Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

i am working modulo 26 in the alphabet $A,B,C,\cdots,Z$ where $0=A,1=B$ etc. I may see the encryptions (using the Hill cipher) of $[3,0,6]; [7,14,8]$ and $[13,14,20]$. Is it now possible for me to find de encryption matrix in $GL_3(\mathbb Z /26 \mathbb Z)$ ?

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

The third input vector is the second plus twice the first. Thus you get no new information from the third encryption and don't have enough information to reconstruct the encryption matrix.

share|improve this answer
    
yes. thanks - didn't see that. –  André Nov 20 '12 at 16:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.