
This is exactly $C^TC=I$, proving that the <matrix product> $AB$ is orthogonal. The <Kronecker delta> contracts the two middle indices without changing the multiplication order of the <matrices>.
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