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Is the point of a known plaintext attack on a Hill Cipher to get the matrix key?

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Indeed. If the matrix is $n$ by $n$, and you have $n$ many known plaintext blocks with corresponding ciphertext, you get $n^2$ linear equations in $n^2$ unknowns (the matrix elements) (modulo the alphabet size), which very often can be solved uniquely.

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  • $\begingroup$ @Connor Glad I could help! $\endgroup$ – Henno Brandsma Feb 22 '15 at 18:25

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