Past exam of the mathematical methods I course of the University of Oxford / 2015 resit / 11 / a / Solution

Multiplication by each column of $B$ givesSince $B$ is invertible, its columns are an eigenbasis and the similarity transformation givesThere is an important source variant: the alternative scanned copy supplies an erroneous inverse with an extra factor $1/6$ and a different top-right entry. Calling that supplied matrix $C$, direct multiplication gives $(CB{)}_{22}=1/6$, so it cannot be $B^{-1}$. Inverting $B$ instead givesThis is the inverse in the cleaner copy used here. Using the scanned erroneous matrix would make $CAB$ nondiagonal, so correcting the inverse is necessary before applying the stated diagonalization. The inverse-matrix verification test catches the error without assuming the claimed outcome.