| Line 6: | Line 6: | ||
:<math> | :<math> | ||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
\begin{bmatrix} | \begin{bmatrix} | ||
A & B & C \\ | A & B & C \\ | ||
| Line 17: | Line 11: | ||
G & H & I | G & H & I | ||
\end{bmatrix} | \end{bmatrix} | ||
| + | cdot | ||
| + | \begin{bmatrix} | ||
| + | 1 & 0 & 4 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 1 & 0 & 1 | ||
| + | \end{bmatrix} | ||
| + | \ | ||
= | = | ||
\begin{bmatrix} | \begin{bmatrix} | ||
Revision as of 14:21, 18 September 2008
1. Bob needs to calculate the inverse of the secret matrix, and multiply it by the code given by Alice to get a vector. Then replaces each three entries by its corresponding letter in the alphabet.
2.Eve can get the secret matrix through calculation.
- $ \begin{bmatrix} A & B & C \\ D & E & F \\ G & H & I \end{bmatrix} cdot \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \ = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} $
Thus we have
A+4G=2
B+4H=0
C+4I=0
D=0
E=1
F=0
A+G=0
B+H=0
C+I=3
and so
D=0
E=1
F=0
A=B=-2/3
G=H=2/3
I=-1
C=4
i.e.
- $ \begin{bmatrix} -\frac{2}{3} & -\frac{2}{3} & 4 \\ 0 & 1 & 0 \\ \frac{2}{3} & \frac{2}{3} & -1 \end{bmatrix} $
3. The inverse matrix is
- $ \begin{bmatrix} \frac{1}{2} & -1 & 2 \\ 0 & 1 & 0 \\ \frac{1}{3} & 0 & \frac{1}{3} \end{bmatrix} $
So(2,23,3) --> (1,23,1) --> AWE
