Exercise 2.4.5

Answers

This is because that ${(A^{- 1})}^{t} A^{t} = {(A A^{- 1})}^{t} = I$ and $A^{t} {(A^{- 1})}^{t} = {(A^{- 1} A)}^{t} = I$ .

jephianlin

2011-06-27 00:00