(Part C: Application of Linearity)
(Part C: Application of Linearity)
Line 4: Line 4:
 
<br>
 
<br>
 
<br>
 
<br>
2.  No.  <math>[Secret Message]*[Secret Matrix]=[Encoded Message]\!</math>.  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.
+
2.  Yes.  <math>[Secret Message]*[Secret Matrix]=[Encoded Message]\!</math>.  <br>
 +
<br>
 +
3.

Revision as of 11:34, 18 September 2008

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. Yes. $ [Secret Message]*[Secret Matrix]=[Encoded Message]\! $.

3.

Alumni Liaison

Prof. Math. Ohio State and Associate Dean
Outstanding Alumnus Purdue Math 2008

Jeff McNeal