Deciphering using Affine Ciphers and Hill ciphers.

Outline to decipher: $$

Convert $XKYALG$ to two $3 \: X \: 1$ blocks: $$

$B_{1} = \left[ \begin{array}{ccc} 23 \\ 10 \\ 24 \\ \ \end{array} \right] \mod{26}$ and $B_{2} = \left[ \begin{array}{ccc} 0 \\ 11 \\ 6 \\ \ \end{array} \right] \mod{26}.$ $$

Using $A * X{j} \equiv B_{j} \mod 26$ we solve obtaining $X_{j} \equiv A^{-1} * B_{j},$ where $(1 <= j <= 2)$ and $X_{1} = \left[ \begin{array}{ccc} x_{11} \\ x_{12} \\ x_{13} \\ \ \end{array} \right] \mod{26}$ and $X_{2} = \left[ \begin{array}{ccc} x_{21} \\ x_{22} \\ x_{23} \\ \ \end{array} \right] \mod{26}$ , and list the numbers : $$

$x_{11} \: x_{12} \: x_{13} \: x_{21} \: x_{22} \: x_{23}$ .

Now using the above numbers we decipher the list above using $P_{jk} \equiv \bar{a} * (x_{jk} - b) \mod{26},$ where $(1 <= j <= 2)$ and $(1 <= k <= 3),$ and $\bar{a}$ is the inverse of $a$ modulo $26.$ Last convert each number $P_{jk}$ to it's corresponding letter $L_{jk}$ .

$$

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

$B = A|I = \left[ \begin{array}{ccc|ccc} 16 & 1 & 3 & 1 & 0 & 0 \\ 5 & 1 & 10 & 0 & 1 & 0 \\ 23 & 1 & 7 & 0 & 0 & 1 \\ \ \end{array} \right] \mod{26}$
$Row_{1} \leftrightarrow Row_{2} \implies$ $$

$B = \left[ \begin{array}{ccc|ccc} 5 & 1 & 10 & 0 & 1 & 0 \\ 16 & 1 & 3 & 1 & 0 & 0 \\ 23 & 1 & 7 & 0 & 0 & 1 \\ \ \end{array} \right] \mod{26}$
$21 * Row_{1} \implies$ $$

$B = \left[ \begin{array}{ccc|ccc} 1 & 21 & 2 & 0 & 21 & 0 \\ 16 & 1 & 3 & 1 & 0 & 0 \\ 23 & 1 & 7 & 0 & 0 & 1 \\ \ \end{array} \right] \mod{26}$
$10 * Row_{1} + Row_{2} \implies$ $$

$B = \left[ \begin{array}{ccc|ccc} 1 & 21 & 2 & 0 & 21 & 0 \\ 0 & 3 & 23 & 1 & 2 & 0 \\ 23 & 1 & 7 & 0 & 0 & 1 \\ \ \end{array} \right] \mod{26}$
$9 * Row_{2} \implies$ $$

$B = \left[ \begin{array}{ccc|ccc} 1 & 21 & 2 & 0 & 21 & 0 \\ 0 & 1 & 25 & 9 & 18 & 0 \\ 23 & 1 & 7 & 0 & 0 & 1 \\ \ \end{array} \right] \mod{26}$
$3 * Row_{1} + Row_{3} \implies$ $$

$B = \left[ \begin{array}{ccc|ccc} 1 & 21 & 2 & 0 & 21 & 0 \\ 0 & 1 & 25 & 9 & 18 & 0 \\ 0 & 12 & 13 & 0 & 11 & 1 \\ \ \end{array} \right] \mod{26}$
$14 * Row_{2} + Row_{3} \implies$ $$

$B = \left[ \begin{array}{ccc|ccc} 1 & 21 & 2 & 0 & 21 & 0 \\ 0 & 1 & 25 & 9 & 18 & 0 \\ 0 & 0 & 25 & 22 & 3 & 1 \\ \ \end{array} \right] \mod{26}$
$5 * Row_{2} + Row_{1} \implies$ $$

$B = \left[ \begin{array}{ccc|ccc} 1 & 0 & 23 & 19 & 7 & 0 \\ 0 & 1 & 25 & 9 & 18 & 0 \\ 0 & 0 & 25 & 22 & 3 & 1 \\ \ \end{array} \right] \mod{26}$
$25 * Row_{3} \implies$ $$

$B = \left[ \begin{array}{ccc|ccc} 1 & 0 & 23 & 19 & 7 & 0 \\ 0 & 1 & 25 & 9 & 18 & 0 \\ 0 & 0 & 1 & 4 & 23 & 25 \\ \ \end{array} \right] \mod{26}$
$Row_{3} + Row_{2} \implies$ $$

$B = \left[ \begin{array}{ccc|ccc} 1 & 0 & 23 & 19 & 7 & 0 \\ 0 & 1 & 0 & 13 & 15 & 25 \\ 0 & 0 & 1 & 4 & 23 & 25 \\ \ \end{array} \right] \mod{26}$
$3 * Row_{3} + Row_{1} \implies$ $$

$B = \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 5 & 24 & 23 \\ 0 & 1 & 0 & 13 & 15 & 25 \\ 0 & 0 & 1 & 4 & 23 & 25 \\ \ \end{array} \right] \mod{26}$
$\implies$ $$

$A^{-1} = \left[ \begin{array}{ccc} 5 & 24 & 23 \\ 13 & 15 & 25 \\ 4 & 23 & 25 \\ \ \end{array} \right] \mod{26}$

Using $B_{1} = \left[ \begin{array}{ccc} 23 \\ 10 \\ 24 \\ \ \end{array} \right] \mod{26}$ and $B_{2} = \left[ \begin{array}{ccc} 0 \\ 11 \\ 6 \\ \ \end{array} \right] \mod{26}$ $$ $\implies$

$X \equiv \left[ \begin{array}{ccc} 5 & 24 & 23 \\ 13 & 15 & 25 \\ 4 & 23 & 25 \\ \ \end{array} \right] * \left[ \begin{array}{ccc} 23 & 0 \\ 10 & 11 \\ 24 & 6 \\ \ \end{array} \right] \equiv \left[ \begin{array}{ccc} 23 & 12 \\ 9 & 3 \\ 12 & 13 \\ \ \end{array} \right] \mod{26}$ $$

Our list of numbers to decipher are: $$

$23 \: 9 \: 12 \: 12 \: 3 \: 13$ $$

$25 * P + 16 \equiv C \mod 26 \implies P \equiv 25 * (C - 16) \mod{26}$ $$

$C_{1} = 23 \implies P_{1} \equiv 175 \mod{26} \equiv \boxed{19 \mod{26}}$ $$

$C_{2} = 9 \implies P_{2} \equiv -175 \mod{26} \equiv -19 \mod{26} \equiv \boxed{7 \mod{26}}$ $$

$C_{3} = 12 \implies P_{3} \equiv -100 \mod{26} \equiv -22 \mod{26} \equiv \boxed{4 \mod{26}}$ $$

$C_{4} = 12 \implies P_{4} \equiv \boxed{4 \mod{26}}$ $$

$C_{5} = 3 \implies P_{5} \equiv -325 \mod{26} \equiv -13 \mod{26} \equiv \boxed{13 \mod{26}}$ $$

$C_{6} \implies P_{6} \equiv -75 \mod{26} \equiv -23 \mod{26} \equiv \boxed{3 \mod{26}}$

$$

$\therefore$ The message reads: $THE\: END$ . $$

This is an appropriate message since this is my last problem on ciphers. $$

As a string of integers we have: $\boxed{19744133}$ .