Advanced Matrices and Cryptography Question Help

Oct 2018
3
0
USA
Hello, I was stuck on the following question.

'Your task it to crack the following code and find the encrypted word.

To make your task easier, the following information about the encoding matrix is given:
  • Position 1,1 in the encoding matrix is an even number.
  • The decoding matrix only contains integers.
  • Position 1,1 in the encoding matrix is the negative of the number in position 1,1 in the decoding matrix.
  • Modulo arithmetic hasn't been used.

Your coded message received is below. Explain in detail how you cracked the code.

91, -38, 137, -59, 145, -59

Thank You!
 

Debsta

MHF Helper
Oct 2009
1,361
633
Brisbane
Do you know the dimension of the encoding and decoding matrices?
 
Oct 2018
3
0
USA
Yes. It is a 2x2 matrix for the encoding and decoding matrices.
 

Debsta

MHF Helper
Oct 2009
1,361
633
Brisbane
Hello, I was stuck on the following question.

'Your task it to crack the following code and find the encrypted word.

To make your task easier, the following information about the encoding matrix is given:
  • Position 1,1 in the encoding matrix is an even number.
  • The decoding matrix only contains integers.
  • Position 1,1 in the encoding matrix is the negative of the number in position 1,1 in the decoding matrix.
  • Modulo arithmetic hasn't been used.

Your coded message received is below. Explain in detail how you cracked the code.

91, -38, 137, -59, 145, -59

Thank You!
Can you please double check the info you have given us? In particular, are all the positions you've given us 1,1 or should one be position 2,2 ??
 
Oct 2018
3
0
USA
Can you please double check the info you have given us? In particular, are all the positions you've given us 1,1 or should one be position 2,2 ??
Yes, I double checked just now mate and that is exactly how it is worded in the paper. Any suggestions please! It would be greatly appreciated.
 

Debsta

MHF Helper
Oct 2009
1,361
633
Brisbane
So I assume the input numbers are from 1 to 26 where A=1, B=2 etc as is usual for these types of codes.


I let the encoding matrix (E) be:


\(\displaystyle \left( \begin{smallmatrix} a&b\\ c&d \end{smallmatrix} \right)\)


Assuming ad - bc =1 (for now - we can adjust that later by using a multiplier out the front of the matrix if we need to). This will also help with the condition that the decoding matrix has integers (if the encoding one does).


Because your output numbers (without using modulo arithmetic) are not too large and these are generated by using multiples of 1-26 added together
I started off setting a=2 (smallest even number).


So E = \(\displaystyle \left( \begin{smallmatrix} 2&b\\ c&d \end{smallmatrix} \right)\)


The decoding matrix D = \(\displaystyle \left( \begin{smallmatrix} d&-b\\ -c&a \end{smallmatrix} \right)\)


Now since "Position 1,1 in the encoding matrix is the negative of the number in position 1,1 in the decoding matrix", then d=-2


So E = \(\displaystyle \left( \begin{smallmatrix} 2&b\\ c&-2 \end{smallmatrix} \right)\)


For the det to be equal to 1 ie -4 - bc =1, bc=-5. Keeping to integers there are a limited number of combinations,

b=1 and c=-5, b=-1 and c=5, b=5 and c=-1, b=-5 and c=1


So D could equal =\(\displaystyle \left( \begin{smallmatrix} -2&1\\ -5&2 \end{smallmatrix} \right)\) or one of the other 3 combinations (or a multiple thereof)


If you multiply this matrix by (91, -38) {written as a column}you should expect output between 1 and 26, but you don't get that. Since modulo arithmetic is not used, then that wasn't the correct D.
So try another combination. Eventually one of them worked (YAY!!) and I successfully decoded the word. Have a go and let me know what you get.
 
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