| Line 26: | Line 26: | ||
| − | <math>\begin{pmatrix} | + | <math>\begin{pmatrix} <math>*</math> <math>\begin{pmatrix} |
| − | + | 1/2 & 0 & 1/3 \\ 2\\ | |
| − | 2 & 0 & 1/3 | + | 0 & 1 & 0 \\ 23\\ |
| − | \end{pmatrix}</math> | + | 2 & 0 & 1/3 3 |
| + | \end{pmatrix}</math> \end{pmatrix}</math> | ||
Revision as of 15:17, 19 September 2008
1) Bob can easily decrypt the message. All he has to do is to multiply the inverse of the secret matrix( which Alice has told him) by the first three entries of the vector and then by the other three and lastly by the remaining three vectors. Then he has to associate the numbers obatined with the corresponding alphabetical order , and use a space for a zero.
2)
No, eve cannot find the encrypted code without finding out the inverse of the secret matrix. However, she can easiy find out the secret matrix by using the basic linear algebra, as she knows both of the encrypted and the decrypted code.
3) The secret matrix is $ \begin{pmatrix} -2/3 & 0 & 2/3 \\ 0 & 1 & 0 \\ 4 & 0 & -1 \end{pmatrix} $
The inverse of the secret matrix is
$ \begin{pmatrix} 1/2 & 0 & 1/3 \\ 0 & 1 & 0 \\ 2 & 0 & 1/3 \end{pmatrix} $
In order to get the message multiply the inverse of the secret matrix with the given code of ( 2, 23, 3)
$ \begin{pmatrix} <math>* $ $ \begin{pmatrix} 1/2 & 0 & 1/3 \\ 2\\ 0 & 1 & 0 \\ 23\\ 2 & 0 & 1/3 3 \end{pmatrix} $ \end{pmatrix}</math>
