Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 2 / Section B / 7

True. Multiplicativity of the determinant and the identity $\det (B^{-1})=1/\det B$ give $\det (B^{-1}AB)=\det (B^{-1})\det A\det B=\det A$. This is invariance under matrix similarity.

False. The cyclic property of the trace permits $ABC\mapsto BCA$, but not an arbitrary interchange. Even with $A$ and $B$ both invertible matrices, takeThen $\mathrm{Tr}(ABC)=1$ and $\mathrm{Tr}(BAC)=2$, providing a counterexample.

True. Distribute the matrix product while retaining its order:Only addition is commuted; no commutation of $A$ and $B$ is assumed.

The permutation matrix cycles the three coordinate vectors, so three applications give $C^3=I$. Consequently $C^{2}C=C{C}^2=I$, proving $C^{-1}=C^2$. Since $16=5\cdot 3+1$,

Expansion of the determinant gives $\det M=-\mu -\mu =-2\mu$. The matrix inverse exists precisely when

For $\mu \ne 0$, the adjugate matrix formula, or elimination on the three unit right-hand sides, givesMultiplication by $M$ gives the identity matrix, checking all signs and coefficients.

Subtracting the first and third equations gives $\mu (x-z)=0$. Since $\mu \ne 0$, $x=z$, and the second equation gives $x=z=1$. The first then gives $y=1-\mu$, so the unique solution of the system of linear equations is

At $\mu =0$, the first and third equations both reduce to $y=1$, while the second gives $x+z=2$. All solutions form the straight lineThe limit of the nonzero-$\mu$ solution selects only the point $(1,1,1)$ on this whole line; setting a parameter to its singular value can enlarge the solution set.