Line 6: Line 6:
 
Can Eve decrypt the message without finding the inverse of the secret matrix?<br>
 
Can Eve decrypt the message without finding the inverse of the secret matrix?<br>
 
The asnwer is "no." She can find the inverse of the secret matrix from the intercepted message.<br>
 
The asnwer is "no." She can find the inverse of the secret matrix from the intercepted message.<br>
The coded and decrypted message can be arranged in a 3-by-3 matrix form. <br>
+
The coded and decrypted message can be arranged in a 3-by-3 matrix form. <br><br>
:<math> \begin{bmatrix}
+
:<math> Coded = \begin{bmatrix}
 +
1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\
 +
\end{bmatrix} \ </math>
 +
<br>
 +
:<math> Decrypted = \begin{bmatrix}
 
2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \\  
 
2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \\  
 
\end{bmatrix} \ </math>
 
\end{bmatrix} \ </math>
 
+
<br>
  
 
:<math> \begin{bmatrix}
 
:<math> \begin{bmatrix}

Revision as of 07:25, 19 September 2008

Part 1

How can Bob decrypt the message?

Bob can decrypt the message by multiplying the inverse of the 3-by-3 secret matrix with the coded message.

Part 2

Can Eve decrypt the message without finding the inverse of the secret matrix?
The asnwer is "no." She can find the inverse of the secret matrix from the intercepted message.
The coded and decrypted message can be arranged in a 3-by-3 matrix form.

$ Coded = \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix} \ $


$ Decrypted = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \\ \end{bmatrix} \ $


$ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix}^{-1} = \frac{1}{ad - bc} \begin{bmatrix} \,\,\,d & \!\!-b \\ -c & \,a \\ \end{bmatrix}. $

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett