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Encrypt the message “GTU EXAMINATION” using the hill cipheralgorithm with the key matrix 5 174 15

Question

Encrypt the message GTU EXAMINATION using the hill cipher algorithm with the key matrix

5  17  4
1  5  15
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Solution

1. Break Down the Problem

We need to encrypt the message "GTU EXAMINATION" using the Hill cipher algorithm with the provided key matrix. The key matrix is given as:

K=(517415) K = \begin{pmatrix} 5 & 17 \\ 4 & 15 \end{pmatrix}

We will follow these steps:

  1. Convert the message into numerical form.
  2. Divide the message into vectors that match the dimensions of the key matrix.
  3. Multiply each vector by the key matrix and reduce modulo 26.
  4. Convert the numerical results back to letters.

2. Relevant Concepts

  1. Numerical Representation: The letters A-Z are represented by numbers 0-25 (e.g., A=0, B=1, ..., Z=25).
  2. Matrix Multiplication: The Hill cipher uses matrix multiplication for encryption.
  3. Modulo Operation: After multiplication, we need to reduce the results using modulo 26.

3. Analysis and Detail

  1. Convert Message to Numerical Form:

    • G = 6, T = 19, U = 20, E = 4, X = 23, A = 0, M = 12, I = 8, N = 13, A = 0, T = 19, I = 8, O = 14, N = 13
    • The numerical representation of "GTU EXAMINATION" is: [6,19,20,4,23,0,12,8,13,0,19,8,14,13] [6, 19, 20, 4, 23, 0, 12, 8, 13, 0, 19, 8, 14, 13]
  2. Divide into Vectors:

    • We'll create vectors of size 2 (since our key matrix is 2x2).
    • The vectors will be: [6,19],[20,4],[23,0],[12,8],[13,0],[19,8],[14,13] [6, 19], [20, 4], [23, 0], [12, 8], [13, 0], [19, 8], [14, 13]
  3. Matrix Multiplication:

    • Using the key matrix K K :

    • For each vector V V , calculate C=KVmod26 C = K \cdot V \mod 26 .

    • For vector [6,19] [6, 19] : C=K(619)=(517415)(619)=((56+1719)(46+1519))=(120351) C = K \cdot \begin{pmatrix} 6 \\ 19 \end{pmatrix} = \begin{pmatrix} 5 & 17 \\ 4 & 15 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 19 \end{pmatrix} = \begin{pmatrix} (5 \cdot 6 + 17 \cdot 19) \\ (4 \cdot 6 + 15 \cdot 19) \end{pmatrix} = \begin{pmatrix} 120 \\ 351 \end{pmatrix}

    • Reducing modulo 26: Cmod26=(120mod26351mod26)=(1415) C \mod 26 = \begin{pmatrix} 120 \mod 26 \\ 351 \mod 26 \end{pmatrix} = \begin{pmatrix} 14 \\ 15 \end{pmatrix}

    • Repeat this process for each vector:

      • For [20,4] [20, 4] : C=[2,16] C = [2, 16]
      • For [23,0] [23, 0] : C=[17,8] C = [17, 8]
      • For [12,8] [12, 8] : C=[8,29mod26=3] C = [8, 29 \mod 26 = 3]
      • For [13,0] [13, 0] : C=[65mod26=13,52mod26=0] C = [65 \mod 26 = 13, 52 \mod 26 = 0]
      • For [19,8] [19, 8] : C=[52mod26=0,96mod26=18] C = [52 \mod 26 = 0, 96 \mod 26 = 18]
      • For [14,13] [14, 13] : C=[65mod26=13,54mod26=2] C = [65 \mod 26 = 13, 54 \mod 26 = 2]
  4. Convert Numerical Results Back to Letters:

    • The resulting numerical arrays:
    • [14,15] [14, 15] -> OP
    • [2,16] [2, 16] -> CQ
    • [17,8] [17, 8] -> RI
    • [8,3] [8, 3] -> IH
    • [13,0] [13, 0] -> NA
    • [0,18] [0, 18] -> AS
    • [13,2] [13, 2] -> NC

4. Verify and Summarize

  • The encrypted message using the Hill cipher with the key matrix is made up of the letter combinations obtained above.

Final Answer

The encrypted message is "OP CQ RI IH NA AS NC".

This problem has been solved

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