Explanation Calculation: Consider the matrix A = − 5 4 − 3 10 − 7 6 8 − 6 5 and the cryptogram is 145 − 105 92 264 − 188 160 23 − 16 15 129 − 84 78 − 9 8 − 5 159 − 118 100 219 − 152 133 370 − 265 225 − 105 84 − 63 . Use the elementary row operations to compute A − 1 , as A ⋮ I ⇒ I ⋮ A − 1 . Write the augmentation of the matrix A with the identity matrix of order 3 × 3 . A ⋮ I = − 5 4 − 3 10 − 7 6 8 − 6 5 ⋮ ⋮ ⋮ 1 0 0 0 1 0 0 0 1 Apply the row operations to reduce matrix A to I . Make the leading entry of column 1 to 1 , multiply row R 1 by − 1 5 . − 1 5 R 1 → 1 − 4 5 3 5 10 − 7 6 8 − 6 5 ⋮ ⋮ ⋮ − 1 5 0 0 0 1 0 0 0 1 Make the elements below the leading element of the first column 0 by multiplying row R 1 by − 10 and adding the result to R 2 , and multiplying row R 1 by − 8 and adding the result to R 3 . − 10 R 1 + R 2 → − 8 R 1 + R 3 → 1 − 4 5 3 5 0 1 0 0 2 5 1 5 ⋮ ⋮ ⋮ − 1 5 0 0 2 1 0 8 5 0 1 Make the elements below the leading element of the second column 0 by multiplying row R 2 by − 2 5 and adding the result to R 3 . − 2 5 R 2 + R 3 → 1 − 4 5 3 5 0 1 0 0 0 1 5 ⋮ ⋮ ⋮ − 1 5 0 0 2 1 0 4 5 − 2 5 1 Make the leading entry of column 3 to 1 , multiply row R 3 by 5 . 5 R 3 → 1 − 4 5 3 5 0 1 0 0 0 1 ⋮ ⋮ ⋮ − 1 5 0 0 2 1 0 4 − 2 5 Make the elements above the leading element of the third column 0 by multiplying row R 3 by − 3 5 and adding the result to R 1 . − 3 5 R 3 + R 1 → 1 − 4 5 0 0 1 0 0 0 1 ⋮ ⋮ ⋮ − 13 5 6 5 − 3 2 1 0 4 − 2 5 Make the elements above the leading element of the second column 0 by multiplying row R 2 by 4 5 and adding the result to R 1 . 4 5 R 2 + R 1 → 1 0 0 0 1 0 0 0 1 ⋮ ⋮ ⋮ − 1 2 − 3 2 1 0 4 − 2 5 The above matrix results I ⋮ A − 1 . Therefore, the inverse of the matrix A is A − 1 = − 1 2 − 3 2 1 0 4 − 2 5 . Now, partition the cryptogram into groups of three to form the coded row matrices. 145 − 105 92 264 − 188 160 23 − 16 15 129 − 84 78 − 9 8 − 5 159 − 118 100 219 − 152 133 370 − 265 225 − 105 84 − 63 Obtain the decoded matrices for the coded row matrices by multiplying the coded row matrices by the matrix A − 1 = − 1 2 − 3 2 1 0 4 − 2 5

10th Edition

ISBN: 8220103599528

Author: Larson

Publisher: CENGAGE L

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Chapter 7, Problem 120RE

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To calculate: The decoded message of the cryptogram 145 −105 92 264 −188 160 23 −16 15 129 −84 78 −9 8 −5 159 −118 100 219 −152 133 370 −265 225 −105 84 −63 using A−1, if the matrix is A=−54−310−768−65.

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Chapter 7, Problem 119RE

Chapter 7, Problem 121RE

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Chapter 7 Solutions

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The decoded message of the cryptogram 145 − 105 92 264 − 188 160 23 − 16 15 129 − 84 78 − 9 8 − 5 159 − 118 100 219 − 152 133 370 − 265 225 − 105 84 − 63 using A − 1 , if the matrix is A = − 5 4 − 3 10 − 7 6 8 − 6 5 .

Solution Summary: The author calculates the decoded message of the cryptogram using A-1.

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To calculate: The decoded message of the cryptogram 145 −105 92 264 −188 160 23 −16 15 129 −84 78 −9 8 −5 159 −118 100 219 −152 133 370 −265 225 −105 84 −63 using A−1, if the matrix is A=−54−310−768−65.

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Chapter 7 Solutions