Explanation Given: The given matrix A is, A = [ -5 4 -3 10 -7 6 8 -6 5 ] The given code is, 145 − 105 92 264 − 188 160 23 − 16 15 129 − 84 78 − 9 8 − 5 159 − 118 100 219 − 152 133 370 − 265 255 − 105 84 − 63 Approach: A cryptogram is a message written according to a secret code. (The Greek word kryptos means “hidden.”) Matrix multiplication can be used to encode and decode messages. To begin, assign a number to each letter in the alphabet (with 0 assigned to a blank space), as listed below, 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 Then convert the message to number and partition the number into uncoded row matrices, each having n entities. The general 3 × 3 matrix is given as, A = [ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ] Then, the adjoint of the matrix A is given as, Finally A − 1 is calculated with the of the formula, A − 1 = 1 | A | ( a d j A ) ... ( 1 ) Calculation: Consider the given matrix A , A = [ -5 4 -3 10 -7 6 8 -6 5 ] The determinant of A is given as, | A | = | − 5 4 − 3 10 − 7 6 8 − 6 5 | = ( − 5 ) | − 7 6 − 6 5 | − ( 4 ) | 10 6 8 5 | + ( − 3 ) | 10 − 7 8 − 6 | = − 5 ( − 35 + 36 ) − 4 ( 50 − 48 ) − 3 ( − 60 + 56 ) = − 5 ( 1 ) − 4 ( 2 ) − 3 ( − 4 ) Further solve the above equation. | A | = − 5 − 8 + 12 = − 13 + 12 = − 1 Since, | A | ≠ 0 . So, A − 1 exists. The value of adjoint A is, a d j A = [ + | − 7 6 − 6 5 | − | 10 6 8 5 | + | 10 − 7 8 − 6 | − | 4 − 3 − 6 5 | + | − 5 − 3 8 5 | − | − 5 4 8 − 6 | + | 4 − 3 − 7 6 | − | − 5 − 3 10 6 | + | − 5 4 10 − 7 | ] T = [ 1 − 2 − 4 − 2 − 1 2 3 0 − 5 ] T = [ 1 − 2 3 − 2 − 1 0 − 4 2 − 5 ] So, the value of A − 1 is given as,

Bundle: Precalculus, Loose-leaf Version, 10th + WebAssign Printed Access Card for Larson's Precalculus, 10th Edition, Single-Term

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ISBN: 9781337605090

Author: Ron Larson

Publisher: Cengage Learning

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Chapter 8.CR, Problem 122CR

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The decoded cryptogram based on the inverse of the given matrix A=[-54-310-768-65].

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Chapter 8.CR, Problem 121CR

Chapter 8.CR, Problem 123CR

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Chapter 8 Solutions

Bundle: Precalculus, Loose-leaf Version, 10th + WebAssign Printed Access Card for Larson's Precalculus, 10th Edition, Single-Term

Ch. 8.1 - Prob. 1CP Ch. 8.1 - Prob. 2CP Ch. 8.1 - Prob. 3CP Ch. 8.1 - Prob. 4CP Ch. 8.1 - Prob. 5CP Ch. 8.1 - Prob. 6CP Ch. 8.1 - Prob. 7CP Ch. 8.1 - Prob. 8CP Ch. 8.1 - Prob. 9CP Ch. 8.1 - Prob. 1E

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The decoded cryptogram based on the inverse of the given matrix A = [ -5 4 -3 10 -7 6 8 -6 5 ] .

Solution Summary: The author explains how matrix multiplication can be used to encode and decode messages.

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The decoded cryptogram based on the inverse of the given matrix A=[-54-310-768-65].

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Chapter 8 Solutions