Chapter 8, Problem 17PS

a.

To determine

Find the value of the given variables using given data.

Answer to Problem 17PS

The value of $w=1,x=-2,y=1\text{and}z=-3$ .

Explanation of Solution

Given:

The given encoded message is

  $\begin{array}{l}45-3538-3018-1835-3081-60\\ 42-2875-552-222-2115-10\end{array}$

The inverse of the encoding matrix $A$ is

  $A^{-1}=\left[\begin{array}{cc}w& x\\ y& z\end{array}\right]$

Calculation:

  $\begin{array}{l}\left[\begin{array}{cc}45& -35\end{array}\right]A^{-1}=\left[\begin{array}{cc}10& 15\end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}45& -35\end{array}\right]\left[\begin{array}{cc}w& x\\ y& z\end{array}\right]=\left[\begin{array}{cc}10& 15\end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}45w-35y& 45x-35z\end{array}\right]=\left[\begin{array}{cc}10& 15\end{array}\right]\end{array}$

  $\begin{array}{l}\left[\begin{array}{cc}38& -30\end{array}\right]A^{-1}=\left[\begin{array}{cc}8& 14\end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}38& -30\end{array}\right]\left[\begin{array}{cc}w& x\\ y& z\end{array}\right]=\left[\begin{array}{cc}8& 14\end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}38w-30y& 38x-30z\end{array}\right]=\left[\begin{array}{cc}8& 14\end{array}\right]\end{array}$

From the above result.

  $45w-35y=10---(i)$

  $45x-35z=15---(ii)$

  $38w-30y=8---(iii)$

  $38x-30z=14---(iv)$

Subtract $(iii)\times 35$ from $(i)\times 30$

  $\begin{array}{l}\left[45w-35y=10\right]\times 30\\ \Rightarrow 1350w-1050y=300\\ \left[38w-30y=8\right]\times 35\\ \Rightarrow 1330w-1050y=280\\ \underset{\_}{\begin{array}{l}1350w-1050y=300\\ 1330w-1050y=280\\ -+-\end{array}}\\ 20w+0=20\\ \Rightarrow \frac{20}{20}w=\frac{20}{20}\\ \Rightarrow w=1\end{array}$

Substitute the value of $w=1$ in the equation $\left(i\right)$ .

  $\begin{array}{l}45w-35y=10\\ \Rightarrow 45\left(1\right)-35y=10\\ \Rightarrow 45-35y=10\\ \Rightarrow 45-35y-45=10-45\\ \Rightarrow -35y=-35\\ \Rightarrow y=1\end{array}$

Now,

Subtract $(iv)\times 35$ from $(ii)\times 30$

  $\begin{array}{l}\left[45x-35z=15\right]\times 30\\ \Rightarrow 1350x-1050z=450\\ \left[38x-30z=14\right]\times 35\\ \Rightarrow 1330x-1050x=490\\ \underset{\_}{\begin{array}{l}1350x-1050z=450\\ 1330x-1050z=490\\ -+-\end{array}}\\ 20x+0=-40\\ \Rightarrow \frac{20}{20}x=-\frac{40}{20}\\ \Rightarrow x=-2\end{array}$

Substitute the value of $x=-2$ in the equation $\left(i\right)$ .

  $\begin{array}{l}45x-35z=15\\ \Rightarrow 45\left(-2\right)-35z=15\\ \Rightarrow -90-35z=15\\ \Rightarrow -90-35z+90=15+90\\ \Rightarrow -35z=105\\ \Rightarrow z=-3\end{array}$

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

Hence the value of $w=1,x=-2,y=1\text{and}z=-3$ .

b.

To determine

Decode the message using given data.

Answer to Problem 17PS

The result is “ JOHN RETURN TO BASE”.

Explanation of Solution

Given:

The given encoded message is

  $\begin{array}{l}45-3538-3018-1835-3081-60\\ 42-2875-552-222-2115-10\end{array}$

The inverse of the encoding matrix $A$ is

  $A^{-1}=\left[\begin{array}{cc}w& x\\ y& z\end{array}\right]$

Calculation:

The inverse of the encoding matrix $A$ is

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

For decode the matrix, partition the code into groups of two and multiply by the inverse matrix.

  $\begin{array}{l}\left[\begin{array}{cc}18& -18\end{array}\right]\left[\begin{array}{cc}1& -2\\ 1& -3\end{array}\right]=\left[\begin{array}{cc}0& 18\end{array}\right]\\ \left[\begin{array}{cc}35& -30\end{array}\right]\left[\begin{array}{cc}1& -2\\ 1& -3\end{array}\right]=\left[\begin{array}{cc}5& 20\end{array}\right]\\ \left[\begin{array}{cc}81& -60\end{array}\right]\left[\begin{array}{cc}1& -2\\ 1& -3\end{array}\right]=\left[\begin{array}{cc}21& 18\end{array}\right]\\ \left[\begin{array}{cc}42& -28\end{array}\right]\left[\begin{array}{cc}1& -2\\ 1& -3\end{array}\right]=\left[\begin{array}{cc}14& 0\end{array}\right]\\ \left[\begin{array}{cc}75& -55\end{array}\right]\left[\begin{array}{cc}1& -2\\ 1& -3\end{array}\right]=\left[\begin{array}{cc}20& 15\end{array}\right]\end{array}$

  $\begin{array}{l}\left[\begin{array}{cc}2& -2\end{array}\right]\left[\begin{array}{cc}1& -2\\ 1& -3\end{array}\right]=\left[\begin{array}{cc}0& 2\end{array}\right]\\ \left[\begin{array}{cc}22& -21\end{array}\right]\left[\begin{array}{cc}1& -2\\ 1& -3\end{array}\right]=\left[\begin{array}{cc}1& 19\end{array}\right]\\ \left[\begin{array}{cc}15& -10\end{array}\right]\left[\begin{array}{cc}1& -2\\ 1& -3\end{array}\right]=\left[\begin{array}{cc}5& 0\end{array}\right]\end{array}$

Now, the decoded row matrices are

  $\left[\begin{array}{cc}0& 18\end{array}\right]$

  $\left[\begin{array}{cc}5& 20\end{array}\right]$

  $\left[\begin{array}{cc}21& 18\end{array}\right]$

  $\left[\begin{array}{cc}14& 0\end{array}\right]$

  $\left[\begin{array}{cc}20& 15\end{array}\right]$

  $\left[\begin{array}{cc}0& 2\end{array}\right]$

  $\left[\begin{array}{cc}1& 19\end{array}\right]$

  $\left[\begin{array}{cc}5& 0\end{array}\right]$

Use the assigned number for each letter.

Hence the result is “ JOHN RETURN TO BASE”.