(New page: == Question 1 == Bod knows a 3 by 3 matrix and encrypted message. Then Bob is able to get encrypted message by multiplying inversed matrix by encrypted message. == Question 2 == == Qu...)
 
(Question 3)
 
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== Question 1 ==
 
== Question 1 ==
  
Bod knows a 3 by 3 matrix and encrypted message. Then Bob is able to get encrypted message by multiplying inversed matrix by encrypted message.   
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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 ==
 
== Question 2 ==
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No. Eve has to find inverse of the secret matrix to decrypt the message.
  
 
== Question 3 ==
 
== Question 3 ==
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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>
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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}</math>.<br>
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Then,  <math>\begin{matrix} - x \\ y \\ z \end{matrix} =  \begin{matrix} - 2 \\ 23 \\ 5 \end{matrix}</math>.<br>
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As a text, it is BWE.

Latest revision as of 08:56, 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} $.


Then, $ \begin{matrix} - x \\ y \\ z \end{matrix} = \begin{matrix} - 2 \\ 23 \\ 5 \end{matrix} $.

As a text, it is BWE.

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009