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HW09-solutions - geyer(dag2798 HW09 rusin(55735 This...

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geyer (dag2798) – HW09 – rusin – (55735) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points By evaluating the determinant, express f ( x ) = v v v v v v 1 x x 2 4 2 1 4 1 3 v v v v v v as a quadratic Function in x . 1. f ( x ) = 5 4 x 8 x 2 2. f ( x ) = 5 + 4 x + 8 x 2 3. f ( x ) = 5 8 x 4 x 2 4. f ( x ) = 5 + 8 x + 4 x 2 correct 5. f ( x ) = 5 8 x + 4 x 2 6. f ( x ) = 5 + 8 x 4 x 2 Explanation: ±or this 3 × 3 determinant, use expansion by minors along the top row: f ( x ) = v v v v 2 1 1 3 v v v v v v v v 4 1 4 3 v v v v x + v v v v 4 2 4 1 v v v v x 2 . Evaluating the 2 × 2 determinants, we thus see that f ( x ) = 5 + 8 x + 4 x 2 . keywords: matrix, determinant, quadratic Function, expansion by minors 002 10.0 points ±ind the value oF the determinant D = v v v v v v v 3 2 1 2 1 3 x y z v v v v v v v . 1. D = 7 x + 11 y z 2. D = 7 x 11 y + z 3. D = 7 x + 11 y z correct 4. D = 7 x 11 y z 5. D = 7 x + 11 y + z 6. D = 7 x 11 y + z Explanation: ±or any 3 × 3 determinant v v v v v v v A B C a 1 b 1 c 1 a 2 b 2 c 2 v v v v v v v = A v v v v v b 1 c 1 b 2 c 2 v v v v v B v v v v v a 1 c 1 a 2 c 2 v v v v v + C v v v v v a 1 b 1 a 2 b 2 v v v v v . Thus D = v v v v v v v 3 2 1 2 1 3 x y z v v v v v v v = 3 v v v v v 1 3 y z v v v v v + 2 v v v v v 2 3 x z v v v v v v v v v v 2 1 x y v v v v v = 3 (1 z + 3 y ) + 2 ( 2 z + 3 x ) ( 2 yx ) . Consequently, D = 7 x + 11 y z . keywords: determinant 003 10.0 points ±ind the cross product oF the vectors a = i 2 j 3 k , b = 3 i 2 j + 3 k . 1. a × b = 11 i + 6 j 9 k
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geyer (dag2798) – HW09 – rusin – (55735) 2 2. a × b = 11 i + 6 j 8 k 3. a × b = 12 i + 6 j 8 k correct 4. a × b = 12 i + 4 j 8 k 5. a × b = 12 i + 3 j 9 k 6. a × b = 11 i + 4 j 9 k Explanation: One way of computing the cross product ( i 2 j 3 k ) × ( 3 i 2 j + 3 k ) is to use the fact that i × j = k , j × k = i , k × i = j , while i × i = 0 , j × j = 0 , k × k = 0 . For then a × b = 12 i + 6 j 8 k . Alternatively, we can use the de±nition a × b = v v v v v v i j k 1 2 3 3 2 3 v v v v v v = v v v v 2 3 2 3 v v v v i v v v v 1 3 3 3 v v v v j + v v v v 1 2 3 2 v v v v k to determine a × b . 004 10.0 points Which of the following statements are true for all vectors a , b n = 0 ? A. if a × b = 0, then a b b , B. | a · b | 2 − | a × b | 2 = | a | 2 | b | 2 , C. a × b + b × a = 0. 1. A only 2. C only 3. A and C only correct 4. all of them 5. B only 6. A and B only 7. none of them 8. B and C only Explanation: A. TRUE: if θ, 0 θ π , is the angle between a , and b , then | a × b | = | a || b | sin θ , so if b a b n = 0 and b b b n = 0, then | a × b | = 0 = sin θ = 0 . Thus θ = 0 , π . In this case, a is parallel to b .
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