Answer to Question #342204 in Combinatorics | Number Theory for Riyajawahar

Question #342204

You intercept the message "KVW?TA!KJB?FVR."(The blanks after? and R are part of the message,but the final is not.You know that a linear enciphering transformation is being used with a 30- letter alphabet, in which A-Z have numerical equivalents 0-25, blank=26, ?=27, !=28, .=29.You further know that the first six letters of the plain text are "C.I.A". Find the deciphering matrix A inverse and the full plain text message.


1
Expert's answer
2022-05-23T16:04:27-0400

Here the ciphertext is"\\begin{pmatrix}\n\n 10 & 22 & 26 & 0 & 10 & 1 & 5 & 17 \\\\\n\n 21 & 27 & 19 & 28 & 9 & 27 & 21 & 26\n\n\\end{pmatrix}"

and the first three columns of plaintext are "\\begin{pmatrix} 2 & 8 & 0 \\\\ 29 & 29 & 29 \\end{pmatrix}" In attempting to use

A-1 = PC-1 ,note that the matrix formed from the first two digraphs of C has determinant whose g.c.d. with 30 is 6. Using the 

1st and 3rd digraphs improves the situation: det "\\begin{pmatrix} 10 & 26 \\\\ 21 & 19 \\end{pmatrix} =4" , and 

g.e.d.(4,30) = 2. Use this matrix for C and work modulo 15 to find 

that A-1 = "\\begin{pmatrix} 2 & 2 \\\\ 8 & 4 \\end{pmatrix} + 15A"1 , where "A \\in M2(Z\/2Z)" Use the fact that

A-1 "\\begin{pmatrix} 10 & 22 & 26 \\\\ 21 & 27 & 19 \\end{pmatrix}=\\begin{pmatrix} 2 & 8 & 0 \\\\ 29 & 29 & 29 \\end{pmatrix}" and the fact that det(A-1 ) is odd to show that either

A-1 ="\\begin{pmatrix} 17& 2 \\\\ 8 & 19 \\end{pmatrix} or \\begin{pmatrix} 17 & 2 \\\\ 23 & 19 \\end{pmatrix} ." The first possibility gives the plaintext massage "C.I.A. WILLLHTLA;" the second possibility gives "C.I.A. WILL HELP.

ANSWER: "C.I.A. WILL HELP."

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