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Matrix cipher is defined by:

a) Encryption: $C = K \times Z$, where $C$ is a cryptogram vector, $Z$ is a message vector and $K$ is encryption matrix (encryption key)

b) Decryption: $Z = K^{-1} \times C$, where $K^{-1}$ is decryption matrix (decryption key), which is inverse to matrix $K$.

Is the matrix cipher symmetric or asymmetric cryptosystem?

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  • $\begingroup$ Given $K$, how difficult is it to compute $K^{-1}$ (and hence $K^{-1} \times C$)? $\endgroup$
    – poncho
    Commented Sep 24, 2020 at 19:09
  • $\begingroup$ For an asymmetric cryptosystem, you need two different keys that should not be computable from each other. $\endgroup$
    – kelalaka
    Commented Sep 24, 2020 at 19:13
  • $\begingroup$ The calculation of the inverse matrix has a polynomial complexity - it is relatively fast even for very large matrices. So when I take this into account, we are talking about an symmetric cryptosystem? Thank you for reassurance. $\endgroup$
    – Filip CZ
    Commented Sep 24, 2020 at 19:19
  • $\begingroup$ See Hill Cipher. $\endgroup$
    – kelalaka
    Commented Sep 24, 2020 at 19:25

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