Hill Cipher for the New Year!

Note: $\: \det(A) = 3$ and $\gcd(3,26) = 1$ , so for each block $X_{j}$ the system $A X_{j} = B_{j} \bmod{26}$ has a unique solution, where $(1 \leq j \leq 6)$ .

Using six blocks for $VVIXHUIEYCRI$ we obtain: $$

$21\: 21$ $$

$8 \: 23$ $$

$7 \: 20$ $$

$8 \: 4$ $$

$24 \: 2$ $$

$17 \: 8$ $$

$A * X_{j} = B_{j} \bmod{26} \implies X_{j} = A^{-1} * B_{j} \bmod{26}$ $$

Using $A^{-1} \textbf{:}$ $$

$C = A|I = \left [\begin{array}{cc|cc} 3 & 1 & 1 & 0 \\ 3 & 2 & 0 & 1 \\ \ \end{array} \right] \bmod{26}$

$$

Using the row operations: $9 * Row_{1}$ and $23 * Row_{1} + Row_{2} \implies$

$C = \left [\begin{array}{cc|cc} 1 & 9 & 9 & 0 \\ 0 & 1 & 25 & 1 \\ \ \end{array} \right] \bmod{26}$

$$

Using the row operation: $17 * Row_{2} + Row_{1} \implies$

$C = \left [\begin{array}{cc|cc} 1 & 0 & 18 & 17\\ 0 & 1 & 25 & 1 \\ \ \end{array} \right] \bmod{26}$

$$

$\implies$

$A^{-1} = \left [\begin{array}{cc|c} 18 & 17\\ 25 & 1 \\ \ \end{array} \right] \bmod{26}$

$$

Note: $A * A^{-1} = I$

Deciphering the first block $\left [\begin{array}{cc|c} 21 \\ 21 \\ \ \end{array} \right] \textbf{:}$ $$

$X_{1} = A^{-1} * B_{1} \bmod{26} = \left [\begin{array}{cc|c} 18 & 17\\ 25 & 1 \\ \ \end{array} \right] * \left [\begin{array}{cc|c} 21 \\ 21 \\ \ \end{array} \right] \bmod{26} = \left [\begin{array}{cc|c} 7 \\ 0 \\ \ \end{array} \right] \bmod{26}$ : $$

Deciphering the second block $\left [\begin{array}{cc|c} 8 \\ 23 \\ \ \end{array} \right] \textbf{:}$ $$ $$

$X_{2} = A^{-1} * B_{2} \bmod{26} = \left [\begin{array}{cc|c} 18 & 17\\ 25 & 1 \\ \ \end{array} \right] * \left [\begin{array}{cc|c} 8 \\ 23 \\ \ \end{array} \right] \bmod{26} = \left [\begin{array}{cc|c} 15 \\ 15 \\ \ \end{array} \right] \bmod{26}$ $$

Repeat the same procedure for the remaining four blocks. $$

Our deciphered blocks are: $$

$7 \: 0$ $$

$15 \: 15$ $$

$24 \: 13$ $$

$4 \: 22$ $$

$24 \: 4$ $$

$0 \: 17$ $$

Our plain text message is:

$HAPPYNEWYEAR$

$$

As a string of integers we have: $7015152413422244017$ .

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