Explanation Given information: . The cryptogram is 5 2 25 11 − 2 − 7 − 15 − 15 32 14 − 8 − 13 38 19 − 19 − 19 37 16 . The last word of the message is − SUE . Calculation: Assign a number to each letter in the alphabet (with 0 assigned to a blank space, but ignoring punctuation): 0 = − 9 = I 18 = R 1 = A 10 = J 19 = S 2 = B 11 = K 20 = T 3 = C 12 = L 21 = U 4 = D 13 = M 22 = V 5 = E 14 = N 23 = W 6 = F 15 = O 24 = X 7 = G 16 = P 25 = Y 8 = H 17 = Q 26 = Z The message is encoded with a 2 × 2 matix A . Convert the last word of message − RON into group of two letters and then into the decoded row matrix. Message [ − S ] [ U E ] Decoded row [ 0 19 ] [ 21 5 ] The cryptogram is 5 2 25 11 − 2 − 7 − 15 − 15 32 14 − 8 − 13 38 19 − 19 − 19 37 16 . Now, to decode the message, partition the cryptogram into groups of two to form the coded row matrices [ 5 2 ] [ 25 11 ] [ − 2 − 7 ] [ − 15 − 15 ] [ 32 14 ] [ − 8 − 13 ] [ 38 19 ] [ − 19 − 19 ] [ 37 16 ] . To find the A − 1 , multiply each coded row matrix by A − 1 on the right side and equate to the last two decoded rows, where A − 1 = [ a b c d ] . The decoded row matrix for the coded row matrix [ − 19 − 19 ] is [ 0 19 ] . The decoded row matrix for the coded row matrix [ 37 16 ] is [ 21 5 ] . Coded Row Decoding Decoded Matrix Matrix A -1 Row matrix [ − 19 − 19 ] [ a b c d ] = [ 0 19 ] ⇒ [ − 19 a − 19 c − 19 b − 19 d ] = [ 0 19 ] ⇒ − 19 a − 19 c = 0 and − 19 b − 19 d = 19 . ⇒ a + c = 0 and b + d = − 1 ⇒ a = − c and b = − 1 − d ...(1) [ 37 16 ] [ a b c d ] = [ 21 5 ] ⇒ [ 37 a + 16 c 37 b + 16 d ] = [ 21 5 ] ⇒ 37 a + 16 c = 21 and 37 b + 16 d = 5 ...(2) Substitute the value a = − c and b = − 1 − d in equation (2), ⇒ − 37 c + 16 c = 21 ⇒ − 21 c = 21 ⇒ c = − 1 And ⇒ 37 ( − 1 − d ) + 16 d = 5 ⇒ − 37 − 37 d + 16 d = 5 ⇒ − 37 − 21 d = 5 ⇒ − 21 d = 42 ⇒ d = − 2 Substitute the value of c = − 1 and d = − 2 in equation (1), Therefore, a = 1 and b = 1 . Thus, the A − 1 = [ a b c d ] = [ 1 1 − 1 − 2 ]

Bundle: Elementary Linear Algebra, Loose-leaf Version, 8th + MindTap Math, 1 term (6 months) Printed Access Card

8th Edition

ISBN: 9781337131216

Author: Ron Larson

Publisher: Cengage Learning

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Chapter 2.6, Problem 8E

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What is the message for the cryptogram 522511−2−7−15−153214−8−133819−19−193716 encoded with a 2×2 matix and the last word of the message is _SUE.

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Chapter 2.6, Problem 7E

Chapter 2.6, Problem 9E

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What is the message for the cryptogram 5 2 25 11 − 2 − 7 − 15 − 15 32 14 − 8 − 13 38 19 − 19 − 19 37 16 encoded with a 2 × 2 matix and the last word of the message is _ SUE .

Solution Summary: The author explains the cryptogram's message, which is encoded with a 2times 2 matix.

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What is the message for the cryptogram 522511−2−7−15−153214−8−133819−19−193716 encoded with a 2×2 matix and the last word of the message is _SUE.

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