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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 $GL_3(\mathbb Z /26 \mathbb Z)$ ?

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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.

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yes. thanks - didn't see that. – Andre Nov 20 '12 at 16:17

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