Line 20: Line 20:
  
 
== 3. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.) ==
 
== 3. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.) ==
Using MATLAB to do some matrix math, I found the secret matrix to be <math> [X] = \begin{matrix} - \frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 &0 & -1 \end{matrix} </math>.
+
Using MATLAB to do some matrix math, I found the secret matrix to be:
 +
<math>
 +
[X] = \left[ \begin{array}{ccc}
 +
\ \frac{1}{2} & 0 & \frac{1}{3} \\
 +
0 & 1 & 0 \\
 +
2 & 0 & \frac{1}{3} \end{array} \right]
 +
</math>.
  
Now, just multiply the secret matrix with (2,23,3)
+
Now, just multiply the secret matrix with the column vector [2,23,3]:
<math>[X] \times \left[ \begin{array}{ccc}
+
 
 +
<math>
 +
[X]^{-1} \times \left[ \begin{array}{ccc}
 
2\\
 
2\\
 
23\\
 
23\\
Line 30: Line 38:
 
2\\
 
2\\
 
23\\
 
23\\
5\end{array} \right]</math>
+
5\end{array} \right]
 +
</math>
 +
 
 +
Which translates to the letters BWE.

Latest revision as of 14:35, 18 September 2008

1. How can Bob decrypt the message?

According the problem, there is a secret matrix, X, that is known to Bob. Bob has also been sent the encrypted matrix as well. If you set it up as an equation as follows:

$ \left[ \begin{array}{ccc} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array} \right] \times \left[ \begin{array}{ccc} X \end{array} \right] = \left[ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{array} \right] $

Then, Bob can simply just find the inverse of the secret matrix (X) and multiply it with the encrypted matrix.

2. Can Eve decrypt the message without finding the inverse of the secret matrix?

I would say no, but it should be possible if she knows the properties of linearity well. I guess it depends on if she is an electrical engineer.

3. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.)

Using MATLAB to do some matrix math, I found the secret matrix to be: $ [X] = \left[ \begin{array}{ccc} \ \frac{1}{2} & 0 & \frac{1}{3} \\ 0 & 1 & 0 \\ 2 & 0 & \frac{1}{3} \end{array} \right] $.

Now, just multiply the secret matrix with the column vector [2,23,3]:

$ [X]^{-1} \times \left[ \begin{array}{ccc} 2\\ 23\\ 3\end{array} \right] = \left[ \begin{array}{ccc} 2\\ 23\\ 5\end{array} \right] $

Which translates to the letters BWE.

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood