Revision as of 15:27, 18 September 2008 by Huang122 (Talk)

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

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva