I have the plaintext
ｂｒｅａｔｈｔａｋｉｎｇ
Which encrypts to the cipher text
ＲＵＰＯＴＥＮＴＯＩＦＶ
I know that it was encrypted with a hill cipher
(m not specified) but I have no clue what i'm
doing wrong trying to decipher the text.
I set up the vectors for the plaintext
m1 = (1,17,4)
m2=(0,19,7)
m3=(19,0,10)
And the vectors for the ciphertext
c1 = (17,20,16)
c2 = (14,19,4)
c3 = (13,19,14)
I know to solve for K (the key matrix) in (mod 26)
I need to solve
K * M = C
Where M is the matrix of the plaintext in (mod 26)
and C is the matrix w/ the ciphertext in (mod 26)
So, I start off by trying to find the inverse of the
plaintext matrix
\begin{bmatrix}
1 & 0 & 19 \\
17 & 19 & 0 \\
4 & 7 & 10 \\
\end{bmatrix}
But I end up gettng 13 and 13 doesn't have an inverse
mod 26.. So I have no idea how to solve this.
Best Answer
You cannot invert it because the matrix is singular. Try to use the encryption of "ing" (vector m4) with two of the encryptions of "bre" "ath" and "tak" (vectors m1, m2, m3), in order to get a non-singular matrix (try all the combinations: m1, m2, m4 and m1, m3, m4, and m2, m3, m4). There is a reason why your plaintext/cipertext is 12 characters long instead of 9.