Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 2 / 14

Let $N=\sqrt{8}A$ be the unscaled matrix. Expansion along its third row givesScaling a three-by-three matrix by $1/\sqrt{8}$ scales its determinant by $(\sqrt{8}{)}^{-3}$. Therefore

The three columns of $N=\sqrt{8}A$ are $(\sqrt{3},1,2)$, $(-\sqrt{2},\sqrt{6},0)$, and $(-\sqrt{3},-1,2)$. Each has squared Euclidean norm eight. Their pairwise dot products areHence $N^{\mathsf{T}}N=8I$, so $\boxed{A^{\mathsf{T}}A=I}$. Thus $A$ is an orthogonal matrix, with $A^{-1}=A^{\mathsf{T}}$. The determinant from (i) places it in the determinant-one component.

The transpose rule gives $(BC{)}^{\mathsf{T}}=C^{\mathsf{T}}{B}^{\mathsf{T}}$. If $C$ is orthogonal, $C^{-1}=C^{\mathsf{T}}$, and the identity becomesMultiplying on the left by $C$ yields $\boxed{B^{\mathsf{T}}=-B}$. This is precisely the defining condition for an antisymmetric matrix.

The inverse appearing in the assumption requires $\det C\ne 0$. Taking determinants of $(BC{)}^{\mathsf{T}}=-C^{-1}B$ giveswhere the minus sign comes from the odd matrix order. Multiplying by $\det C$ gives $\det B((\det C{)}^2+1)=0$. Since $C$ is real, the second factor is positive. Thus $\boxed{\det B=0}$ and $B$ is a singular matrix. This is the singularity forced by an odd-dimensional transpose-inverse identity, with the roles of the two matrix letters exchanged.