Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 17 / a

The transpose reverses the product order, so the definition isIf $A$ and $B$ are orthogonal matrices, then $A^{T}A=A{A}^T=I$ and $B^{T}B=B{B}^T=I$. HenceThus $C$ is orthogonal. The original PDF includes the transpose of the entire product $AB$, which is essential to the expression used here.

Use Einstein notation, with repeated suffixes summed. The matrix multiplication and transpose giveInterchange the free indices, then relabel the dummy indices $k,\ell$:The negative sign uses $B_{k\ell }=-B_{\ell k}$. Therefore $C$ is skew-symmetric. No orthogonality of $A$ is needed for this part: matrix congruence preserves skew-symmetry.