| Line 37: | Line 37: | ||
4G+I=3 <br> | 4G+I=3 <br> | ||
and so | and so | ||
| + | A=-\frac{2}{3} <br> | ||
| + | C=\frac{8}{3} <br> | ||
D=0 <br> | D=0 <br> | ||
| − | |||
F=0 <br> | F=0 <br> | ||
| − | + | G=1<br> | |
| − | G= | + | |
I=-1 <br> | I=-1 <br> | ||
| − | |||
i.e. | i.e. | ||
Revision as of 15:27, 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+C=2
B=0
4A+C=0
D+F=0
E=1
4D+F=0
G+I=0
H=0
4G+I=3
and so
A=-\frac{2}{3}
C=\frac{8}{3}
D=0
F=0
G=1
I=-1
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
