Exercise 5.2.16

Answers

Directly calculate that

{(CY)}_{ij}^{'} = {(\sum_{k = 1}^{n} C_{ik} Y_{kj})}^{'}

= \sum_{k = 1}^{n} C_{ik} Y_{kj}^{'} = C Y^{'} .

jephianlin

2011-06-27 00:00