Editing Hill cipher (section)

==Example==
Let

:<math>K= \begin{pmatrix} 3 & 3 \\ 2 & 5 \end{pmatrix}</math>

be the key and suppose the plaintext message is 'HELP'. Then this plaintext is represented by two pairs

:<math>HELP \to \begin{pmatrix} H \\ E \end{pmatrix} , \begin{pmatrix} L \\ P \end{pmatrix} \to \begin{pmatrix} 7 \\ 4 \end{pmatrix} , \begin{pmatrix} 11 \\ 15 \end{pmatrix}</math>

Then we compute

:<math>\begin{pmatrix} 3 & 3 \\ 2 & 5 \end{pmatrix} \begin{pmatrix} 7 \\ 4 \end{pmatrix} \equiv \begin{pmatrix} 7 \\ 8 \end{pmatrix} \pmod{26},</math> and

:<math>\begin{pmatrix} 3 & 3 \\ 2 & 5 \end{pmatrix} \begin{pmatrix} 11 \\ 15 \end{pmatrix} \equiv \begin{pmatrix} 0 \\ 19 \end{pmatrix}\pmod{26}</math>

and continue encryption as follows:

:<math>\begin{pmatrix} 7 \\ 8 \end{pmatrix}, \begin{pmatrix} 0 \\ 19 \end{pmatrix} \to \begin{pmatrix} H \\ I \end{pmatrix}, \begin{pmatrix} A \\ T \end{pmatrix}</math>

The matrix ''K'' is invertible, hence <math>K^{-1}</math> exists such that <math>KK^{-1}=K^{-1}K=I_2</math>.
The inverse of ''K'' can be computed by using the [[Invertible matrix#Inversion of 2×2 matrices|formula]]
<math>\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1}=(ad-bc)^{-1}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}</math>

This formula still holds after a modular reduction if a [[modular multiplicative inverse]] is used to compute {{nowrap|<math>(ad-bc)^{-1}</math>.}} Hence in this case, we compute

:<math>K^{-1} \equiv 9^{-1} \begin{pmatrix} 5 & 23 \\ 24 & 3 \end{pmatrix} \equiv 3 \begin{pmatrix} 5 & 23 \\ 24 & 3 \end{pmatrix} \equiv \begin{pmatrix} 15 & 17 \\ 20 & 9 \end{pmatrix}\pmod{26}</math>

:<math>HIAT \to \begin{pmatrix} H \\ I \end{pmatrix}, \begin{pmatrix} A \\ T \end{pmatrix} \to \begin{pmatrix} 7 \\ 8 \end{pmatrix}, \begin{pmatrix} 0 \\ 19 \end{pmatrix}</math>

Then we compute

:<math>\begin{pmatrix} 15 & 17 \\ 20 & 9 \end{pmatrix}\begin{pmatrix} 7 \\ 8 \end{pmatrix} = \begin{pmatrix} 241 \\ 212 \end{pmatrix} \equiv \begin{pmatrix} 7 \\ 4 \end{pmatrix}\pmod{26},</math> and

:<math>\begin{pmatrix} 15 & 17 \\ 20 & 9 \end{pmatrix}\begin{pmatrix} 0 \\ 19 \end{pmatrix} = \begin{pmatrix} 323 \\ 171 \end{pmatrix} \equiv \begin{pmatrix} 11 \\ 15 \end{pmatrix}\pmod{26}</math>

Therefore,

:<math>\begin{pmatrix} 7 \\ 4 \end{pmatrix}, \begin{pmatrix} 11 \\ 15 \end{pmatrix} \to \begin{pmatrix} H \\ E \end{pmatrix}, \begin{pmatrix} L \\ P \end{pmatrix} \to HELP</math>.