102_solutions6

102_solutions6 - MATH 102 SOLUTIONS TO HW#6 Section 4.2...

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Unformatted text preview: MATH 102 SOLUTIONS TO HW #6 Section 4.2, problem 2. a) Recall that in general for an n × n matrix A , and any constant c , we have det ( cA ) = c n det ( A ). Thus, in this case we have that det ( 1 2 A ) =- 1 8 . Also, we have det (- A ) = (- 1) 3 det ( A ) = 1. b) Recall also the general rules det ( A 2 ) = [ det ( A )] 2 , and det ( A- 1 ) = [ det ( A )]- 1 . Therefore, in this case we have that det ( A 2 ) = 1 and det ( A- 1 ) =- 1. Section 4.2, problem 12. To deal with this problem, it is necessary to break down into two cases. Suppose first that b = a . Then it is easy to see that the first and second rows of the Vandermonde matrix A are in fact equal to each other. Thus, det ( A ) = 0 in this case which is the same as ( b- a )( c- a )( c- b ) when b = a . In the case where ( b- a ) 6 = 0 we can perform the following two Gaussian elimi- nation steps: (1)- 1 × Row 1 + Row 2 ⇒ Row 2. (2)- 1 × Row 1 + Row 3 ⇒ Row 3. (3)- ( c- a ) ( b- a ) × Row 2 + Row 3 ⇒ Row 3. Notice that by using the identities b 2- a 2 = ( b + a )( b- a ) and c 2- a 2 = ( c + a )( c- a ) we have:- ( c- a ) ( b- a ) · ( b 2- a 2 ) + ( c 2- a 2 ) = ( c- a )( c- b ) . Therefore, the reduced matrix in this case is: R = 1 a a 2 ( b- a ) ( b 2- a 2 ) ( c- a )( c- b ) ....
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This note was uploaded on 08/02/2011 for the course MATH 102 taught by Professor Szypowsk during the Fall '08 term at UCSD.

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102_solutions6 - MATH 102 SOLUTIONS TO HW#6 Section 4.2...

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