# 012 50points the set h of all polynomials p x a x 3 a

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012 5.0points The set H of all polynomials p ( x ) = a + x 3 , a in R , is a subspace of the vector space P 6 of all polynomials of degree at most 6. True or False? 1. FALSE correct 2. TRUE Explanation: The zero polynomial p ( x ) = 0 + 0 x 3 does not belong to H . Consequently, the statement is FALSE . 013 10.0points Find the solution of the differential equa- tion d u dt = A u ( t ) , u (0) = bracketleftbigg 2 6 bracketrightbigg , when A is a 2 × 2 matrix with eigenvalues 2 , 1 2 and corresponding eigenvectors v 1 = bracketleftbigg 1 1 bracketrightbigg , v 2 = bracketleftbigg 1 1 bracketrightbigg . 1. u ( t ) = 2 e 2 t v 1 + 2 e t/ 2 v 2 2. u ( t ) = 4 e 2 t v 1 + 4 e t/ 2 v 2 3. u ( t ) = 4 e 2 t v 1 4 e t/ 2 v 2 4. u ( t ) = 2 e 2 t v 1 + 4 e t/ 2 v 2 correct 5. u ( t ) = 2 e 2 t v 1 2 e t/ 2 v 2 6. u ( t ) = 4 e 2 t v 1 + 2 e t/ 2 v 2 Explanation: Since v 1 , v 2 are eigenvectors corresponding to distinct eigenvalues of A , they form an eigenbasis for R 2 . Thus u (0) = c 1 v 1 + c 2 v 2 . To compute c 1 , c 2 we apply row reduction to the augmented matrix [ v 1 v 2 u (0) ] = bracketleftbigg 1 1 2 1 1 6 bracketrightbigg bracketleftbigg 1 0 2 0 1 4 bracketrightbigg , for then c 1 = 2 , c 2 = 4 and u (0) = 2 v 1 + 4 v 2 . But v 1 , v 2 are eigenvectors corresponding to respective eigenvalues 2 , 1 2 so set u ( t ) = 2 e 2 t v 1 + 4 e t/ 2 v 2 . Then u (0) is the given initial value and A u ( t ) = 2 e 2 t A v 1 + 4 e t/ 2 A v 2 = 2(2 e 2 t ) v 1 + 4 parenleftbigg 1 2 e t/ 2 parenrightbigg v 2 = d u ( t ) dt , so u ( t ) = 2 e 2 t v 1 + 4 e t/ 2 v 2 solves the differential equation. 014 10.0points Find the area of the triangle Δ ABC having vertices A = ( 2 , 2) , B = (3 , 3) , C = ( 3 , 2) . 1. area = 13 2. area = 12
Version 059 – FINAL – gilbert – (56540) 8 3. area = 27 2 4. area = 25 2 correct 5. area = 14 Explanation: The area of Δ ABC is given by 1 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle det 2 2 1 3 3 1 3 2 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle . But by properties of determinants, det 2 2 1 3 3 1 3 2 1 = det 2 2 1 5 5 0 1 4 0 = vextendsingle vextendsingle vextendsingle vextendsingle 5 5 1 4 vextendsingle vextendsingle vextendsingle vextendsingle . Consequently, Δ ABC has area = 25 2 . 015 10.0points Determine the unique solution x 1 of the matrix equation A x = 3 1 2 6 2 5 9 5 6 x 1 x 2 x 3 = 3 9 17 when A = LU = 1 0 0 2 1 0 3 2 1 3 1 2 0 4 1 0 0 2 1. x 1 = 1 2. x 1 = 0 3. x 1 = 2 correct 4. x 1 = 3 5. x 1 = 1 Explanation: Set y = U x . Then A x = L y = b , and so y = L 1 b . But L 1 = 1 0 0 2 1 0 1 2 1 , in which case A x = b reduces to U x = L 1 b = 1 0 0 2 1 0 1 2 1 3 9 17 = 3 3 2 .
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