2.3 - MATH321 – HOMEWORK SOLUTIONS HOMEWORK#4 Section 2.3...

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Unformatted text preview: MATH321 – HOMEWORK SOLUTIONS HOMEWORK #4 Section 2.3: Problems 1, 2(b), 3, 4(a)(c), 9, 11, 12, 13 Krzysztof Galicki Problem 2.3.1 (See Answers to Selected Exercises ). Problem 2.3.2 (a) A (2 B + 3 C ) = parenleftbigg 1 3 2 − 1 parenrightbiggparenleftbigg 2 parenleftbigg 1 − 3 4 1 2 parenrightbigg + 3 parenleftbigg 1 1 4 − 1 − 2 parenrightbiggparenrightbigg = = parenleftbigg 1 3 2 − 1 parenrightbiggparenleftbigg 5 3 6 5 − 4 4 parenrightbigg = parenleftbigg 20 − 9 18 5 10 8 parenrightbigg . ( AB ) D = parenleftbiggparenleftbigg 1 3 2 − 1 parenrightbiggparenleftbigg 1 − 3 4 1 2 parenrightbiggparenrightbigg 2 − 2 3 = = parenleftbigg 13 3 3 − 2 − 1 − 8 parenrightbigg 2 − 2 3 = parenleftbigg 29 − 26 parenrightbigg . ( AB ) D = parenleftbigg 1 3 2 − 1 parenrightbiggparenleftbiggparenleftbigg 1 − 3 4 1 2 parenrightbigg 2 − 2 3 parenrightbigg = = parenleftbigg 1 3 2 − 1 parenrightbiggparenleftbigg − 7 12 parenrightbigg = parenleftbigg 29 − 26 parenrightbigg . (b) A t = parenleftbigg 2 − 3 4 5 1 2 parenrightbigg A t B = parenleftbigg 2 − 3 4 5 1 2 parenrightbigg 3 − 2 1 − 1 4 5 5 3 = parenleftbigg 23 19 26 − 1 10 parenrightbigg . BC t = 3 − 2 1 − 1 4 5 5 3 4 3 = 12 16 29 . CB = ( 4 3 ) 3 − 2 1 − 1 4 5 5 3 = ( 27 7 9 ) . CA = ( 4 3 ) 2 5 − 3 1 4 2 = ( 20 26 ) ....
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This note was uploaded on 11/30/2009 for the course MATH 115A taught by Professor Liu during the Winter '07 term at UCLA.

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2.3 - MATH321 – HOMEWORK SOLUTIONS HOMEWORK#4 Section 2.3...

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