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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.
 Fall '08
 SZYPOWSK
 Math, Linear Algebra, Algebra

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