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Let $k=\left(\begin{array}{cc}k_0 & k_1\\k_2 & k_3\end{array}\right)$ be the key. And I'll assume the transformation $a=0$, $b=1$, and so on. So you know $k\left(\begin{array}{c}0\\1\end{array}\right)=\left(\begin{array}{c}c_0\\c_1\end{array}\right)$ where you know $c_0$ and $c_1$. Thus $c_0=0k_0+1k_1$ and $c_1=0k_2+1k_3$. So that is two equations, ...


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Since it's a linear cipher, you should be wary about a guessable padding, otherwise if your last block is only one char long, you will reveal almost your whole matrix on this last block. If you're too afraid of mangling the last word, use something like 'Z'+(random chars). But I really would not use any predictable padding with such a cipher.


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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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Updated answer The question was changed, so here is my updated answer. This scheme is not secure against known-plaintext attacks. It is no better than an ordinary Hill cipher. If you iterate the recurrence relation you listed, we find that $$x_k = Bx_0 - Cx_{-1}$$ and $$x_{k-1} = Cx_0 - Dx_{-1}$$ where $B,C,D$ are matrices given by $B=f(A)$, ...



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