Break your message into groups of three letters, using a dash (-) to represent a space. Create a column matrix for the groups, using the correspondence:a -> 1b -> 2...z -> 26space -> 27Now create a 3x3 matrix (M) that has an inverse. Encode your message by multiplying by M, following example 7. This will give you a series of 3x1 column matrices.

[tex]\begin{gathered} M=\begin{pmatrix}1 & 1 & -3 \\ -2 & 1 & 0 \\ 0 & 0 & 3\end{pmatrix} \\ M^{-1}=\begin{pmatrix}\frac{1}{3} & -\frac{1}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ 0 & 0 & \frac{1}{3}\end{pmatrix} \end{gathered}[/tex]

Matrix M has an inverse M^-1

Now, we need to mutiply each column matrix by M

IRE = [9;18;5]

[tex]\begin{gathered} \begin{pmatrix}1 & 1 & -3 \\ -2 & 1 & 0 \\ 0 & 0 & 3\end{pmatrix}\times\begin{pmatrix}9 \\ 18 \\ 5\end{pmatrix} \\ =\begin{pmatrix}1\cdot\: 9+1\cdot\: 18+\mleft(-3\mright)\cdot\: 5 \\ \mleft(-2\mright)\cdot\: 9+1\cdot\: 18+0\cdot\: 5 \\ 0\cdot\: 9+0\cdot\: 18+3\cdot\: 5\end{pmatrix} \\ =\begin{pmatrix}12 \\ 0 \\ 15\end{pmatrix} \end{gathered}[/tex]

NE- = [14;5;27]

[tex]\begin{gathered} \begin{pmatrix}1 & 1 & -3 \\ -2 & 1 & 0 \\ 0 & 0 & 3\end{pmatrix}\begin{pmatrix}14 \\ 5 \\ 27\end{pmatrix} \\ =\begin{pmatrix}1\cdot\: 14+1\cdot\: 5+\mleft(-3\mright)\cdot\: 27 \\ \mleft(-2\mright)\cdot\: 14+1\cdot\: 5+0\cdot\: 27 \\ 0\cdot\: 14+0\cdot\: 5+3\cdot\: 27\end{pmatrix} \\ =\begin{pmatrix}-62 \\ -23 \\ 81\end{pmatrix} \end{gathered}[/tex]

HAS = [8;1;19]

[tex]\begin{gathered} \begin{pmatrix}1 & 1 & -3 \\ -2 & 1 & 0 \\ 0 & 0 & 3\end{pmatrix}\begin{pmatrix}8 \\ 1 \\ 19\end{pmatrix} \\ =\begin{pmatrix}1\cdot\: 8+1\cdot\: 1+\mleft(-3\mright)\cdot\: 19 \\ \mleft(-2\mright)\cdot\: 8+1\cdot\: 1+0\cdot\: 19 \\ 0\cdot\: 8+0\cdot\: 1+3\cdot\: 19\end{pmatrix} \\ =\begin{pmatrix}-48 \\ -15 \\ 57\end{pmatrix} \end{gathered}[/tex]

-A- = [27;1;27]

[tex]\begin{gathered} \begin{pmatrix}1 & 1 & -3 \\ -2 & 1 & 0 \\ 0 & 0 & 3\end{pmatrix}\begin{pmatrix}27 \\ 1 \\ 27\end{pmatrix} \\ =\begin{pmatrix}1\cdot\: 27+1\cdot\: 1+\mleft(-3\mright)\cdot\: 27 \\ \mleft(-2\mright)\cdot\: 27+1\cdot\: 1+0\cdot\: 27 \\ 0\cdot\: 27+0\cdot\: 1+3\cdot\: 27\end{pmatrix} \\ =\begin{pmatrix}-53 \\ -53 \\ 81\end{pmatrix} \end{gathered}[/tex]

PAR = [16;1;18]

[tex]\begin{gathered} \begin{pmatrix}1 & 1 & -3 \\ -2 & 1 & 0 \\ 0 & 0 & 3\end{pmatrix}\begin{pmatrix}16 \\ 1 \\ 18\end{pmatrix} \\ =\begin{pmatrix}1\cdot\: 16+1\cdot\: 1+\mleft(-3\mright)\cdot\: 18 \\ \mleft(-2\mright)\cdot\: 16+1\cdot\: 1+0\cdot\: 18 \\ 0\cdot\: 16+0\cdot\: 1+3\cdot\: 18\end{pmatrix} \\ =\begin{pmatrix}-37 \\ -31 \\ 54\end{pmatrix} \end{gathered}[/tex]

TY- = [20;25;27]

[tex]\begin{gathered} \begin{pmatrix}1 & 1 & -3 \\ -2 & 1 & 0 \\ 0 & 0 & 3\end{pmatrix}\begin{pmatrix}20 \\ 25 \\ 27\end{pmatrix} \\ =\begin{pmatrix}1\cdot\: 20+1\cdot\: 25+\mleft(-3\mright)\cdot\: 27 \\ \mleft(-2\mright)\cdot\: 20+1\cdot\: 25+0\cdot\: 27 \\ 0\cdot\: 20+0\cdot\: 25+3\cdot\: 27\end{pmatrix} \\ =\begin{pmatrix}-36 \\ -15 \\ 81\end{pmatrix} \end{gathered}[/tex]

Now let's write this codes as a string of numbers:

12 0 15 -62 -23 81 -48 -15 57 -53 -53 81 -37 -31 54 -36 -15 81