Chapter 7, Problem 138CR

To determine

To find: The decoded message for a given coded message.

Answer to Problem 138CR

The required decoded message is- I WILL BE BACK.

Explanation of Solution

Given: The coded message is the following array of numbers-

  $32$ $-46$ $37$ $9$ $-48$ $15$ $3$ $-14$ $10$ $-1$ $-6$ $2$ $-8$ $-22$ $-3$

The encoding matrix is $A=\left[\begin{array}{ccc}1& 0& -1\\ -1& -2& 0\\ 1& -2& 2\end{array}\right]$ .

Concept Used: In cryptography, each letter starting from A to Z are assigned number 1-26 respectively and blank space is assigned to 0. As a result, any message can be partitioned and converted into an array of numbers from which groups of three numbers can be used to produce matrices of order $1\times 3$ which are called the uncoded matrices. The uncoded matrix is then multiplied to a given encoding matrix which gives the coded matrix. Finally all the numbers in the coded matrix can be written in an array forming a coded message.

To decode a message, the inverse of the encoding matrix is determined which is then multiplied to the coded matrices to obtain the uncoded matrices. Each number is reverted back to its letter to obtain the required decoded message.

The inverse of a matrix can be found using the formula-

  $A^{-1}=\frac{adj(A)}{\left|A\right|}$ , where $adj(A)$ is the matrix obtained by transposing the cofactors.

Calculations: The encoding matrix is-

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

The cofactors of this matrix are-

  $C_{11}=\left|\begin{array}{cc}-2& 0\\ -2& 2\end{array}\right|=-4$

  $C_{12}=-\left|\begin{array}{cc}-1& 0\\ 1& 2\end{array}\right|=2$

  $C_{13}=\left|\begin{array}{cc}-1& -2\\ 1& -2\end{array}\right|=4$

  $C_{21}=-\left|\begin{array}{cc}0& -1\\ -2& 2\end{array}\right|=2$

  $C_{22}=\left|\begin{array}{cc}1& -1\\ 1& 2\end{array}\right|=3$

  $C_{23}=-\left|\begin{array}{cc}1& 0\\ 1& -2\end{array}\right|=2$

  $C_{31}=\left|\begin{array}{cc}0& -1\\ -2& 0\end{array}\right|=-2$

  $C_{32}=-\left|\begin{array}{cc}1& -1\\ -1& 0\end{array}\right|=1$

  $C_{33}=\left|\begin{array}{cc}1& 0\\ -1& -2\end{array}\right|=-2$

Hence, the adjoint of the encoding matrix is-

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

Now,

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

  $=1\times C_{11}+0\times C_{12}+(-1)\times C_{13}$

  $=1\times (-4)+0+(-1)\times 4$

  $=-8$

Thus, the inverse of the encoding matrix is-

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

  $=\frac{-1}{8}\left[\begin{array}{ccc}-4& 2& -2\\ 2& 3& 1\\ 4& 2& -2\end{array}\right]$

The coded message is the following array of numbers-

  $32$ $-46$ $37$ $9$ $-48$ $15$ $3$ $-14$ $10$ $-1$ $-6$ $2$ $-8$ $-22$ $-3$

Partitioning it into groups of three gives the following coded matrices-

  $\left[\begin{array}{ccc}32& -46& 37\end{array}\right]$$\left[\begin{array}{ccc}9& -48& 15\end{array}\right]$$\left[\begin{array}{ccc}3& -14& 10\end{array}\right]$$\left[\begin{array}{ccc}-1& -6& 2\end{array}\right]$$\left[\begin{array}{ccc}-8& -22& -3\end{array}\right]$

The uncoded matrices are obtained by multiplying the coded matrices to the inverse of the encoding matrix as follows-

  $\frac{-1}{8}\left[\begin{array}{ccc}32& -46& 37\end{array}\right]\left[\begin{array}{ccc}-4& 2& -2\\ 2& 3& 1\\ 4& 2& -2\end{array}\right]=\frac{-1}{8}\left[\begin{array}{ccc}-72& 0& -184\end{array}\right]=\left[\begin{array}{ccc}9& 0& 23\end{array}\right]$

  $\frac{-1}{8}\left[\begin{array}{ccc}9& -48& 15\end{array}\right]\left[\begin{array}{ccc}-4& 2& -2\\ 2& 3& 1\\ 4& 2& -2\end{array}\right]=\frac{-1}{8}\left[\begin{array}{ccc}-72& -96& -96\end{array}\right]=\left[\begin{array}{ccc}9& 12& 12\end{array}\right]$

  $\frac{-1}{8}\left[\begin{array}{ccc}3& -14& 10\end{array}\right]\left[\begin{array}{ccc}-4& 2& -2\\ 2& 3& 1\\ 4& 2& -2\end{array}\right]=\frac{-1}{8}\left[\begin{array}{ccc}0& -16& -40\end{array}\right]=\left[\begin{array}{ccc}0& 2& 5\end{array}\right]$

  $\frac{-1}{8}\left[\begin{array}{ccc}-1& -6& 2\end{array}\right]\left[\begin{array}{ccc}-4& 2& -2\\ 2& 3& 1\\ 4& 2& -2\end{array}\right]=\frac{-1}{8}\left[\begin{array}{ccc}0& -16& -8\end{array}\right]=\left[\begin{array}{ccc}0& 2& 1\end{array}\right]$

  $\frac{-1}{8}\left[\begin{array}{ccc}-8& -22& -3\end{array}\right]\left[\begin{array}{ccc}-4& 2& -2\\ 2& 3& 1\\ 4& 2& -2\end{array}\right]=\frac{-1}{8}\left[\begin{array}{ccc}-24& -88& 0\end{array}\right]=\left[\begin{array}{ccc}3& 11& 0\end{array}\right]$

Assigning letters to the uncoded matrices gives the required message which is-

  $\left[\begin{array}{ccc}9& 0& 23\end{array}\right]$$\left[\begin{array}{ccc}9& 12& 12\end{array}\right]$$\left[\begin{array}{ccc}0& 2& 5\end{array}\right]$$\left[\begin{array}{ccc}0& 2& 1\end{array}\right]$$\left[\begin{array}{ccc}3& 11& 0\end{array}\right]$

I W I L L B E B A C K

Hence, the required uncoded message is I WILL BE BACK.