(Question 3)
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== Question 3 ==
 
== Question 3 ==
  
The secret matrix is <math> \begin{matrix} - \frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 &0 & -1 \end{matrix} </math>.
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The secret matrix is <math> \begin{matrix} - \frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 &0 & -1 \end{matrix} </math>.<br>
 
So, <math> \begin{matrix} - \frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 &0 & -1 \end{matrix} * \begin{matrix} - x \\ y \\ z \end{matrix} =  \begin{matrix} - 2 \\ 23 \\ 2 \end{matrix}
 
So, <math> \begin{matrix} - \frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 &0 & -1 \end{matrix} * \begin{matrix} - x \\ y \\ z \end{matrix} =  \begin{matrix} - 2 \\ 23 \\ 2 \end{matrix}

Revision as of 08:54, 16 September 2008

Question 1

Bod knows the 3 by 3 secret matrix and encrypted message. Then Bob is able to get encrypted message by multiplying inversed matrix by encrypted message.

Question 2

No. Eve has to find inverse of the secret matrix to decrypt the message.

Question 3

The secret matrix is $ \begin{matrix} - \frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 &0 & -1 \end{matrix} $.
So, $ \begin{matrix} - \frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 &0 & -1 \end{matrix} * \begin{matrix} - x \\ y \\ z \end{matrix} = \begin{matrix} - 2 \\ 23 \\ 2 \end{matrix} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood