Explanation Given: A = [ − 1 0 2 1 ] and B = [ 1 1 0 2 ] Calculation: For L.H.S Compute ( A + B ) ( A + B ) = [ − 1 0 2 1 ] + [ 1 1 0 2 ] = [ − 1 + 1 0 + 1 2 + 0 1 + 2 ] = [ 0 1 2 3 ] Compute ( A − B ) ( A − B ) = [ − 1 0 2 1 ] − [ 1 1 0 2 ] = [ − 1 − 1 0 − 1 2 − 0 1 − 2 ] = [ − 2 − 1 2 − 1 ] Compute ( A + B ) ( A − B ) L .H .S = ( A + B ) ( A − B ) = [ 0 1 2 3 ] [ − 2 − 1 2 − 1 ] = [ 0 × ( − 2 ) + 1 × 2 0 × ( − 1 ) + 1 × ( − 1 ) 2 × ( − 2 ) + 3 × 2 2 × ( − 1 ) + 3 × ( − 1 ) ] = [ 2 − 1 2 − 5 ] For R.H.S, Compute A 2 . A 2 = AA = [ − 1 0 2 1 ] [ − 1 0 2 1 ] = [ ( − 1 ) × ( − 1 ) + 0 × 2 ( − 1 ) × 0 + 0 × 1 2 × ( − 1 ) + 1 × 2 2 × 0 + 1 × 1 ] = [ 1 0 0 1 ] Compute B 2

College Algebra Plus Mylab Math With Pearson Etext -- Access Card Package (5th Edition) (beecher, Penna, & Bittinger, The College Algebra Series, 5th Edition)

5th Edition

ISBN: 9780321981769

Author: Judith A. Beecher, Judith A. Penna, Marvin L. Bittinger

Publisher: PEARSON

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Question

Chapter 6.4, Problem 53E

To determine

To check:

(A+B)(A−B)=A2+BA−AB−B2 for matrices,

A=[−1021] and B=[1102]

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Chapter 6.4, Problem 52E

Chapter 6.4, Problem 54E

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Chapter 6 Solutions

College Algebra Plus Mylab Math With Pearson Etext -- Access Card Package (5th Edition) (beecher, Penna, & Bittinger, The College Algebra Series, 5th Edition)

Ch. 6.1 - Prob. 1VG Ch. 6.1 - Prob. 2VG Ch. 6.1 - Prob. 3VG Ch. 6.1 - Prob. 4VG Ch. 6.1 - Prob. 5VG Ch. 6.1 - Prob. 6VG Ch. 6.1 - Prob. 7VG Ch. 6.1 - Prob. 8VG Ch. 6.1 - Prob. 9VG Ch. 6.1 - Prob. 10VG

Knowledge Booster

To check: ( A + B ) ( A − B ) = A 2 + BA − AB − B 2 for matrices, A = [ − 1 0 2 1 ] and B = [ 1 1 0 2 ]

To check: (A+B)(A−B)=A2+BA−AB−B2 for matrices, A=[−1021] and B=[1102]

Want to see the full answer?

Chapter 6 Solutions

To check:

(A+B)(A−B)=A2+BA−AB−B2 for matrices,

A=[−1021] and B=[1102]