Line 1: Line 1:
| 3 3 3 |
 
| 3 3 3 |
 
| 3 3 3 |
 
 
 
 
 
1. How can Bob decrypt the message?   
 
1. How can Bob decrypt the message?   
  
 
Bob can use matrix algebra to find the inverse of the 3-by-3 matrix, which he can then multiply the encrypted vector by in order to obtain the 1-by-3 decrypted vector that corresponds to the numerical place of each letter of the alphabet.  
 
Bob can use matrix algebra to find the inverse of the 3-by-3 matrix, which he can then multiply the encrypted vector by in order to obtain the 1-by-3 decrypted vector that corresponds to the numerical place of each letter of the alphabet.  
  
that the message (1,0,4,0,1,0,1,0,1) yields the encrypted vector (2,0,0,0,1,0,0,0,3).
 
 
2. Can Eve decrypt the message without finding the inverse of the secret matrix?
 
2. Can Eve decrypt the message without finding the inverse of the secret matrix?
  
Line 15: Line 8:
  
 
3. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.)
 
3. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.)
 +
 +
By solving the system equations given by the vector of decrypted numbers and vector of encrypted numbers one can derive the original matrix [A] where
 +
 +
a11 = -2/3
 +
a12 = 0
 +
a13 = 2/3
 +
 +
a21 = 0
 +
a22 = 1
 +
a23 = 0
 +
 +
a31 = 4
 +
a32 = 0
 +
a33 = -1
 +
 +
which can be used as matrix [A] in matlab to find the inverse matrix [B] which, when multiplied with the encrypted vector [2, 23, 3] yields
 +
 +
 +
A =
 +
 +
  -0.6667        0    0.6667
 +
        0    1.0000        0
 +
    4.0000        0  -1.0000
 +
 +
 +
B =
 +
 +
    0.5000        0    0.3333
 +
        0    1.0000        0
 +
    2.0000        0    0.3333
 +
 +
 +
Message =
 +
 +
    2
 +
    23
 +
    5
 +
 +
which corresponds to the letters of the alphabet B W and E.
 +
 +
Answer = BWE

Latest revision as of 15:25, 19 September 2008

1. How can Bob decrypt the message?

Bob can use matrix algebra to find the inverse of the 3-by-3 matrix, which he can then multiply the encrypted vector by in order to obtain the 1-by-3 decrypted vector that corresponds to the numerical place of each letter of the alphabet.

2. Can Eve decrypt the message without finding the inverse of the secret matrix?

Yes. Eve can solve the system of equations with the vectors she has in order to decrypt the message.

3. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.)

By solving the system equations given by the vector of decrypted numbers and vector of encrypted numbers one can derive the original matrix [A] where

a11 = -2/3 a12 = 0 a13 = 2/3

a21 = 0 a22 = 1 a23 = 0

a31 = 4 a32 = 0 a33 = -1

which can be used as matrix [A] in matlab to find the inverse matrix [B] which, when multiplied with the encrypted vector [2, 23, 3] yields


A =

  -0.6667         0    0.6667
        0    1.0000         0
   4.0000         0   -1.0000


B =

   0.5000         0    0.3333
        0    1.0000         0
   2.0000         0    0.3333


Message =

    2
   23
    5

which corresponds to the letters of the alphabet B W and E.

Answer = BWE

Alumni Liaison

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

Francisco Blanco-Silva