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One possibility for what you might be missing: normally the same key (the same matrix) is re-used to encrypt many messages. So now try counting the total entropy in $M$ length-$N$ messages, and the entropy in a $N\times N$ matrix, and compare what happens when $M$ gets large.... Another possibility you might be missing is the consequences of the fact that ...


2

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, ...


2

You would need (at least) 3 pairs of vectors in order to determine the 3*3 matrix.


1

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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Linearizing should work fine. You'll have to make a minor adjustment to deal with the fact that we are working modulo 26, but either of the following two simple tweaks should work fine: You could use generalized Gaussian elimination, generalized to work over $\mathbb{Z}/26\mathbb{Z}$ (standard Gaussian elimination assumes we are working over a field, but ...


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Both representations are essentially equivalent. If $$ \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} k_{11} & k_{12} & k_{13} \\ k_{21} & k_{22} & k_{23} \\ k_{31} & k_{32} & k_{33} \end{bmatrix} \cdot \begin{bmatrix} p_1 \\ p_2 \\ p_3 \end{bmatrix}, $$ then, equivalently $$ \begin{bmatrix} p_1 & p_2 & ...



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