Linear Algebra Matrices- Cryptography

Question

Please advise what am I doing wrong here? Thanks. The question is: And this is what I have done so far:

Answer 1

You don't use your encoding scheme at all, your encryption matrix

$$E = \begin{bmatrix} 3 & 4 & 0\\ 2 & 2 & 1 \\ 1 & 4 & 3 \\ \end{bmatrix}$$

is one modulo $5$ (!).

So DECIDE becomes, using the encoding scheme:

$$(0,3), (0,4), (0,2), (1,3), (0,3), (0,4)$$

which we can see as $3$-vectors modulo $5$, by regrouping:

$$\begin{bmatrix}0\\3\\0\\ \end{bmatrix}, \begin{bmatrix}4\\0\\2\\\end{bmatrix}, \begin{bmatrix}1\\3\\0\\ \end{bmatrix}, \begin{bmatrix}3\\0\\4\\ \end{bmatrix}$$

and which we can each encrypt by $E$, e.g. the first:

$$E = \begin{bmatrix} 3 & 4 & 0\\ 2 & 2 & 1 \\ 1 & 4 & 3 \\ \end{bmatrix} \begin{bmatrix}0\\3\\0\\ \end{bmatrix} = \begin{bmatrix}12\\6\\12\\ \end{bmatrix} = \begin{bmatrix}2\\1\\2\\ \end{bmatrix}$$

where the last step is due to modulo $5$ reduction.

You can do the others too, and then write them out in a sequence of pairs again and use the encoding scheme. So the first ciphertext letter will be from $(2,1)$ which is M. And so on.

You're working according to the "one letter = one number" system modulo $26$ which is not what this question is about. modulo $5$ (a prime) is easier to work with because it's a field. Using this scheme makes it a bit harder to use single letter frequencies, as letters are split into pairs, and some letters are then in two different vectors etc. It creates some more diffusion.

FWIW, my interpretation gives a valid English word as a solution to the decryption part (I computed the inverse for fun and checking). So I’m convinced my interpretation of this exercise is correct.