Latest revision as of 12:40, 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 20: / Line 20: @@
 A*B=C
-A'*A*B=A*C,  I*B=A'*C,   B=A'*C
+<math>A^{-1}*A*B=A*C</math>
+<math>I*B=A^{-1}*C</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}
@@ Line 29: / Line 38: @@
+A=<math>\left[ \begin{matrix}2 & 23 & 3 \end{matrix} \right]</math> * <math>B^{-1}</math>
-C= <math>\left[ \begin{matrix}1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1\end{matrix}
-\right]</math> <br>
+A=<math>\left[ \begin{matrix}2 & 23 & 5 \end{matrix} \right]</math>
+So, it is "BWE"