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

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.

A power of a diagonal matrix acts entrywise. ThusFrom $A=BD{B}^{-1}$, adjacent $B^{-1}B$ factors cancel in repeated multiplication, givingThis is the matrix power rule for a similarity transformation, proved directly rather than by taking powers of separate noncommuting factors.

All three eigenvalues are nonzero, so $D^{-1}=\mathrm{diag}(-1,1/2,1)$. The inverse matrix is thereforeMultiplying this matrix by $A$ gives $I$, independently checking both the sign and the fractions.

The determinant is multiplicative and unchanged by a similarity transformation. HenceFor $n=1$, expansion of the original determinant givesin agreement with the product of its eigenvalues.