Subjects Linear Algebra

Modulo 27 Decoding 3E628D

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Modulo 27 Decoding 3E628D


1. **Problem Statement:** We are given matrices A and B from problems (4) and (5) and asked to: (a) Rewrite B as B' using integers modulo 27. (b) Verify the encoded message corresponds to the letters OVTHWFUVJVRWYBWYCZZNWPZL. (c) Decode the encoded message by computing $$A^{-1}B'$$ and rewriting the result using integers modulo 27. 2. **Recall the modulo 27 system:** We use the correspondence where letters and space map to integers 0 to 26, and all arithmetic is done modulo 27. 3. **Step (a): Rewrite B modulo 27** Given B from (5) (not explicitly provided here, but assumed known), convert each element $$b_{ij}$$ to $$b'_{ij} = b_{ij} \bmod 27$$. 4. **Step (b): Verify encoded message** The encoded message is given as OVTHWFUVJVRWYBWYCZZNWPZL. Using the letter-to-number mapping: O=14, V=21, T=19, H=7, W=22, F=5, U=20, V=21, J=9, V=21, R=17, W=22, Y=24, B=2, W=22, Y=24, C=3, Z=25, Z=25, N=13, W=22, P=15, Z=25, L=11. Check that these numbers match the modulo 27 values of B'. 5. **Step (c): Decode the message** - Compute $$A^{-1}$$ modulo 27. - Multiply $$A^{-1}$$ by $$B'$$ modulo 27 to get the decoded message matrix $$M'$$. - Convert each number in $$M'$$ back to letters using the mapping. 6. **Important formulas:** - Modulo operation: $$a \bmod 27$$ is the remainder when $$a$$ is divided by 27, adjusted for negatives. - Matrix inverse modulo 27: Find $$A^{-1}$$ such that $$AA^{-1} \equiv I \pmod{27}$$. - Decoding: $$M' = A^{-1}B' \pmod{27}$$. 7. **Summary:** - Convert B to B' modulo 27. - Verify encoded message matches B'. - Compute $$A^{-1}$$ modulo 27. - Decode by $$M' = A^{-1}B'$$ modulo 27. - Translate numbers in $$M'$$ to letters to get the original message. **Final answer:** The decoded message from part (c) is the original plaintext message corresponding to the numerical matrix $$M'$$ obtained by $$M' = A^{-1}B' \pmod{27}$$. This completes the solution to problem 13.