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

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.

Collecting the coefficients in the given equation order yields the system of linear equations in matrix form

The determinant is $\det L=4-{\lambda }^2$, so there is exactly one solution whenever $\lambda \ne \pm 2$. A zero determinant does not alone distinguish inconsistency from infinitely many solutions, so check those values separately.

Eliminate $x$ and $z$ using $x=1-\lambda y$ and $z=1-4y$. The remaining equation isFor $\lambda =2$ this reads $0=1$, and for $\lambda =-2$ it reads $0=5$. Both are inconsistent. Consequentlyand each permitted value has a unique solution.

For $\lambda \ne \pm 2$, the reduced equation gives $y=(3-\lambda )/(4-{\lambda }^2)$. Back substitution into $x=1-\lambda y$ and $z=1-4y$ gives all solutions:The nonzero determinant proves that there are no other solutions at these parameter values.

Let $Y=\left(\begin{array}{cc}p& q\\ r& s\end{array}\right)$. The commutativity condition givesEquality is equivalent to $s=p$ and $q=-2r$. ThereforeEach displayed matrix commutes with $X$, since it is a polynomial in $X$. The centralizer of a nonscalar two by two matrix explains why this two-parameter form is typical.

The matrix trace and determinant of the matrices in part (i) areSince $\mathrm{Tr}X=0$ and $\det X=2$, the extra conditions force $\alpha =0$ and ${\beta }^2=1$. HenceBoth choices are real, commute with $X$, and satisfy the required trace and determinant conditions.