| Line 53: | Line 53: | ||
\end{bmatrix} | \end{bmatrix} | ||
</math> | </math> | ||
| + | |||
| + | <math>=\begin{bmatrix} | ||
| + | 2 \\ | ||
| + | 23 \\ | ||
| + | 5 | ||
| + | \end{bmatrix} | ||
| + | </math> | ||
| + | |||
| + | if we translate to alphabet... | ||
| + | |||
| + | I got '''BWE''' | ||
Latest revision as of 06:44, 18 September 2008
1. How can Bob decrypt the message?
since bob has a 3by3 matrix, he just need to multiplying by inverse of the encryption matrix. then he has to translate to alphabet from what he got.
2. Can Eve decrypt the message without finding the inverse of the secret matrix?
no, i don't think so. if there is no inverse of the secret matrix we can't calculate the matrix therefore eve can't decrypt.
3. What is the decrypted message corresponding to (2,23,3)?
$ \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0& 3 \end{bmatrix} * $ $ \begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 4 & 1 & 1 \end{bmatrix}^{-1}= $ $ \begin{bmatrix} \frac{-2}{3} & 0 & \frac{-2}{3} \\ 0 & 1 & 0 \\ 4 & 0 & -1 \end{bmatrix} $
if we inverse above matrix we got..
$ \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{3} \\ 0 & 1 & 0 \\ 2 & 0 & \frac{1}{3} \end{bmatrix} $
above matrix is decrypt matrix so we have to multiplying by secret matrix...
$ \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{3} \\ 0 & 1 & 0 \\ 2 & 0 & \frac{1}{3} \end{bmatrix} $ $ *\begin{bmatrix} 2 \\ 23 \\ 3 \end{bmatrix} $
$ =\begin{bmatrix} 2 \\ 23 \\ 5 \end{bmatrix} $
if we translate to alphabet...
I got BWE
