Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 18

Put $d=c-1$. The matrix is real and symmetric, so orthogonality means $A^{\mathsf{T}}A=A^2=I$. Multiplication gives the scalar conditionsIf $d=0$, then $c=1$ and $a=\pm 1$; if $a=0$, then $d=\pm 1$ and $c=0$ or $2$. In both cases $b=\pm 1$. The eight parameter triples are thereforeThese conditions are sufficient as well as necessary, since they make every entry of $A^{\mathsf{T}}A$ equal to the corresponding entry of the identity matrix.

A three-dimensional rotation matrix is an orthogonal matrix with determinant $+1$. Here $\det A=b(a^2-d^2)$. For the branch $d=0$, the determinant condition gives $b=1$; for the branch $a=0$, it gives $b=-1$. HenceThe first is the identity rotation; the remaining three are half-turns. Their fixed axes are respectively the $y$ axis, the direction $(1,0,-1)$ and the direction $(1,0,1)$. The identity is included under the usual rotation matrix convention.

A square homogeneous linear system has a nonzero solution exactly when its determinant vanishes. Expanding along the second row givesConsequentlyFor an explicit check, the null space contains $(3,-1,1{)}^{\mathsf{T}}$ at $\lambda =3$ and $(3,1,-1{)}^{\mathsf{T}}$ at $\lambda =-3$.

The original PDF requires $B$ itself to be an orthogonal skew-symmetric matrix; it does not require $C$ to be diagonal. A real two-dimensional skew-symmetric matrix has the form $B=\left(\begin{array}{cc}0& u\\ -u& 0\end{array}\right)$. The orthogonal matrix condition gives $u^2=1$, so choose $u=1$. Then $C=B^{\mathsf{T}}AB$ is forced:Direct multiplication verifies $B^{\mathsf{T}}=-B$, $B^{\mathsf{T}}B=I$ and $BC{B}^{\mathsf{T}}=A$. Choosing $-B$ gives the same $C$. Replacing $B$ by a matrix of eigenvectors to make $C$ diagonal would generally fail the printed antisymmetry requirement.

For the $C$ obtained above, the characteristic polynomial isThus the eigenvalues are $5$ and $-5$. For $\mu =5$, $(C-5I)\mathbf{v}=0$ gives $v_2=-3v_1$; for $\mu =-5$, it gives $v_1=3v_2$. Normalizing the eigenvectors yieldsBoth are unit vectors, their dot product is zero, and either overall sign is allowed for each eigenvector.