Latest revision as of 13:40, 19 September 2008

Application of linearity

Assuming that Matrix A is what Alice wants to say, and matix B is a 3-by-3 matrix to encrypt Alice's message, and matrix C is the encoded messange.

It can be expressed A*B=C.

In order to find A, Bob needs to find the inverse matrix of B.

A*B*B`= C*B`, A*I=C*B'

After that, he needs to calculate C*B`. By finding its corresponding order in the alphabet, He can figure our what she wants to say.

No, There is no way to decrypt the message without finding the inverse of the secret matirx.

== What is the decrypted message corresponding to (2, 23, 3)?

A*B=C

$A^{-1}*A*B=A*C$

$I*B=A^{-1}*C$

$B=A^{-1}*C$

A= $\left[ \begin{matrix}2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3\end{matrix} \right]$

C= $\left[ \begin{matrix}1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1\end{matrix} \right]$

B = $\left[ \begin{matrix}-2/3 & 0 & 4 \\ 0 & 1 & 0 \\ 2/3 & 0 & -1\end{matrix} \right]$

A= $\left[ \begin{matrix}2 & 23 & 3 \end{matrix} \right]$ * $B^{-1}$

A= $\left[ \begin{matrix}2 & 23 & 5 \end{matrix} \right]$

So, it is "BWE"

@@ Line 14: / Line 14: @@
 == Can Eve decrypt the message without finding the inverse of the secret matrix? ==
 No, There is no way to decrypt the message without finding the inverse of the secret matirx.
-== What is the decrypted message corresponding to (2,23,3)==
+== What is the decrypted message corresponding to (2, 23, 3)?
-First of all, we need to find out the secret matrix to encrypt the message. A(message)*B(secret)=C(decrypted)
-To find out B, A'*A*B= A'*C, we need to calculate the inverse matrix of A.
+A*B=C
-A= <math>\left[ \begin{matrix}2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3\end{matrix} \right]</math> <br>
+<math>A^{-1}*A*B=A*C</math>
-C= <math>\left[ \begin{matrix}1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1\end{matrix} \right]</math> <br>
+<math>I*B=A^{-1}*C</math>
-B = <math>\left[\begin{array}{ccc}-\frac{2}{3} & 0 & \frac{2}{3} \\0 & 1 & 0 \\4 & 0 & -1 \end{array}\right]\!</math>
+<math>B=A^{-1}*C</math>
+A= <math>\left[ \begin{matrix}2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3\end{matrix}
+\right]</math> <br>
+C= <math>\left[ \begin{matrix}1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1\end{matrix}
+\right]</math> <br>
+B = <math>\left[ \begin{matrix}-2/3 & 0 & 4 \\ 0 & 1 & 0 \\ 2/3 & 0 & -1\end{matrix}
+\right]</math> <br>
+A=<math>\left[ \begin{matrix}2 & 23 & 3 \end{matrix} \right]</math> * <math>B^{-1}</math>
+A=<math>\left[ \begin{matrix}2 & 23 & 5 \end{matrix} \right]</math>
+So, it is "BWE"