Chapter 7, Problem 198RE

To determine

To decode: the cryptogram by evaluating $A^{-1}$ .

Answer to Problem 198RE

The decoding massage is CAN YOU HEAR ME NOW.

Explanation of Solution

Given information:

The given cryptogram:

  $30-7305108037341640-738-383616-15823460$

The encoding matrix is:

  $A=\left[\begin{array}{ccc}1& 2& 0\\ -1& 1& 2\\ 2& -1& 2\end{array}\right]$

Formula used:

The un-coded message is obtained by multiplying the coded matrix and the decoding matrix.

The decoding matrix is obtained by the inverse of the encoding matrix.

Inverse of a matrix A is

  $A^{-1}=\frac{1}{\left|A\right|}\left(adjA\right)$

Calculation:

The given cryptogram:

  $30-7305108037341640-738-383616-15823460$

So, the coded matrix is:

  $\left[\begin{array}{ccc}30& -7& 30\end{array}\right]\left[\begin{array}{ccc}5& 10& 80\end{array}\right]\left[\begin{array}{ccc}37& 34& 16\end{array}\right]\left[\begin{array}{ccc}40& -7& 38\end{array}\right]\left[\begin{array}{ccc}-3& 8& 36\end{array}\right]\left[\begin{array}{ccc}16& -1& 58\end{array}\right]\left[\begin{array}{ccc}23& 46& 0\end{array}\right]$

The encoding matrix is:

  $A=\left[\begin{array}{ccc}1& 2& 0\\ -1& 1& 2\\ 2& -1& 2\end{array}\right]$

So, the decoding matrix is given by:

  $A^{-1}=\frac{1}{\left|A\right|}\left(adjA\right)$

Here,

  $\begin{array}{l}\left|A\right|=\left|\begin{array}{ccc}1& 2& 0\\ -1& 1& 2\\ 2& -1& 2\end{array}\right|\\ =\left(1\right)\left|\begin{array}{cc}1& 2\\ -1& 2\end{array}\right|+\left(-2\right)\left|\begin{array}{cc}-1& 2\\ 2& 2\end{array}\right|\\ =4+12\\ \left|A\right|=16\end{array}$

And,

  $adj\left(A\right)=\left[\begin{array}{ccc}4& -4& 4\\ 6& 2& -2\\ -1& 5& 3\end{array}\right]$

So,

  $\begin{array}{l}{A}^{-1}=\frac{1}{\left|A\right|}\left(adjA\right)\\ =\frac{1}{16}\left[\begin{array}{ccc}4& -4& 4\\ 6& 2& -2\\ -1& 5& 3\end{array}\right]\\ A^{-1}=\left[\begin{array}{ccc}\frac{1}{4}& -\frac{1}{4}& \frac{1}{4}\\ \frac{3}{8}& \frac{1}{8}& -\frac{1}{8}\\ -\frac{1}{16}& \frac{5}{16}& \frac{3}{16}\end{array}\right]\end{array}$

Now, the un-coded message is obtained by multiplying the coded matrix and the decoding matrix

And

  $\left[\begin{array}{ccc}23& 46& 0\end{array}\right]\left[\begin{array}{ccc}\frac{1}{4}& -\frac{1}{4}& \frac{1}{4}\\ \frac{3}{8}& \frac{1}{8}& -\frac{1}{8}\\ -\frac{1}{16}& \frac{5}{16}& \frac{3}{16}\end{array}\right]=\left[\begin{array}{ccc}23& 0& 0\end{array}\right]$

So, the un-coded matrices are:

  $\left[\begin{array}{ccc}3& 1& 4\end{array}\right]\left[\begin{array}{ccc}0& 25& 15\end{array}\right]\left[\begin{array}{ccc}21& 0& 8\end{array}\right]\left[\begin{array}{ccc}5& 1& 18\end{array}\right]\left[\begin{array}{ccc}0& 13& 5\end{array}\right]\left[\begin{array}{ccc}0& 14& 15\end{array}\right]\left[\begin{array}{ccc}23& 0& 0\end{array}\right]$

So, the message in Cryptography can be written as:

  $\begin{array}{ccc}3& 1& 4\end{array}\begin{array}{ccc}0& 25& 15\end{array}\begin{array}{ccc}21& 0& 8\end{array}\begin{array}{ccc}5& 1& 18\end{array}\begin{array}{ccc}0& 13& 5\end{array}\begin{array}{ccc}0& 14& 15\end{array}\begin{array}{ccc}23& 0& 0\end{array}$

Hence, the original message is CAN YOU HEAR ME NOW.