4.1--4.2 - MATH 431 HOMEWORK 6 SOLUTIONS 1. Section 4.1...

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Unformatted text preview: MATH 431 HOMEWORK 6 SOLUTIONS 1. Section 4.1 (4d) For the following pair of vectors u and v in R 2 , compute the area of the parallelogram determined by u and v . u = (3 , 4) and v = (2 ,- 6) A u v = det u v = det 3 4 2- 6 = |- 18- 8 | = |- 26 | = 26 . (9) Prove that det( AB ) = det( A ) det( B ) for any A, B M 2 2 ( F ). Proof. Let A = a b c d and B = v w x y where a, b, c, d, v, w, x, y F . Then AB = a b c d v w x y = av + bx aw + by cv + dx cw + dy . So det( AB ) = ( av + bx )( cw + dy )- ( aw + by )( cv + dx ) = avcw + avdy + bxcw + bxdy- awcv- awdx- bycv- bydx = advy + bcwx + adwx + bcvy = ad ( vy- wx )- bc ( vy- wx ) = ( ad- bc )( vy- wx ) = det( A ) det( B ) . (10) The classical adjoint of a 2 2 matrix A M 2 2 ( F ) is the matrix C = A 22- A 12- A 21 A 11 . (a) Claim: CA = AC = [det( A )] I. Date : October 17, 2006. 1 2 MATH 431 HW 6 Proof. Let A = a b c d . Then C = d- b- c a . Notice CA = d- b- c a a b c d = da- bc db- bd- ca + ac- cb + ad = ad- bc ad- bc , AC = a b c d d- b- c a = ad- bc- ab + ba cd- dc- cb + da = ad- bc ad- bc , and [det( A )] I = ( ad- bc ) 1 1 = ad- bc ad- bc . Hence CA = AC = [det( A )] I . (d) Claim: If A is invertible, then A- 1 = [det( A )]- 1 C . Proof. Let A = a b c d . Then by Theorem 4.2, A- 1 = 1 det( A ) d- b- c a = [det( A )]- 1 C....
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This note was uploaded on 02/02/2012 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.

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4.1--4.2 - MATH 431 HOMEWORK 6 SOLUTIONS 1. Section 4.1...

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