 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-2/14/b/i/solution

The transpose rule gives

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

. If

C

C^{- 1} = C^{T}

, and the identity becomes

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

Multiplying on the left by

C

yields

B^{T} = - B

. This is precisely the defining condition for an antisymmetric matrix.

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