Explanation Given: A = [ 1 2 0 − 1 ] , B = [ 0 − 1 1 1 ] , C = [ − 2 0 0 1 ] . That is to verify, A ( B + C ) = A B + A C And ( A + B ) C = A C + B C Let A = [ 1 2 0 − 1 ] , B = [ 0 − 1 1 1 ] and C = [ − 2 0 0 1 ] . First, consider A ( B + C ) , That is, A ( B + C ) = [ 1 2 0 − 1 ] ( [ 0 − 1 1 1 ] + [ − 2 0 0 1 ] ) = [ 1 2 0 − 1 ] ( [ 0 − 2 − 1 + 0 1 + 0 1 + 1 ] ) = [ 1 2 0 − 1 ] [ − 2 − 1 1 2 ] Now find the product as follows: [ 1 2 0 − 1 ] [ − 2 − 1 1 2 ] = [ 1 ( − 2 ) + 2 ( 1 ) 1 ( − 1 ) + 2 ( 2 ) 0 ( − 2 ) − 1 ( 1 ) 0 ( − 1 ) − 1 ( 2 ) ] = [ − 2 + 2 − 1 + 4 0 − 1 0 − 2 ] = [ 0 3 − 1 − 2 ] Therefore, A ( B + C ) = [ 0 3 − 1 − 2 ] Next find, A B + A C . For this consider, A B = [ 1 2 0 − 1 ] [ 0 − 1 1 1 ] = [ 1 ( 0 ) + 2 ( 1 ) 1 ( − 1 ) + 2 ( 1 ) 0 ( 0 ) − 1 ( 1 ) 0 ( − 1 ) − 1 ( 1 ) ] = [ 0 + 2 − 1 + 2 0 − 1 0 − 1 ] = [ 2 1 − 1 − 1 ] And, A C = [ 1 2 0 − 1 ] [ − 2 0 0 1 ] = [ 1 ( − 2 ) + 2 ( 0 ) 1 ( 0 ) + 2 ( 1 ) 0 ( − 2 ) − 1 ( 0 ) 0 ( 0 ) − 1 ( 1 ) ] = [ − 2 + 0 0 + 2 0 − 0 0 − 1 ] = [ − 2 2 0 − 1 ] Now add A B and A C as follows: A B + A C = [ 2 1 − 1 − 1 ] + [ − 2 2 0 − 1 ] = [ 2 − 2 1 + 2 − 1 + 0 − 1 − 1 ] = [ 0 3 − 1 − 2 ] This verifies that A ( B + C ) = A B + A C = [ 0 3 − 1 − 2 ]

8th Edition

ISBN: 9780136922148

Author: Blitzer

Publisher: RENT PEARS

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Chapter 6.3, Problem 47E

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To Verify: The distributive property of the matrices.

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To Verify: The distributive property of the matrices.

Solution Summary: The author explains that the distributive property of the matrices holds.

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