Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 6C (Vectors and Matrices)

Since $P=P^{\prime }=0$, it suffices to map the two edge vectors. Let the columns of the required matrix be $u_1,u_2,u_3$. The condition $MQ=Q^{\prime }$ givesThe condition $MR=R^{\prime }$ then givesand henceThese are orthonormal. Taking $u_3=u_1\times u_2=(0,-1,0{)}^T$ produces the orthogonal matrixIts columns are orthonormal, and direct substitution verifies $MP=P^{\prime }$, $MQ=Q^{\prime }$, and $MR=R^{\prime }$.

Write $a=(a_1,a_2,a_3,a_4{)}^T$. The reflection in a hyperplane orthogonal to $a$ sendssoIt acts as $-1$ on $\mathrm{span}\{a\}$ and as $1$ on the three-dimensional hyperplane $a^{\perp }$. Therefore

For $a=(a_1,a_2,a_3{)}^T$, the cross-product matrix isIts diagonal is zero, so $\mathrm{tr}A=0$. Also $Aa=a\times a=0$, and $a\ne 0$, so $A$ has a nontrivial kernel. Hence

Expanding $\det A^{(n)}$ along its last row gives the last diagonal contribution $a_n{d}_{n-1}$. The only other nonzero entry is $c_{n-1}$; expanding its minor along the last column contributesThus the tridiagonal determinant recurrence isso

The recurrence becomesIts characteristic polynomial isTherefore $d_n=(\alpha +\beta n)(-1{)}^n$. The initial conditions give $\alpha =1$ and $\beta =1$, so