Chapter 8.5, Problem 59E

To determine

The decoding of cryptogram using ${\text{A}}^{\text{-1}}$

Answer to Problem 59E

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

  $\text{CLASS IS CANCELED}$

Explanation of Solution

Given information:

To decode the following cryptogram:

  $\begin{array}{ccc}9& -1& -9\end{array}\begin{array}{ccc}38& -19& -19\end{array}\begin{array}{ccc}28& -9& -19\begin{array}{ccc}-80& 25& 41\end{array}\end{array}\begin{array}{ccc}-64& 21& 31\end{array}\begin{array}{ccc}9& -5& -4\end{array}$

  $A=\left[\begin{array}{ccc}1& -1& 0\\ 1& 0& -1\\ -6& 2& 3\end{array}\right]$

Formula used:

Matrix multiplication is used.

Calculation:

Dividing them into $1\times 3$ matrices we get,

  $\left[\begin{array}{ccc}9& -1& -9\end{array}\right]\left[\begin{array}{ccc}38& -19& -19\end{array}\right]\left[\begin{array}{ccc}28& -9& -19\end{array}\right]\left[\begin{array}{ccc}-80& 25& 41\end{array}\right]\left[\begin{array}{ccc}-64& 21& 31\end{array}\right]\left[\begin{array}{ccc}9& -5& -4\end{array}\right]$

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

To find $A^{-1}$

  $\left[\begin{array}{ccc}1& -1& 0\\ 1& 0& -1:\\ -6& 2& 3\end{array}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Using $R_2\to R_2-R_1\text{and }{R}_3\to R_3-6R_1$

  $\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -1:\\ 0& 0& 1\end{array}\begin{array}{ccc}1& 0& 0\\ -1& 1& 0\\ -6& 0& 1\end{array}\right]$

Using $R_1\to R_1+R_2\text{and }{R}_3\to R_3-4R_2$

  $\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& -1:\\ 0& 0& 1\end{array}\begin{array}{ccc}0& 1& 0\\ -1& 1& 0\\ -2& -4& 1\end{array}\right]$

Using $R_1\to R_1+R_3\text{and }{R}_2\to R_2+R_3$

  $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0:\\ 0& 0& 1\end{array}\begin{array}{ccc}-2& -3& 1\\ -3& -3& 1\\ -2& -4& 1\end{array}\right]$

Thus,

  $A^{-1}=\left[\begin{array}{ccc}-2& -3& 1\\ -3& -3& 1\\ -2& -4& 1\end{array}\right]$

Now to get the uncoded row matrices,

Coded matrix Decoding matrix Uncoded Matrix

  $\begin{array}{l}\left[\begin{array}{ccc}9& -1& -9\end{array}\right]\left[\begin{array}{ccc}-2& -3& 1\\ -3& -3& 1\\ -2& -4& 1\end{array}\right]\left[\begin{array}{ccc}3& 12& 1\end{array}\right]\\ \left[\begin{array}{ccc}38& -19& -19\end{array}\right]\left[\begin{array}{ccc}-2& -3& 1\\ -3& -3& 1\\ -2& -4& 1\end{array}\right]\left[\begin{array}{ccc}19& 19& 0\end{array}\right]\\ \left[\begin{array}{ccc}28& -9& -19\end{array}\right]\left[\begin{array}{ccc}-2& -3& 1\\ -3& -3& 1\\ -2& -4& 1\end{array}\right]\left[\begin{array}{ccc}9& 19& 0\end{array}\right]\end{array}$

Continuing further we get

  $\begin{array}{l}\left[\begin{array}{ccc}-80& 25& 41\end{array}\right]\left[\begin{array}{ccc}-2& -3& 1\\ -3& -3& 1\\ -2& -4& 1\end{array}\right]\left[\begin{array}{ccc}3& 1& 14\end{array}\right]\\ \left[\begin{array}{ccc}-64& 21& 31\end{array}\right]\left[\begin{array}{ccc}-2& -3& 1\\ -3& -3& 1\\ -2& -4& 1\end{array}\right]\left[\begin{array}{ccc}3& 5& 12\end{array}\right]\\ \left[\begin{array}{ccc}9& -5& -4\end{array}\right]\left[\begin{array}{ccc}-2& -3& 1\\ -3& -3& 1\\ -2& -4& 1\end{array}\right]\left[\begin{array}{ccc}5& 4& 0\end{array}\right]\end{array}$

We get the following uncoded row matrices,

  $\left[\begin{array}{ccc}3& 12& 1\end{array}\right]\left[\begin{array}{ccc}19& 19& 0\end{array}\right]\left[\begin{array}{ccc}9& 19& 0\end{array}\right]\left[\begin{array}{ccc}3& 1& 14\end{array}\right]\left[\begin{array}{ccc}3& 5& 12\end{array}\right]\left[\begin{array}{ccc}5& 4& 0\end{array}\right]$

Splitting them we get,

  $\text{3}\text{12}\text{1}\text{19}\text{19}\text{0}\text{9}\text{19}\text{0}\text{3}\text{1}\text{14}\text{3}\text{5}\text{12}\text{5}\text{4}\text{0}$

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

  $\text{CLASS IS CANCELED}$

Conclusion:

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

  $\text{CLASS IS CANCELED}$