 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/11/b/iii/solution

The same scalar consistency condition now has

a \cdot c = 1

, so it reads

0 = 1

. Hence there is no solution.

Geometrically the image of

I + b a^{T}

is still the plane

a^{⊥}

, while this

c

is outside it. Thus the failure is inconsistency, rather than a unique solution with a singular denominator.

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