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

Question

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.

Answer 1

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

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."