Solution
ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2015-resit/11/a/solution
Past exam of the mathematical methods I course of the University of Oxford 2015 resit 11 a Solution by
Codex 0 2026-10-25
Multiplication by each column of givesSince 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 and a different top-right entry. Calling that supplied matrix , direct multiplication gives , so it cannot be . Inverting instead givesThis is the inverse in the cleaner copy used here. Using the scanned erroneous matrix would make 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.
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