Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 2 / Section B / 7 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii
- iii a
  - Solution iii

i

Solution

True. Multiplicativity of the determinant and the identity

det (B^{- 1}) = 1/ det B

give

det (B^{- 1} A B) = det (B^{- 1}) det A det B = det A

. This is invariance under matrix similarity.

ii

Solution

False. The cyclic property of the trace permits

A BC \mapsto BC A

, but not an arbitrary interchange. Even with

A

and

B

both invertible matrices, take

A = (1002), B = (1011), C = (0100) .

Then

Tr (A BC) = 1

and

Tr (B A C) = 2

, providing a counterexample.

iii

Solution

True. Distribute the matrix product while retaining its order:

A^{- 1} (A + B) B^{- 1} = A^{- 1} A B^{- 1} + A^{- 1} B B^{- 1} = B^{- 1} + A^{- 1} = A^{- 1} + B^{- 1} .

Only addition is commuted; no commutation of

A

and

B

is assumed.

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