 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-1/11e/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

A

(r_{C} \circ f \circ l_{C}) (A) = r_{C} ((C A)^{- 1}) = A^{- 1} C^{- 1} C = A^{- 1} = f (A) .

l_{C}

and

r_{C}

are linear. Differentiate this identity at

A = I

and use the chain rule and part (a):

r_{C} (D f (C) (C h)) = - h .

Thus

D f (C) (C h) C = - h

. Given an arbitrary increment

k

h = C^{- 1} k

D f (C) (k) = - C^{- 1} k C^{- 1} .

This also proves differentiability at every

C \in V

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