Revision as of 12:31, 18 September 2008 by Pjcannon (Talk)

Part C: Application of Linearity

1. Bob can decrypt the message by multiplying it (in groups of 3 numbers) by the inverse of the 3-by-3 secret matrix.

2. No. $ [Secret Message]*[Secret Matrix]=[Encoded Message]\! $. Thus the only way to solve for the secret message if the encoded message were known would be to multiply both sides by the inverse of the 3-by-3 secret matrix.

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood