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So I don't want the answer but somewhere to start with this problem, first I want to know if my logic and thinking is on the right path before I dive right into computing the decryption matrix so here is the question:

Recall the encryption matrix for AES is $E = \begin{bmatrix}1&x^2\\x^2&1\end{bmatrix}$ over the finite field $\mathbb{F}_{16}$ with irreducible polynomial $x^4 + x + 1$. Compute the decryption matrix $D = E^{-1}$

My thought is find the inverse of each individual polynomial in the matrix but I think that's way to simple to do but, calculating the inverse of each polynomial in the matrix we would just divide by the irreducible polynomial right?

Another thought would be using the same steps as the hill cipher?

please let me if I am on the right track or I'm not even close to the start of it all

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    $\begingroup$ "Recall the encryption matrix for AES is..."; well, no, I don't recall that at all. Might this question be about perhaps the MDS matrix in some dumbed-down version of AES (which, of course, isn't AES)? $\endgroup$ – poncho Apr 16 '18 at 17:04
  • $\begingroup$ Well this is more SAES since we are only working with small computations of the actual AES $\endgroup$ – AceNinja1101 Apr 16 '18 at 17:17
  • $\begingroup$ github.com/bpdegnan/aes/blob/master/aes-sbox/documentation/… $\endgroup$ – b degnan Apr 16 '18 at 21:35
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First of all, this is not "AES", but something else. Someone might claim that it's somewhat similar to AES, but it's still not AES.

However, to answer your question:

please let me if I am on the right track or I'm not even close to the start of it all

Actually, you already got your answer:

Compute the decryption matrix $D = E^{-1}$

That is, compute the matrix $D$ such that $D$ and $E$, multiplied together using standard matrix multiplication, gives:

$$\begin{bmatrix}d_{00}&d_{01}\\d_{10}&d_{11}\end{bmatrix}\begin{bmatrix}1&x^2 \\x^2&1\end{bmatrix}= \begin{bmatrix}1&0\\0&1\end{bmatrix}$$

This is a standard matrix inversion problem, and all the standard tools used to invert matrices over $\mathbb{R}$ work (and as a bonus, you don't have to worry about numeric stability). Now, you do have to do your internal computations in $\mathbb{F}_{16}$, but that's pretty easy.

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  • $\begingroup$ Thank you! So basically I'm going to calculate the determinant then take the inverse which in modular arithmetic we don't have any negative numbers we simply add the value of the irreducible polynomial in our case until its positive then we switch diagonally $\endgroup$ – AceNinja1101 Apr 18 '18 at 1:30

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