Chapter 8.5, Problem 58E

To determine

Decode the cryptogram.

Answer to Problem 58E

  $201502501518014152002015025$

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

TO BE OR NOT TO BE

Explanation of Solution

Given:

  $A=\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right]$

  $8512068101584117425690125608030451926$

Let's decode the following cryptogram:

  $8512068101584117425690125608030451926$

Using $A=\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right]$

Dividing them into 1×2 matrices ,

  $\left[85120\right]\left[68\right]\left[1015\right]\left[84117\right]\left[4256\right]\left[6080\right]\left[3045\right]\left[1926\right].$

These have to be decoded using $A^{-1}$

If $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

Then,

  $\begin{array}{l}{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}\text{d}& -b\\ -c& a\end{array}\right]\hfill \\ A^{-1}=\frac{1}{(4\times 2)-(3\times 3)}\left[\begin{array}{cc}4& -3\\ -3& 2\end{array}\right]\hfill \\ A^{-1}=\frac{1}{8-9}\left[\begin{array}{cc}4& -3\\ -3& 2\end{array}\right]\hfill \\ A^{-1}=-1\left[\begin{array}{cc}4& -3\\ -3& 2\end{array}\right]\hfill \\ A^{-1}=\left[\begin{array}{cc}-4& 3\\ 3& -2\end{array}\right]\hfill \end{array}$ .

Now to get the unconded row matrices,

Coded matrix Decoding matrix Uncoded Matrix

  $\begin{array}{l}[85120]\left[\begin{array}{cc}-4& 3\\ 3& -2\end{array}\right][2015]\hfill \\ \left[\begin{array}{cc}6& 8\end{array}\right]\left[\begin{array}{cc}-4& 3\\ 3& -2\end{array}\right]\left[\begin{array}{ll}0\hfill & 2\hfill \end{array}\right]\hfill \end{array}$

  $$ .

  $\begin{array}{ll}\left[\begin{array}{l}1015\hfill \end{array}\right]\left[\begin{array}{ll}-4\hfill & 3\hfill \\ 3\hfill & -2\hfill \end{array}\right]\hfill & \left[\begin{array}{ll}5\hfill & 0\hfill \end{array}\right]\hfill \\ \left[\begin{array}{ll}84\hfill & 117\hfill \end{array}\right]\left[\begin{array}{cc}-4& 3\\ 3& -2\end{array}\right]\left[\begin{array}{l}1518\hfill \end{array}\right]\hfill & \hfill \\ \left[\begin{array}{ll}42\hfill & 56\hfill \end{array}\right]\left[\begin{array}{ll}-4\hfill & 3\hfill \\ 3\hfill & -2\hfill \end{array}\right]\hfill & \left[\begin{array}{ll}0\hfill & 14\hfill \end{array}\right]\hfill \end{array}$

  $\begin{array}{l}\left[\begin{array}{c}90125\end{array}\right]\left[\begin{array}{cc}-4& 3\\ 3& -2\end{array}\right]\left[\begin{array}{c}1520\end{array}\right]\hfill \\ \left[\begin{array}{cc}60& 80\end{array}\right]\left[\begin{array}{cc}-4& 3\\ 3& -2\end{array}\right]\left[\begin{array}{cc}0& 20\end{array}\right]\hfill \end{array}$

  $\begin{array}{ll}\left[\begin{array}{c}3045\end{array}\right]\left[\begin{array}{cc}-4& 3\\ 3& -2\end{array}\right]\left[\begin{array}{c}150\end{array}\right]\hfill & \hfill \\ \left[\begin{array}{c}1926\end{array}\right]\left[\begin{array}{cc}-4& 3\\ 3& -2\end{array}\right]\hfill & \left[\begin{array}{c}25\end{array}\right]\hfill \end{array}$

The following uncoded row matrices,

  $\left[2015\right]\left[02\right]\left[50\right]\left[1518\right]\left[014\right]\left[1520\right]\left[020\right]\left[150\right]\left[25\right]$

Splitting them,

  $201502501518014152002015025$

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

TO BE OR NOT TO BE