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Unformatted text preview: Homework 7, Due April 29 Adam Gamzon Related reading: sections 6.2, 6.3, 7.1  7.4 from the text 1. (a) Find the volume of the parallelpiped in R 3 defined by the vectors 1 2 , 1 , 1 1 . Solution: The volume is given by det 1 0 1 0 1 0 2 0 1 = 1 . (b) Find the area of the parallelogram with vertices 4 3 , 10 1 , 5 7 , and 11 5 . (Hint: first translate the parallelogram so that one of the vertices is the origin.) Solution: First move the parallelogram so that one of the vertices is at the origin by subtracting the vector 4 2 from all of the vertices. Doing this gives another parallelogram whose vertices are , 6 2 , 1 4 , and 7 2 . This parallelogram is defined by the vectors 6 2 and 1 4 . Therefore, its area is det 6 1 2 4 = 26 . 2. Find the volume of the parallelpiped in R 4 defined by the vectors 1 , 1 1 1 1 , 1 2 3 4 . Solution: Let A = 1 1 1 0 1 2 0 1 3 0 1 4 . Then A T A = 1 0 0 0 1 1 1 1 1 2 3 4 1 1 1 0 1 2 0 1 3 0 1 4 = 1 1 1 1 4 10 1 10 30 , 1 so the volume is q det( A T A ) = v u u u t det 1 1 1 1 4 10 1 10 30 = 6 ....
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This note was uploaded on 11/22/2011 for the course MATH 255 taught by Professor Staff during the Spring '10 term at UMass (Amherst).
 Spring '10
 STAFF
 Math, Linear Algebra, Algebra, Vectors

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