 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-2/14/b/i/solution

(CB)^{T} = B^{T} C^{T}

. If

B

B^{- 1} = B^{T}

and the given identity becomes

B^{T} (C^{T} + C) = 0.

Multiply on the left by

B

to obtain

C^{T} + C = 0

. Therefore $C$ is antisymmetric, so it is an antisymmetric matrix, namely

C^{T} = - C

 New to topics? Read the docs here!