Chapter 8.5, Problem 60E

To determine

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

Answer to Problem 60E

The message is

  $\text{CANCEL ORDERS SUE}$

Explanation of Solution

Given information:

  $\begin{array}{l}522511-2-7-15-153214\\ -8-133819-193716\end{array}$

Formula used:

Matrix multiplication is used.

Calculation:

We have to decode the following cryptogram:

5 2 25 11 -2 -7 -15 -15 32 14 -8 -13 38 19 -19 -19 37 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,

  $\left[\begin{array}{cc}5& 2\end{array}\right]\left[\begin{array}{cc}25& 11\end{array}\right]\left[\begin{array}{cc}-2& -7\end{array}\right]\left[\begin{array}{cc}-15& -15\end{array}\right]\left[\begin{array}{cc}32& 14\end{array}\right]\left[\begin{array}{cc}-8& -13\end{array}\right]\left[\begin{array}{cc}38& 19\end{array}\right]\left[\begin{array}{cc}-19& -19\end{array}\right]\left[\begin{array}{cc}37& 16\end{array}\right]$

The last word of the message is _SUE.

So,

  $\begin{array}{l}\left[\begin{array}{cc}-19& -19\end{array}\right]\times A^{-1}=\left[\begin{array}{cc}-19& -19\end{array}\right]\times \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\\ =\left[\begin{array}{cc}-19a-19c& -19b-19d\end{array}\right]\end{array}$

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

  $\begin{array}{l}-19a-19c=0\\ a=-c\\ \text{And}\\ -19b-19d=19\\ b+d=-1\end{array}$

Also we have

  $\begin{array}{l}\left[\begin{array}{cc}37& 16\end{array}\right]\times A^{-1}=\left[\begin{array}{cc}37& 16\end{array}\right]\times \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\\ =\left[\begin{array}{cc}37a+16c& 37b+16d\end{array}\right]\end{array}$

The last two letters of the last word are U &. E. So,

Using $a=-c\text{and}b+d=-1$ we get,

  $\begin{array}{l}37a-16a=21\\ 21a=21\\ a=1\end{array}$

Thus,

  $c=-1$

And we get

  $\begin{array}{l}37\left(-1-d\right)+16d=5\\ -37-37d+16d=5\\ 21d=-42\\ d=-2\end{array}$

Thus, $b=1$

Thus, we have

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

Now to get the uncoded row matrices,

Coded matrix Decoding matrix Uncoded Matrix

  $\begin{array}{l}\left[\begin{array}{cc}5& 2\end{array}\right]\left[\begin{array}{cc}1& 1\\ -1& -2\end{array}\right]\left[\begin{array}{cc}3& 1\end{array}\right]\\ \left[\begin{array}{cc}25& 11\end{array}\right]\left[\begin{array}{cc}1& 1\\ -1& -2\end{array}\right]\left[\begin{array}{cc}14& 3\end{array}\right]\\ \left[\begin{array}{cc}-2& -7\end{array}\right]\left[\begin{array}{cc}1& 1\\ -1& -2\end{array}\right]\left[\begin{array}{cc}5& 12\end{array}\right]\\ \left[\begin{array}{cc}-15& -15\end{array}\right]\left[\begin{array}{cc}1& 1\\ -1& -2\end{array}\right]\left[\begin{array}{cc}0& 15\end{array}\right]\\ \left[\begin{array}{cc}32& 14\end{array}\right]\left[\begin{array}{cc}1& 1\\ -1& -2\end{array}\right]\left[\begin{array}{cc}18& 4\end{array}\right]\\ \left[\begin{array}{cc}-8& -13\end{array}\right]\left[\begin{array}{cc}1& 1\\ -1& -2\end{array}\right]\left[\begin{array}{cc}5& 18\end{array}\right]\\ \left[\begin{array}{cc}38& 19\end{array}\right]\left[\begin{array}{cc}1& 1\\ -1& -2\end{array}\right]\left[\begin{array}{cc}19& 0\end{array}\right]\end{array}$

  $\begin{array}{l}\left[\begin{array}{cc}-19& -19\end{array}\right]\left[\begin{array}{cc}1& 1\\ -1& -2\end{array}\right]\left[\begin{array}{cc}0& 19\end{array}\right]\\ \left[\begin{array}{cc}37& 16\end{array}\right]\left[\begin{array}{cc}1& 1\\ -1& -2\end{array}\right]\left[\begin{array}{cc}21& 5\end{array}\right]\end{array}$

We get the following uncoded row matrices,

  $\left[\begin{array}{cc}3& 1\end{array}\right]\left[\begin{array}{cc}14& 3\end{array}\right]\left[\begin{array}{cc}5& 12\end{array}\right]\left[\begin{array}{cc}0& 15\end{array}\right]\left[\begin{array}{cc}18& 4\end{array}\right]\left[\begin{array}{cc}5& 18\end{array}\right]\left[\begin{array}{cc}19& 0\end{array}\right]\left[\begin{array}{cc}0& 19\end{array}\right]\left[\begin{array}{cc}21& 5\end{array}\right]$

Splitting them we get,

3 1 14 3 5 12 0 15 18 4 5 18 19 0 0 19 21 5

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

  $\text{CANCEL ORDERS SUE}$

Conclusion:

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

  $\text{CANCEL ORDERS SUE}$