Chapter 8.5, Problem 59E

To determine

The cryptogram decoding using encoded $2\times 2$ matrix.

Answer to Problem 59E

The final message is $\text{MEET ME TONIGHT RON}$

Explanation of Solution

Given information:

  $\begin{array}{l}821-15-10-13-13510525\\ 519-162040-18-18116\end{array}$

Formula used:

Matrix multiplication is used.

Calculation:

To decode the following cryptogram:

8 21 -15 -10 -13 -13 5 10 5 25 5 19 -1 6 20 40 -18 -18 1 16

Using the some unknown decoding matrix $A^{-1}=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

Dividing them into $1\times 2$ matrices we get,

  $\left[\begin{array}{cc}8& 21\end{array}\right]\left[\begin{array}{cc}-15& -10\end{array}\right]\left[\begin{array}{cc}-13& -13\end{array}\right]\left[\begin{array}{cc}5& 10\end{array}\right]\left[\begin{array}{cc}5& 25\end{array}\right]\left[\begin{array}{cc}5& 19\end{array}\right]\left[\begin{array}{cc}-1& 6\end{array}\right]\left[\begin{array}{cc}20& 40\end{array}\right]\left[\begin{array}{cc}-18& -18\end{array}\right]\left[\begin{array}{cc}1& 16\end{array}\right]$

The last word of the message is _RON.

So,

  $\begin{array}{l}\left[\begin{array}{cc}-18& -18\end{array}\right]\times A^{-1}=\left[\begin{array}{cc}-18& 18\end{array}\right]\times \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\\ =\left[\begin{array}{cc}-18a-18c& -18b-18d\end{array}\right]\end{array}$

The first two letters of the last word are _ & R. So,

  $\begin{array}{l}-18a-18c=0\\ a=-c\end{array}$

And

  $\begin{array}{l}-18b-18d=18\\ b+d=-1\end{array}$

Also ,

  $\begin{array}{l}\left[\begin{array}{cc}1& 16\end{array}\right]\times A^{-1}=\left[\begin{array}{cc}1& 16\end{array}\right]\times \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\\ =\left[\begin{array}{cc}a+16c& b+16d\end{array}\right]\end{array}$

The last two letters of the last word are 0 & N. So,

  $a+16c=15\&b+16d=14$

Using $a=-c$ and $b+d=-1$ we get

  $\begin{array}{l}-c+16c=15\\ 15c=15\\ c=1\end{array}$

  $a=-1$

Thus, get

  $\begin{array}{l}-1-d+16d=14\\ 15d=15\\ d=1\end{array}$

  $b=-2$

Thus, get

  $A^{-1}=\left[\begin{array}{cc}-1& -2\\ 1& 1\end{array}\right]$

Now to get the uncoded row matrices,

Coded matrix Decoding matrix Uncoded Matrix

  $\begin{array}{l}\left[\begin{array}{cc}8& 21\end{array}\right]\left[\begin{array}{cc}-1& -2\\ 1& 1\end{array}\right]\left[\begin{array}{cc}13& 5\end{array}\right]\\ \left[\begin{array}{cc}-15& -10\end{array}\right]\left[\begin{array}{cc}-1& -2\\ 1& 1\end{array}\right]\left[\begin{array}{cc}5& 20\end{array}\right]\\ \left[\begin{array}{cc}-13& -13\end{array}\right]\left[\begin{array}{cc}-1& -2\\ 1& 1\end{array}\right]\left[\begin{array}{cc}0& 13\end{array}\right]\\ \left[\begin{array}{cc}5& 10\end{array}\right]\left[\begin{array}{cc}-1& -2\\ 1& 1\end{array}\right]\left[\begin{array}{cc}5& 0\end{array}\right]\end{array}$

  $\begin{array}{l}\left[\begin{array}{cc}5& 25\end{array}\right]\left[\begin{array}{cc}-1& -2\\ 1& 1\end{array}\right]\left[\begin{array}{cc}20& 15\end{array}\right]\\ \left[\begin{array}{cc}5& 19\end{array}\right]\left[\begin{array}{cc}-1& -2\\ 1& 1\end{array}\right]\left[\begin{array}{cc}14& 9\end{array}\right]\\ \left[\begin{array}{cc}-1& 6\end{array}\right]\left[\begin{array}{cc}-1& -2\\ 1& 1\end{array}\right]\left[\begin{array}{cc}7& 8\end{array}\right]\\ \left[\begin{array}{cc}20& 40\end{array}\right]\left[\begin{array}{cc}-1& -2\\ 1& 1\end{array}\right]\left[\begin{array}{cc}20& 0\end{array}\right]\end{array}$

We get the following uncoded row matrices,

  $\left[\begin{array}{cc}13& 5\end{array}\right]\left[\begin{array}{cc}5& 20\end{array}\right]\left[\begin{array}{cc}0& 13\end{array}\right]\left[\begin{array}{cc}5& 0\end{array}\right]\left[\begin{array}{cc}20& 15\end{array}\right]\left[\begin{array}{cc}14& 9\end{array}\right]\left[\begin{array}{cc}7& 8\end{array}\right]\left[\begin{array}{cc}20& 0\end{array}\right]$

Splitting them we get,

13 5 5 20 0 13 5 0 20 15 14 9 7 8 20 0

When the numbers are converted to the alphabets they are assigned to we get the message

  $\text{MEET ME TONIGHT}$

Using the last word of the message, the final message is

  $\text{MEET ME TONIGHT RON}$

Conclusion:

The final message is $\text{MEET ME TONIGHT RON}$