Encrption

[elder barn]

Okay, so for this type of encryption, we know that it works the following way:
Encryption
Cipher Text = (Plain Text * Key) Mod 29

Decryption
Plain Text = (Cipher Text * Inverse Key) Mod 29

So, the first thing you should do here is put "I LOVE MATHS" in a matrix of the appropriate form:

Now we do matrix multiplication with placeholder variables for the key matrix:

Which spits out the equations:

C_11 = 9A + 27D +12G
C_12 = 15B + 22E + 5H
C_13 = 27C + 14F + 1I

But wait, we know the cipher text, so we know what the first three letters here should be!

So we have:
C_11 = 9A + 27D +12G = 25
C_12 = 15B + 22E + 5H = 13
C_13 = 27C + 14F + 1I = 5

You should recognize these individually as linear combinations, and therefore all of them taken together as a system of linear equations.

You should be able to solve for the key matrix now, no problem!

@wanton hornet

Note: Of course, keep in mind that the answers to each linear combination have already been reduced via the modulo operation.

[wanton hornet]

i did this is it correct?

[elder barn]

No, not by a long shot. I'm actually not even sure what you're trying to do, here.

Could you explain your thought process, @wanton hornet ?

Ah, wait. I think I might see where the confusion is happening, here.

Side note: Fuck, I need my coffee. Am slightly loopy atm.

[elder barn]

Anyway, you know how to solve this system of equations, right?

If so, then do that before anything else.

Edit and a note: It occurs to me that I didn't explain enough here that you're going to need more than the first three letters of the ciphertext in order to solve this particular system. I just thought that was fairly obvious, and so forgot to mention it. I'm sorry. 😦

Although looking everything, maybe you need more explicit data. It's all there. Try getting the equations from the rest of the ciphertext using the methods I did.

Lmk if you have further questions.

[wanton hornet]

sorry i dont know can u show me a little bit about it ?

[elder barn]

Sure. One moment.

[elder barn]

Okay, so we have the system:

[elder barn]

Moreover, it's interesting to note that:

for some J, K, L in Z.

This system essentially encodes taking:
I -> Y
[SPACE] -> M
L -> E

Then, for the rest of the ciphertext, we have some transformation:
O -> N
V -> M
E -> N
[SPACE] -> .
M -> D
A -> F
T -> W
H -> T
S -> M

[wanton hornet]

My lecturer say that it should be in 3x1 matrices for cipher and plaintext as it is hill 3 cipher

I did this