hw1solutions_2

hw1solutions_2 - Homework 1 Solutions Josh Hernandez...

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Homework 1 Solutions Josh Hernandez October 25, 2009 1.1 - Introduction 2. Find the equations of the lines through the following pairs of points in space. b. (3 , -2 , 4) and (-5 , 7 , 1) Solution: x = (3 , -2 , 4) + r [(-5 , 7 , 1) - (3 , -2 , 4)] = (3 , -2 , 4) + r (-8 , 9 , -3) . d. (-2 , -1 , 5) and (3 , 9 , 7) Solution: x = (-2 , -1 , 5) + r [(3 , 9 , 7) - (-2 , -1 , 5)] = (-2 , -1 , 5) + r (5 , 10 , 2) . 3. Find the equations of the planes containing the following points in space. b. (3 , -6 , 7), (-2 , 0 , -4), and (5 , -9 , -2). Solution: x = (3 , -6 , 7) + r [(-2 , 0 , -4) - (3 , -6 , 7)] + s [(5 , -9 , -2) - (3 , -6 , 7)] = (3 , -6 , 7) + r (-5 , 6 , -11) + s (2 , 3 , -9) d. (1 , 1 , 1), (5 , 5 , 5), and (-6 , 4 , 2). Solution: x = (1 , 1 , 1) + r [(5 , 5 , 5) - (1 , 1 , 1)] + s [(-6 , 4 , 2) - (1 , 1 , 1)] = (1 , 1 , 1) + r (4 , 4 , 4) + s (-7 , 3 , 1) 7. Prove that the diagonals of a parallelogram bisect each other. Solution: A parallelogram ABCD in R 2 may be translated so that vertex A lies on the origin. After translation, denote the coordinate-vectors of the adjacent vertices B and D by v and w . The four vertices therefore have coordinate-vectors 0 ,v,v + w , and w . By the midpoint rule, segment BD has midpoint at 1 2 ( v + w ), and AC has midpoint 1 2 (0+( v + w )). These midpoints are identical, so BD and AC bisect one another. 1.2 - Vector Spaces 2. Write the zero vector of M 3 × 4 ( F ). 1
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Solution: Consider the matrix 0 := ± 0 0 0 0 0 0 0 0 0 0 0 0 ² . For any matrix A M 3 × 4 , 0 + A = ± 0 0 0 0 0 0 0 0 0 0 0 0 ² + ± a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 ² = ± 0+ a 11 0+ a 12 0+ a 13 0+ a 14 0+ a 21 0+ a 22 0+ a 23 0+ a 24 0+ a 31 0+ a 32 0+ a 33 0+ a 34 ² = ± a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 ² = A Thus 0 is the zero vector of M 3 × 4 ( F ). 12. Prove that the set of even functions from R to R is a vector space over F (under the usual addition and scalar multiplication). Solution: Note that (1) f = g means f ( x ) = g ( x ) for all x R , and (2) the “usual addition and scalar multiplication” means that ( f + g )( x ) = f ( x ) + g ( x ) and ( cf )( x ) = cf ( x ) for all x R . The even functions are those f : R R which satisfy f (- x ) = f ( x ) for all x R . Adding two such functions: ( f + g )(- x ) = f (- x ) + g (- x ) = f ( x ) + g ( x ) = ( f + g )( x ) , so the sum f + g is an even function. As for scalar multiplication, ( cf )(- x ) = cf (- x ) = cf ( x ) = ( cf )( x ) . The remaining properties hold equally well for any set S , field F , scalars a,b F , point x S and functions f,g,h ∈ F ( S, F ). VS1 Commutativity of addition is inherited from F : ( f + g )( x ) = f ( x ) + g ( x ) = g ( x ) + f ( x ) = ( g + f )( x ) .
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This note was uploaded on 02/01/2010 for the course MATH math115a taught by Professor Shalom during the Spring '10 term at UCLA.

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hw1solutions_2 - Homework 1 Solutions Josh Hernandez...

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