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# 4.1--4.2 - MATH 431 HOMEWORK 6 SOLUTIONS 1 Section 4.1(4d...

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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

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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 0 0 ad - bc , AC = a b c d d - b - c a = ad - bc - ab + ba cd - dc - cb + da = ad - bc 0 0 ad - bc , and [det( A )] I = ( ad - bc ) 1 0 0 1 = ad - bc 0 0 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. 2. Section 4.2 (2) Find the value of k that satisfies the following equation det 3 a 1 3 a 2 3 a 3 3 b 1 3 b 2 3 b 3 3 c 1 3 c 2 3 c 3 = k det a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 .
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4.1--4.2 - MATH 431 HOMEWORK 6 SOLUTIONS 1 Section 4.1(4d...

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