Explanation Given: The provided matrices are A = [ − 3 − 9 2 6 ] , B = [ 4 6 2 3 ] . Proof: Consider; A + B = [ − 3 − 9 2 6 ] + [ 4 6 2 3 ] = [ − 3 + 4 − 9 + 6 2 + 2 6 + 3 ] = [ 1 − 3 4 9 ] Now, evaluate the value of A − B . A − B = [ − 3 − 9 2 6 ] − [ 4 6 2 3 ] = [ − 3 − 4 − 9 − 6 2 − 2 6 − 3 ] = [ − 7 − 15 0 3 ] Now, evaluate the value of ( A + B ) ( A − B ) . ( A + B ) ( A − B ) = [ 1 − 3 4 9 ] [ − 7 − 15 0 3 ] = [ ( 1 ) ( − 7 ) + ( − 3 ) ( 0 ) ( 1 ) ( − 15 ) + ( − 3 ) ( 3 ) ( 4 ) ( − 7 ) + ( 9 ) ( 0 ) ( 4 ) ( − 15 ) + ( 9 ) ( 3 ) ] = [ − 7 + 0 − 15 − 9 − 28 + 0 − 60 + 27 ] Therefore, ( A + B ) ( A − B ) = [ − 7 − 24 − 28 − 33 ] …...…..

MyLab Math with Pearson eText -- Standalone Access Card -- for Mathematics with Applications in the Management, Natural, and Social Sciences (12th Edition)

12th Edition

ISBN: 9780134856551

Author: Margaret L. Lial, Thomas W. Hungerford, John P. Holcomb, Bernadette Mullins

Publisher: PEARSON

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Chapter 6.5, Problem 21E

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To prove: The matrices (A+B)(A−B)≠A2−B2.

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Chapter 6.5, Problem 20E

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MyLab Math with Pearson eText -- Standalone Access Card -- for Mathematics with Applications in the Management, Natural, and Social Sciences (12th Edition)

Ch. 6.1 - Checkpoint1 Use the substitution method tosolve...Ch. 6.1 - Checkpoint 2 Use the elimination method to solve...Ch. 6.1 - Checkpoint 3 Solve the system of equations . Draw...Ch. 6.1 - Checkpoint 4 Solve the following system: Ch. 6.1 - Checkpoint 5 Solve the system Draw the graph of...Ch. 6.1 - Prob. 6CP Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Use substitution to solve each system. (See...Ch. 6.1 - Use substitution to solve each system. (See...

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