Decrypt the message “GVPJ” using Hill cipher taking the key as {Row1 = 3 7, Row2= 5 12}.

Question

Decrypt the message &#8220;GVPJ&#8221; using Hill cipher taking the key as  {Row1 = 3  7, Row2= 5  12}.

Suresh Chand · Accepted Answer

Solution: The key is 2 x 2 matrix and key = (begin{pmatrix}3 & 7 5 & 12end{pmatrix}) Step 1: Find the multiplicative inverse of the determinant of key matrix (begin{pmatrix}3 & 7 5 & 12end{pmatrix}) = (3 x 12 - 7 x 5) mod 26 = 1 mod 26 a = 1 Step 2: Find the adj. matrix of key matrix (begin{pmatrix}3 & 7 5 & 12end{pmatrix}) = (begin{pmatrix}12 & -7 -5 & 3end{pmatrix}) mod 26 = (begin{pmatrix}3 & 7 5 & 12end{pmatrix}) Step 3: Multiply the multiplicative inverse of the determinant by adj. matrix = 1 x (begin{pmatrix}12 & -7 -5 & 3end{pmatrix}) mode 26 = (begin{pmatrix}12 & 19 21 & 3end{pmatrix}) Now, cipher text = 'GVPJ' (begin{pmatrix}G Vend{pmatrix} = begin{pmatrix}6 21end{pmatrix}) and (begin{pmatrix}P Jend{pmatrix} = begin{pmatrix}15 9end{pmatrix}) Step 4: Multiply cipher text (c) by the inverse of a matrix = (begin{pmatrix}12 & 19 21 & 3end{pmatrix} begin{pmatrix}6 21end{pmatrix}) mode 26 = (begin{pmatrix}72 + 399 126 + 63end{pmatrix}) mod 26 =  (begin{pmatrix}471 189end{pmatrix}) mod 26 = (begin{pmatrix}3 7end{pmatrix}) = (begin{pmatrix}D Hend{pmatrix}) Similarly, = (begin{pmatrix}12 & 19 21 & 3end{pmatrix} begin{pmatrix}15 9end{pmatrix}) mode 26 = (begin{pmatrix}180 + 171 315 + 27end{pmatrix}) mod 26 =  (begin{pmatrix}351 342end{pmatrix}) mod 26 = (begin{pmatrix}13 4end{pmatrix}) = (begin{pmatrix}N Eend{pmatrix}) Therefore, Plain text = DHNE

Decrypt the message “GVPJ” using Hill cipher taking the key as {Row1 = 3 7, Row2= 5 12}.

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Decrypt the message “GVPJ” using Hill cipher taking the key as {Row1 = 3 7, Row2= 5 12}.

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