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While practicing for Hill Cipher I choose a random Key matrix of $ 2*2 $ given as follows : $ K = \begin{bmatrix}3&2\\1&0\\\end{bmatrix} $

Say the Text to Encrypt is ATTACK By using the Following Equation $ C=K * P \mod 26 $ I got the encrypted Text as MAFTAC, where

$C$ is Cipher Text Matrix

$K$ is Key Matrix

$P$ is Plain Text Matrix

Now while decrypting the Cipher text using equation $ P= K ^{-1} * C \mod 26 $.

I need to find $ K^{-1} = |K|^{-1} Adj A $ But The Multiplicative Inverse $ |K|^{-1}$ exist only if $ 26 $ and $|K|$ are relatively Prime. But In this case $|K|=-2= 24 \mod {26}$.

But 24 and 26 are not relatively Prime. Does That Mean The following Key Can't be used To Encrypt The Text?

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  • $\begingroup$ This is not a research level question, but you could ask it on math.stackexchange.com. $\endgroup$
    – Nate Eldredge
    Commented May 28, 2019 at 0:09
  • $\begingroup$ But a hint: what happens when you try to encrypt the vector $P =\left[\begin{smallmatrix} 0 \\ 13 \end{smallmatrix} \right]$? $\endgroup$
    – Nate Eldredge
    Commented May 28, 2019 at 0:14
  • $\begingroup$ It will give me [A, A] $\endgroup$
    – Gaurav
    Commented May 28, 2019 at 2:35
  • $\begingroup$ Yes, and so will [0,0]. So which of the two should be the result when you decrypt [A,A]? You see the problem? $\endgroup$
    – Nate Eldredge
    Commented May 28, 2019 at 2:38
  • $\begingroup$ I understand That thing, but the question is that i am stuck with decrypting Any Text, because i need multiplicative inverse. and which doesn't exist in this case. $\endgroup$
    – Gaurav
    Commented May 28, 2019 at 2:40

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