Exercise 2.5.11

Answers

1.

This is because

RQ = {[I]}_{β}^{γ} {[I]}_{α}^{β} = {[I]}_{α}^{γ} .

2.

This is because

Q^{- 1} = {({[I]}_{α}^{β})}^{- 1} = {[I^{- 1}]}_{β}^{α} = {[I]}_{β}^{α} .

jephianlin

2011-06-27 00:00