Question

Q1: Use the matrix

9 1

7 2 

(a)  Obtain the Hill cipher for the plaintext message

                                              PAK ARMY

by letting  A=1,B=2,C=3,.......,Y=25 and Z=26

(b)  Decode Hill 2- cipher which was encrypted by this matrix.

Answer 1

To encrypt a message, each block of n letters (considered as an n-component vector) is multiplied by an invertible n × n matrix, against modulus 26. To decrypt the message, each block is multiplied by the inverse of the matrix used for encryption.

a)

$PAK\ ARMY\to \begin{pmatrix} P \\ A \end{pmatrix}, \begin{pmatrix} K \\ A \end{pmatrix}, \begin{pmatrix} R \\ M \end{pmatrix}, \begin{pmatrix} Y \\ X \end{pmatrix}\to\begin{pmatrix} 16 \\ 1 \end{pmatrix}, \begin{pmatrix} 11 \\ 1 \end{pmatrix}, \begin{pmatrix} 18 \\ 13 \end{pmatrix}, \begin{pmatrix} 25 \\ 24 \end{pmatrix}$

$\begin{pmatrix} 9 & 1 \\ 7 & 2 \end{pmatrix}\begin{pmatrix} 16 \\ 1 \end{pmatrix}=\begin{pmatrix} 145 \\ 114 \end{pmatrix}=\begin{pmatrix} 15 \\ 10 \end{pmatrix}(mod\ 26)$

$\begin{pmatrix} 9 & 1 \\ 7 & 2 \end{pmatrix}\begin{pmatrix} 11 \\ 1 \end{pmatrix}=\begin{pmatrix} 100 \\ 79 \end{pmatrix}=\begin{pmatrix} 22 \\ 1 \end{pmatrix}(mod\ 26)$

$\begin{pmatrix} 9 & 1 \\ 7 & 2 \end{pmatrix}\begin{pmatrix} 18 \\ 13 \end{pmatrix}=\begin{pmatrix} 175 \\ 152 \end{pmatrix}=\begin{pmatrix} 19 \\ 22 \end{pmatrix}(mod\ 26)$

$\begin{pmatrix} 9 & 1 \\ 7 & 2 \end{pmatrix}\begin{pmatrix} 25 \\ 24 \end{pmatrix}=\begin{pmatrix} 249 \\ 199 \end{pmatrix}=\begin{pmatrix} 15 \\ 17 \end{pmatrix}(mod\ 26)$

$\begin{pmatrix} 15 \\ 10 \end{pmatrix}, \begin{pmatrix} 22 \\ 1 \end{pmatrix}, \begin{pmatrix} 19 \\ 22 \end{pmatrix}, \begin{pmatrix} 15 \\ 17 \end{pmatrix}\to \begin{pmatrix} O \\ J \end{pmatrix}, \begin{pmatrix} V \\ A \end{pmatrix}, \begin{pmatrix} R \\ M \end{pmatrix}, \begin{pmatrix} X \\ Y \end{pmatrix}$

b)

$\begin{pmatrix} 9 & 1 \\ 7 & 2 \end{pmatrix}^{-1}=\frac{1}{11}\begin{pmatrix} 2 & -1 \\ -7 & 9 \end{pmatrix}\equiv 19\begin{pmatrix} 2 & -1 \\ -7 & 9 \end{pmatrix}=\begin{pmatrix} 12 & 25 \\ 19 & 15 \end{pmatrix}(mod\ 26)$

$\begin{pmatrix} 12 & 25 \\ 19 & 15 \end{pmatrix}\begin{pmatrix} 15 \\ 10 \end{pmatrix}=\begin{pmatrix} 430 \\ 435 \end{pmatrix}=\begin{pmatrix} 16 \\ 1 \end{pmatrix}(mod\ 26)$

$\begin{pmatrix} 12 & 25 \\ 19 & 15 \end{pmatrix}\begin{pmatrix} 22 \\ 1 \end{pmatrix}=\begin{pmatrix} 279 \\ 433 \end{pmatrix}=\begin{pmatrix} 11 \\ 1 \end{pmatrix}(mod\ 26)$

$\begin{pmatrix} 12 & 25 \\ 19 & 15 \end{pmatrix}\begin{pmatrix} 19 \\ 22 \end{pmatrix}=\begin{pmatrix} 778 \\ 691 \end{pmatrix}=\begin{pmatrix} 18 \\ 13 \end{pmatrix}(mod\ 26)$

$\begin{pmatrix} 12 & 25 \\ 19 & 15 \end{pmatrix}\begin{pmatrix} 15 \\ 17 \end{pmatrix}=\begin{pmatrix} 605 \\ 540 \end{pmatrix}=\begin{pmatrix} 25 \\ 24 \end{pmatrix}(mod\ 26)$