In the following, we suppose that we have a ciphertext from an unknown 2×2 Hill Cipher. Let N=[efgh] be the decryption m

[answerhappygod]

In the following, we suppose that we have a ciphertext from anunknown 2×2 Hill Cipher. Let N=[efgh] be thedecryption matrix.
5.) Suppose we suspect that the cipher encrypts "th" as "FR".Use this to find two equations for e in terms of f,modulo 26.
Hint: Say you have 8e≡22−12f(mod26), and need to solve thisfor e.
Since gcd(8,26)=2, and 2∣22−12f, this willhave 2 solutions for e, no matterwhat f is. Once we find one, then the otheris 13 away. But in the notes, getting the first was easy,since we had 2f≡2e+6(mod26) and could "read off"that f=e+3 works. How do we do this here?
Well we want to make 22−12f look like a multipleof 8. So, we add/subtract 26's! But we have two things todeal with. Adding a single 26 to theconstant 22 gives 48, which is indeed a multipleof 8. What about the −12f? Well, we add termsof 26f! Adding a single 26f gives 14f, and thenanother gives 40f. So...
8e≡22−12f(mod26)≡48+40f(mod26)
So here our "easy" solution is e≡6+5f, and then the othermust be e≡19+5f.