# Tag Info

3

No, because matrix inversion can be done efficiently. Namely, if encryption is multiplication by a matrix $A$, then you can define decryption with $A$ as first computing the inverse of $A$ and then multiplying. An essential property of public-key cryptosystems is that it should not be possible to efficiently derive the (private) decryption key from the ...

2

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.

2

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

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