Chapter 8.5, Problem 58E

To determine

The cryptogram decoding by using inverse of A.

Answer to Problem 58E

The message is $\text{RETURN}\text{AT}\text{DAWN}$

Explanation of Solution

Given information:

  $\begin{array}{l}13-9-5961112106-17-73\\ -13111242965144172\end{array}$

Formula used:

The elementary row operations is $\left[I⋮A^{-1}\right]$

Calculation:

Consider the encoding matrix:

  $A=\left[\begin{array}{ccc}1& 2& 2\\ 3& 7& 9\\ -1& -4& -7\end{array}\right]$

The inverse of this matrix is the decoding matrix.

To find the inverse of this matrix, begin by adjoining the identity matrix to this matrix as:

  $\left[A:I\right]=\left[\begin{array}{ccc}1& 2& 2\\ 3& 7& 9\\ -1& -4& -7\end{array}\begin{array}{c}.\\ .\\ .\end{array}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Use elementary row operations to obtain the form $\left[I⋮A^{-1}\right]$ as given below,

  $\begin{array}{l}\left[A⋮I\right]=\left[\begin{array}{ccc}1& 2& 2\\ 3& 7& 9\\ 0& -2& -5\end{array}\begin{array}{c}.\\ .\\ .\end{array}\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]R_3\to R_3+R_1\\ =\left[\begin{array}{ccc}1& 2& 2\\ 0& 1& 3\\ 0& -2& -5\end{array}\begin{array}{c}.\\ .\\ .\end{array}\begin{array}{ccc}1& 0& 0\\ -3& 1& 0\\ 1& 0& 1\end{array}\right]R_2\to R_2+3R_1\\ =\left[\begin{array}{ccc}1& 0& -3\\ 0& 1& 3\\ 0& -2& -5\end{array}\begin{array}{c}.\\ .\\ .\end{array}\begin{array}{ccc}2& 0& 1\\ -3& 1& 0\\ 1& 0& 1\end{array}\right]R_1\to R_1+R_3\end{array}$

  $\begin{array}{l}=\left[\begin{array}{ccc}1& 0& -3\\ 0& 1& 3\\ 0& 0& 1\end{array}\begin{array}{c}.\\ .\\ .\end{array}\begin{array}{ccc}2& 0& 1\\ -3& 1& 0\\ -5& 2& 1\end{array}\right]R_3\to R_3+2R_2\\ =\left[\begin{array}{ccc}1& 0& -3\\ 0& 1& 0\\ 0& 0& 0\end{array}\begin{array}{c}.\\ .\\ .\end{array}\begin{array}{ccc}2& 0& 1\\ 12& -5& -3\\ -5& 2& 1\end{array}\right]R_2\to R_2+3R_3\\ =\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\begin{array}{c}.\\ .\\ .\end{array}\begin{array}{ccc}-13& 6& 4\\ 12& -5& -3\\ -5& 2& 1\end{array}\right]R_1\to R_1+3R_3\end{array}$

  $\left[A⋮I\right]=\left[I⋮A^{-1}\right]$

So, the matrix A is invertible and its inverse is:

  $A^{-1}=\left[\begin{array}{ccc}-13& 6& 4\\ 12& -5& -3\\ -5& 2& 1\end{array}\right]$

Partition the coded messages into groups of three.

Thus, the coded row matrices will be,

  $\left[13-9-59\right]\left[61112106\right]\left[112429\right]\left[65144172\right]$

Multiply each coded row matrix by $A^{-1}$ on the right

Thus,

Coded Matrix Decoding Matrix Uncoded Matrix

  $\begin{array}{l}\left[\begin{array}{ccc}13& -9& -59\end{array}\right]\left[\begin{array}{ccc}-13& 6& 4\\ 12& -5& -3\\ -5& 2& 1\end{array}\right]=\left[\begin{array}{ccc}18& 5& 20\end{array}\right]\\ \left[\begin{array}{ccc}61& 112& 106\end{array}\right]\left[\begin{array}{ccc}-13& 6& 4\\ 12& -5& -3\\ -5& 2& 1\end{array}\right]=\left[\begin{array}{ccc}21& 18& 14\end{array}\right]\\ \left[\begin{array}{ccc}-17& -73& -131\end{array}\right]\left[\begin{array}{ccc}-13& 6& 4\\ 12& -5& -3\\ -5& 2& 1\end{array}\right]=\left[\begin{array}{ccc}0& 1& 20\end{array}\right]\end{array}$

  $\begin{array}{l}\left[\begin{array}{ccc}11& 24& 29\end{array}\right]\left[\begin{array}{ccc}-13& 6& 4\\ 12& -5& -3\\ -5& 2& 1\end{array}\right]=\left[\begin{array}{ccc}0& 4& 1\end{array}\right]\\ \left[\begin{array}{ccc}65& 144& 172\end{array}\right]\left[\begin{array}{ccc}-13& 6& 4\\ 12& -5& -3\\ -5& 2& 1\end{array}\right]=\left[\begin{array}{ccc}23& 14& 0\end{array}\right]\end{array}$

So, the uncoded row matrices are:

  $\left[\begin{array}{ccc}18& 5& 20\end{array}\right]\left[\begin{array}{ccc}21& 18& 14\end{array}\right]\left[\begin{array}{ccc}0& 1& 20\end{array}\right]\left[\begin{array}{ccc}0& 4& 1\end{array}\right]\left[\begin{array}{ccc}23& 14& 0\end{array}\right]$

Finally, removing the matrix notation:

  $\begin{array}{l}18520211814012004123140\\ \text{R}\text{E}\text{T}\text{U}\text{R}\text{N}\text{A}\text{T}\text{D}\text{A}\text{W}\text{N}\end{array}$

Thus, the message is:

  $\text{RETURN}\text{AT}\text{DAWN}$

Conclusion:

The message is $\text{RETURN}\text{AT}\text{DAWN}$