Hill Cipher for the Holidays 2.

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 $X_{j} = A^{-1} B_{j} \bmod{26}$ , where $(1 <= j <= 7)$ .

Using seven blocks for $VEZZMFYQRGTFNRWSSZQIL$ we obtain: $$

$21\: 4 \: 25$ $$

$25 \: 12 \: 5$ $$

$24 \: 16 \: 17$ $$

$6 \: 19 \: 5$ $$

$13 \: 17 \: 22$ $$

$18 \: 18 \: 25$ $$

$16 \: 8 \: 11$ $$

Begin: Find $A^{-1}$ . $$

$B = A|I = \left[ \begin{array}{ccc|ccc} 4 & 2 & 3 & 1 & 0 & 0 \\ 1 & 6 & 5 & 0 & 1 & 0 \\ 3 & 1 & 2 & 0 & 0 & 1 \\ \ \end{array} \right] \mod{26}$

$R_{1} \leftrightarrow R_{2} \rightarrow$

$\left[ \begin{array}{ccc|ccc} 1 & 6 & 5 & 0 & 1 & 0 \\ 4 & 2 & 3 & 1 & 0 & 0 \\ 3 & 1 & 2 & 0 & 0 & 1 \\ \ \end{array} \right] \mod{26}$

$22 * Row_{1} + Row_{2}$ and $23 * Row_{1} + Row_{3} \rightarrow$ $$

$B = \left[ \begin{array}{ccc|ccc} 1 & 6 & 5 & 0 & 1 & 0 \\ 0 & 4 & 9 & 1 & 22 & 0 \\ 0 & 9 & 13 & 0 & 23 & 1 \\ \ \end{array} \right] \mod{26}$

$Row_{2} \leftrightarrow Row_{3} \rightarrow$ $$

$B = \left[ \begin{array}{ccc|ccc} 1 & 6 & 5 & 0 & 1 & 0 \\ 0 & 9 & 13 & 0 & 23 & 1 \\ 0 & 4 & 9 & 1 & 22 & 0 \\ \ \end{array} \right] \mod{26}$

$3 * Row_{2} \rightarrow$ $$

$B = \left[ \begin{array}{ccc|ccc} 1 & 6 & 5 & 0 & 1 & 0 \\ 0 & 1 & 13 & 0 & 17 & 3 \\ 0 & 4 & 9 & 1 & 22 & 0 \\ \ \end{array} \right] \mod{26}$

$20 * Row_{2} + Row_{1}$ and $22 * Row{2} + Row_{3} \rightarrow$ $$

$B = \left[ \begin{array}{ccc|ccc} 1 & 0 & 5 & 0 & 3 & 8 \\ 0 & 1 & 13 & 0 & 17 & 3 \\ 0 & 0 & 9 & 1 & 6 & 14 \\ \ \end{array} \right] \mod{26}$

$3 * Row_{3} \rightarrow$

$B = \left[ \begin{array}{ccc|ccc} 1 & 0 & 5 & 0 & 3 & 8 \\ 0 & 1 & 13 & 0 & 17 & 3 \\ 0 & 0 & 1 & 3 & 18 & 6 \\ \ \end{array} \right] \mod{26}$

$13 * Row_{3} + Row_{2}$ and $21 * Row{3} + Row_{1} \rightarrow$ $$

$B = \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 11 & 17 & 6 \\ 0 & 1 & 0 & 13 & 17 & 3 \\ 0 & 0 & 1 & 3 & 18 & 6 \\ \ \end{array} \right] \mod{26}$

$\implies$

$A^{-1} = \begin{vmatrix}{11} && {17} && {6} \\ {13} && {17} && {3} \\ {3} && {18} && {16}\end{vmatrix} \bmod{26}$ .

Deciphering the first block $\left [\begin{array}{ccc|c} 21 \\ 4 \\ 25 \\ \ \end{array} \right] \textbf{:}$ $$

$X_{1} = A^{-1} * B_{1} \bmod{26} = \left [\begin{array}{ccc|c} 11 & 17 & 6\\ 13 & 17 & 3\\ 3 & 18 & 16\\ \ \end{array} \right] * \left [\begin{array}{ccc|c} 21 \\ 4 \\ 25 \\ \ \end{array} \right] \bmod{26} = \left [\begin{array}{ccc|c} 7 \\ 0 \\ 15 \\ \ \end{array} \right] \bmod{26}$ : $$

Deciphering the second block $\left [\begin{array}{ccc|c} 25 \\ 12 \\ 5 \\ \ \end{array} \right] \textbf{:}$ $$

$X_{2} = A^{-1} * B_{2} \bmod{26} = \left [\begin{array}{ccc|c} 11 & 17 & 6\\ 13 & 17 & 3\\ 3 & 18 & 16\\ \ \end{array} \right] * \left [\begin{array}{ccc|c} 25 \\ 12 \\ 5 \\ \ \end{array} \right] \bmod{26} = \left [\begin{array}{ccc|c} 15 \\ 24 \\ 7 \\ \ \end{array} \right] \bmod{26}$ : $$

Repeating the same procedure for the remaining blocks we obtain:

$14 \: 11 \: 18$ $$

$3 \: 0 \: 24$ $$

$18 \: 4 \: 21$ $$

$4 \: 17 \: 24$ $$

$14 \: 13 \: 4$ $$

The plain text message is:

$HAPPYHOLIDAYSEVERYONE$

As a string of integers we obtain: $\boxed{701515247141183024184214172414134}$

$$

$$

$HAPPY \:\ HOLIDAYS \:\ EVERYONE \:\ !$