Help My 11th Grader Decode Assignment

[kadylake]

My 11th grader was given this assignment. She's an honor student and has no idea how to figure it out. Can anyone help? View attachment 5019

 

[Sudharaka]

My 11th grader was given this assignment. She's an honor student and has no idea how to figure it out. Can anyone help?

Hi kadylake, :)

Welcome to MHB! We don't intentionally provide answers to assignment questions knowing that they constitute to a students final grade. However our helpers will point you in the correct direction if you could write down what you have done so far. :)

 

[kadylake]

Direction would be great! What she has written down is as far as she has gotten. She just isn't sure where to go from there or how to figure it out.

 

[MarkFL]

I would begin by writing:

\(\displaystyle \left[\begin{array}{c}3 & 15 \end{array}\right]\left[\begin{array}{c}a & b \\ c & d \end{array}\right]=\left[\begin{array}{c}-24 & -33 \end{array}\right]\)

\(\displaystyle \left[\begin{array}{c}14 & 4 \end{array}\right]\left[\begin{array}{c}a & b \\ c & d \end{array}\right]=\left[\begin{array}{c}20 & 44 \end{array}\right]\)

This results in the system:

\(\displaystyle 3a+15c=-24\)

\(\displaystyle 3b+15d=-33\)

\(\displaystyle 14a+4c=20\)

\(\displaystyle 14b+4d=44\)

Can your daughter now solve this system to determine the encoding matrix:

\(\displaystyle \left[\begin{array}{c}a & b \\ c & d \end{array}\right]\)

From there, she will need to use Cramer's Rule to find the inverse of the encoding matrix to get the decoding matrix.

 

[kadylake]

Thank you! I will pass this on to her tomorrow and see if this will help her. Much appreciated!

 

[MarkFL]

Thank you! I will pass this on to her tomorrow and see if this will help her. Much appreciated!

I originally made a mistake in what I posted above, but I have corrected it.

edit: After doing some calculations, I realize what I originally posted:

\(\displaystyle \left[\begin{array}{c}-24 & -33 \end{array}\right]\left[\begin{array}{c}a & b \\ c & d \end{array}\right]=\left[\begin{array}{c}3 & 15 \end{array}\right]\)

\(\displaystyle \left[\begin{array}{c}20 & 44 \end{array}\right]\left[\begin{array}{c}a & b \\ c & d \end{array}\right]=\left[\begin{array}{c}14 & 4 \end{array}\right]\)

which results in the system:

\(\displaystyle -24a-33c=3\)

\(\displaystyle -24b-33d=15\)

\(\displaystyle 20a+44c=14\)

\(\displaystyle 20b+44d=4\)

This will directly give you the decoding matrix, without having to compute an inverse. Much simpler. :)

 

[MarkFL]

I found two errors in the encoded message (in bold red)...it should be:

28 65 -12 -4 -26-34 -38 -57 6 24 -2 16 12 27 10 20 24 56 -20 -21 -50 -75 8 17 36 72 -20 -21
-30 -45 28 56 -2 1 -4 -3 22 46 -26 -34 -40 -60 36 77 -12 -4 50 100 28 61 -24 -33 20 44

So you can check your daughter's work:

Spoiler

The decoding matrix is:

\(\displaystyle \left[\begin{array}{c}-\frac{3}{2} & -2 \\ 1 & 1 \end{array}\right]\)

And the decoded message is:

WINTER SOLSTICE THIS YEAR IS ON DECEMBER TWENTY SECOND

 

[kadylake]

She got it! Winter Solstice this year is on December twenty second. Thanks for the help! Much appreciated.