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.