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Recently, I was given three ciphers to crack for my cryptography class. At this point, I have guessed that one of them is likely a Hill cipher (probably 3x3, as that is the most complex we have done so far).

I have made a fairly good guess of two options for the first seven characters in the cipher, but I found myself unable to figure out the key matrix. So far, I have $\bigl(\begin{smallmatrix} a&b&c\\ d&e&f\\ g&h&i \end{smallmatrix} \bigr)$ x $\bigl(\begin{smallmatrix} 14\\ 13\\ 0 \end{smallmatrix} \bigr)$ = $\bigl(\begin{smallmatrix} 10\\ 4\\ 12 \end{smallmatrix} \bigr)$, but since the matrix is modulo 26 there could technically be infinite values for variables a-i (for instance, $14a + 13b + 0c = 10$ could actually be 35, 60, etc.). Is there something that I'm missing?

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  • $\begingroup$ If there is one matrix that solves it, there will be many, since $c$, $f$ and $i$ may have arbitrary values. $\endgroup$ – Artjom B. Sep 6 '14 at 13:03
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You would need (at least) 3 pairs of vectors in order to determine the 3*3 matrix.

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