Chapter 8, Problem 16PS

To determine

To calculate: Use the inverse of A to decode the cryptogram.

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

23 13 -34 31 -34 63 25 -17 61 24 14 -37 41 -17 -8 20 29 40 38 -56 116 13 -11 1 22 3 -6 41 -53 85 28 -32 16

Answer to Problem 16PS

The decoded cryptogram is REMEMBER SEPTEMBER THE ELEVENTH.

Explanation of Solution

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

First find the inverse of given matrix using augmented with a $3\times 3$ matrix and reducing it to row echelon form

Further

  $\begin{array}{l}{R}_3=\frac{3}{11}{R}_3\\ \left[\begin{array}{ccccccc}1& 0& -\frac{4}{3}& \mid & \frac{1}{3}& \frac{2}{3}& 0\\ 0& 1& -\frac{5}{3}& \mid & -\frac{1}{3}& \frac{1}{3}& 0\\ 0& 0& 1& \mid & -\frac{2}{11}& -\frac{1}{11}& \frac{3}{11}\end{array}\right]\\ R_1\to R_1+\frac{4}{3}{R}_3,R_2\to R_2+\frac{5}{3}{R}_3\\ \left[\begin{array}{ccccccc}1& 0& 0& \mid & \frac{1}{11}& \frac{6}{11}& \frac{4}{11}\\ 0& 1& 0& \mid & -\frac{7}{11}& \frac{2}{11}& \frac{5}{11}\\ 0& 0& 1& \mid & -\frac{2}{11}& -\frac{1}{11}& \frac{3}{11}\end{array}\right]\end{array}$

Hence $A^{-1}=\left[\begin{array}{ccc}\frac{1}{11}& \frac{6}{11}& \frac{4}{11}\\ -\frac{7}{11}& \frac{2}{11}& \frac{5}{11}\\ -\frac{2}{11}& -\frac{1}{11}& \frac{3}{11}\end{array}\right]$

Now divide the given code into groups of 3 in given order and multiply it by inverse matrix $A^{-1}$ :

The sequence is 0 18 5 13 5 13 2 5 18 0 19 5 16 20 5 13 2 5 18 0 20 8 5 0 5 12 5 22 5 14 20 8 0 comparing the decoded number with 0 as space and $1,2,3,\cdots ,26$ with $A,B,C,\cdots ,Z$ respectively, the decoded cryptogram i.e.

REMEMBER SEPTEMBER THE ELEVENTH