Q. 39 - Algebra 2

Question

For Exercises 39-41, use the alphabet listed below.

Your friend has sent you a series of messages that were coded with the coding matrix $\mathbf{C}\mathbf{=}\left[\begin{array}{cc}\mathbf{2}& \mathbf{1}\\ \mathbf{1}& \mathbf{1}\\ & \end{array}\right]$. Use the inverse of matrix $\mathbf{C}$ to decode each message.

$\mathbf{50}\mathbf{\left|}\mathbf{36}\mathbf{\right|}\mathbf{51}\mathbf{\left|}\mathbf{29}\mathbf{\right|}\mathbf{18}\mathbf{\left|}\mathbf{18}\mathbf{\right|}\mathbf{26}\mathbf{\left|}\mathbf{13}\mathbf{\right|}\mathbf{33}\mathbf{\left|}\mathbf{26}\mathbf{\right|}\mathbf{44}\mathbf{\left|}\mathbf{22}\mathbf{\right|}\mathbf{48}\mathbf{\left|}\mathbf{33}\mathbf{\right|}\mathbf{59}\mathbf{\left|}\mathbf{34}\mathbf{\right|}\mathbf{61}\mathbf{\left|}\mathbf{35}\mathbf{\right|}\mathbf{4}\mathbf{|}\mathbf{2}$

Step-by-Step Solution

Answer

The decoded message is: $\mathbf{NVVG}\mathbf{\_}\mathbf{RM}\mathbf{\_}\mathbf{GSV}\mathbf{\_}\mathbf{ORYIZIB}\mathbf{\_}$

Step 1. Find the inverse of the coding matrix C.

It is being given that the coding matrix is $C=\left[\begin{array}{cc}2& 1\\ 1& 1\\ & \end{array}\right]$.

The determinant of the coding matrix C is:

$\begin{array}{c}\left|\begin{array}{cc}2& 1\\ 1& 1\end{array}\right|=\left(2\right)\left(1\right)-\left(1\right)\left(1\right)\\ =2-1\\ =1\end{array}$

As the determinant of the coding matrix C  is not equal to zero, therefore the inverse of the matrix C  exist.

The inverse of $2\times 2$ matrix $A=\left[\begin{array}{cc}a& b\\ c& d\\ & \end{array}\right]$ is given by $A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\\ & \end{array}\right]$ and $ad-bc\ne 0$.

Therefore, the inverse of the coding matrix C  is given by:

$\begin{array}{l}\frac{1}{\left(2\right)\left(1\right)-\left(1\right)\left(1\right)}\left[\begin{array}{cc}1& -\left(1\right)\\ -\left(1\right)& 2\end{array}\right]=\frac{1}{2-1}\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]\\ =\frac{1}{1}\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]\\ =\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]\end{array}$

Therefore, the inverse of the coding matrix C is $C^{-1}=\left[\begin{array}{cc}1& -1\\ -1& 2\\ & \end{array}\right]$

Step 2. Description of step.

Consider the coded matrix be matrix D.

The coded matrix D is given by:

$D=\left[\begin{array}{cc}50& 36\\ 51& 29\\ 18& 18\\ 26& 13\\ 33& 26\\ 44& 22\\ 48& 33\\ 59& 34\\ 61& 35\\ 4& 2\end{array}\right]$

Step 3. Description of step.

Multiply the coded matrix $D$ with matrix $C^{-1}$ to decode the message.

$\begin{array}{l}D{C}^{-1}=\left[\begin{array}{cc}50& 36\\ 51& 29\\ 18& 18\\ 26& 13\\ 33& 26\\ 44& 22\\ 48& 33\\ 59& 34\\ 61& 35\\ 4& 2\end{array}\right]\cdot \left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]\\ =\left[\begin{array}{cc}\left(50\right)\left(1\right)+\left(36\right)\left(-1\right)& \left(50\right)\left(-1\right)+\left(36\right)\left(2\right)\\ \left(51\right)\left(1\right)+\left(29\right)\left(-1\right)& \left(51\right)\left(-1\right)+\left(29\right)\left(2\right)\\ \left(18\right)\left(1\right)+\left(18\right)\left(-1\right)& \left(18\right)\left(-1\right)+\left(18\right)\left(2\right)\\ \left(26\right)\left(1\right)+\left(13\right)\left(-1\right)& \left(26\right)\left(-1\right)+\left(13\right)\left(2\right)\\ \left(33\right)\left(1\right)+\left(26\right)\left(-1\right)& \left(33\right)\left(-1\right)+\left(26\right)\left(2\right)\\ \left(44\right)\left(1\right)+\left(22\right)\left(-1\right)& \left(44\right)\left(-1\right)+\left(22\right)\left(2\right)\\ \left(48\right)\left(1\right)+\left(33\right)\left(-1\right)& \left(48\right)\left(-1\right)+\left(33\right)\left(2\right)\\ \left(59\right)\left(1\right)+\left(34\right)\left(-1\right)& \left(59\right)\left(-1\right)+\left(34\right)\left(2\right)\\ \left(61\right)\left(1\right)+\left(35\right)\left(-1\right)& \left(61\right)\left(-1\right)+\left(35\right)\left(2\right)\\ \left(4\right)\left(1\right)+\left(2\right)\left(-1\right)& \left(4\right)\left(-1\right)+\left(2\right)\left(2\right)\end{array}\right]\\ =\left[\begin{array}{cc}50-36& -50+72\\ 51-29& -51+58\\ 18-18& -18+36\\ 26-13& -26+26\\ 33-26& -33+52\\ 44-22& -44+44\\ 48-33& -48+66\\ 59-34& -59+68\\ 61-35& -61+70\\ 4-2& -4+4\end{array}\right]\\ =\left[\begin{array}{cc}14& 22\\ 22& 7\\ 0& 18\\ 13& 0\\ 7& 19\\ 22& 0\\ 15& 18\\ 25& 9\\ 26& 9\\ 2& 0\end{array}\right]\end{array}$

Step 4. Description of step.

Now covert the numbers to letters by using the codes listed in the given table.

Therefore, it can be obtained that:

$14\left|22\right|22\left|7\right|0\left|18\right|13\left|0\right|7\left|19\right|22\left|0\right|15\left|18\right|25\left|9\right|26\left|9\right|2|0\cong NVVG\_RM\_GSV\_ORYIZIB\_$

Step 5. Description of step.

Therefore, the decoded message is $NVVG\_RM\_GSV\_ORYIZIB\_$