# EXAM03-solutions.pdf - Version 101 EXAM03 gilbert(54620...

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This preview shows page 1 out of 7 pages. Unformatted text preview: Version 101 – EXAM03 – gilbert – (54620) This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 002 10.0 points The matrix 2 1 3 A = 0 4 −2 1 0 2 10.0 points Simplify the expression (2u + 4v) · (4u − v) − k2u + vk2 is invertible. True or False? for vectors u, v in Rn . 1. TRUE correct 1. −4kuk2 − 22 u · v − 5kvk2 2. FALSE 2. 4kuk2 + 10 u · v + 5kvk2 2 1 3. −4kuk − 10 u · v − 5kvk Explanation: The matrix A will be invertible if and only if det(A) 6= 0. Now 2 1 3 det(A) = 0 4 −2 1 0 2 0 4 4 −2 0 −2 +3 − = 2 1 0 0 2 1 2 2 4. −4kuk2 − 22 u · v + 5kvk2 5. 4kuk2 + 10 u · v − 5kvk2 correct 6. 4kuk2 − 22 u · v + 5kvk2 Explanation: By linearity, = 2× 4−2− 3×4 = 2. (2u + 4v) · (4u − v) Consequently, the statement is = 2u · (4u − v) + 4v · (4u − v) TRUE . = 8u · u − 2u · v + 16v · u − 4v · v , while 003 10.0 points The vectors u1 and u2 shown in k2u + vk2 = (2u + v) · (2u + v) x2 = 2u · (2u + v) + v · (2u + v) = 4u · u + 2u · v + 2v · u + v · v . 2 u2 But v · u = u · v , u · u = kuk2 , v · v = kvk2 . So after expansion the expression becomes 2 8kuk + 14 u · v − 4kvk = x1 2 2 − (4kuk2 + 4u · v + kvk2 ) 2 u1 4kuk + 10 u · v − 5kvk 2 . are eigenvectors corresponding to eigenvalues λ1 = −3 and λ2 = 1 respectively for a 2 × 2 matrix A. Version 101 – EXAM03 – gilbert – (54620) Which of the following graphs contains the vector A(u1 + u2 )? 2 x2 5. correct 2 −2 1. −2 2 x1 2 x1 x2 6. 2 x2 2. 2 x1 2 x2 3. Explanation: As the graph of u1 , u2 shows,     1 2 . u1 = u1 = 2 1 But then A(u1 + u2 ) = A u1 + A u2 = λ1 u1 + λ2 u2       −5 1 −6 . = + = −3u1 + u2 = −1 2 −3 2 x1 2 4. Consequently, A(u1 + u2 ) is contained in the graph x2 −2 −2 2 2 x1 Version 101 – EXAM03 – gilbert – (54620) 004 10.0 points A subset H of a vector space V is a subspace of V if the following conditions are satisfied: The plane in R3 spanned by u1 , u2 is the subspace W = Span{u1 , u2 }, and each y in R3 has a unique orthogonal decomposition y = projW y + (y − projW y) i) the zero vector of V is in H, ii) the sum u + v is in H for all u, v in H, iii) the scalar multiple cu is in H for all scalars c and u in H. 3 where y − projW y is in the orthogonal complement W ⊥ . But then dist(y, W ) = ky − projW yk. True or False? 1. TRUE correct 2. FALSE Explanation: Property (ii) says that H is closed under vector addition, while property (iii) says that H is closed under scalar multiplication. Thus H is a subspace of V . Consequently, the statement is TRUE . 005 Now u1 , u2 are non-zero othogonal vectors, so form a basis for W such that     y · u1 y · u2 projW y = u1 + u2 . ku1 k2 ku2 k2 But when 3 −2 1 y = −6 , u1 = 2 , u2 = 2 , −9 1 −2 we see that  10.0 points Find the distance from 3 y = −6 −9 to the plane in R3 spanned by −2 1 u1 = 2 , u2 = 2 . 1 −2 1. dist = 7 2. dist = 4 −2 u1 = −3 2 , 1 while  y · u2 ku2 k2  1 u2 = 2 . −2 Consequently, projW −2 1 7 y = −3 2 + 2 = −4 , 1 −2 −5 and so y − projW 3. dist = 6 correct 4. dist = 5 y · u1 ku1 k2  3 7 2 = −6 − −4 = −2 1 . −9 −5 2 Thus dist(y, W ) = 6 . 5. dist = 8 Explanation: 006 10.0 points Version 101 – EXAM03 – gilbert – (54620) Use the fact that 1 A = −1 5 −4 9 −7 2 −4 1 −6 10 7 1 0 −1 5 ∼ 0 −2 5 −6 0 0 0 0 to determine an orthogonal basis for Col(A). 1 1 1. −1 , −4 5 −1 1 2. −1 , 5 −4 3. 2 , −6 −4 4. 2 , −6 thogonal basis: set u1 = a1 and   a2 · u 1 u1 u 2 = a2 − ku1 k2 −4 1 (−36) = 2 − −1 27 −6 5 −4 4/3 −8/3 = 2 + −4/3 = 2/3 . −6 20/3 2/3 Consequently, the set of vectors −4 1 −1 , 1 5 1 −4 1 correct 1 is an orthogonal basis for Col(A). 1 −1 −1 If eigenvectors of an n × n matrix A are a basis for Rn , then A is diagonalizable. True or False? 007 2. FALSE 1 0 −1 5 A ∼ 0 −2 5 −6 0 0 0 0 1 0 −1 5 ∼ 0 1 −5/2 3 . 0 0 0 0 1 a1 = −1 , 5 Explanation: An n × n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. On the other hand, n vectors in Rn are a basis for Rn if and only if the vectors are linearly independent. So if eigenvectors of A form a basis for Rn , they must be linearly independent, in which case A will be diagonalizable. Consequently, the statement is TRUE . 008 Thus the pivot columns of A are 10.0 points 1. TRUE correct 1 −1 5 Explanation: The pivot columns of A provide a basis for Col(A). But by row reduction, 4 −4 a2 = 2 . −6 We apply Gram-Schmidt to produce an or- 10.0 points Given vectors 0 4 1 u1 = −4 , u2 = −2 , u3 = 1 , 4 −2 1 in R3 , determine c1 so that y = c1 u1 + c2 u2 + c3 u3 Version 101 – EXAM03 – gilbert – (54620) 5     when −2 −2 . , v2 = v1 = 0 −5 −2 y = −1 . Let {xk } be a solution of the difference equa3 tion   0 1. No such value of c1 exists. . xk+1 = Axk , x0 = −6 1 2. c1 = Compute x1 . 4   4 3. c1 = 0 1. x1 = −8   1 8 4. c1 = correct 2. x1 = 2 4 1   5. c1 = − −4 2 3. x1 = 8 1   6. c1 = − 4 4 4. x1 = 8   Explanation: 8 Since correct 5. x1 = −4   u1 · u2 = u1 · u3 = u2 · u3 = 0, −8 6. x1 = 4 the vectors u1 , u2, u3 are mutually orthogo3 nal, hence form a basis for R as they are also Explanation: non-zero. Thus by orthogonality, To find x1 we must compute Ax0 . Now, exy = proju1 y + proju2 y + proju3 y press x0 in terms of v1 and v2 . That is, find c1 and c2 such that x0 = c1 v1 + c2 v2 . This y · u2 y · u3 y · u1 u1 + u2 + u3 , = is certainly possible because the eigenvectors u1 · u1 u2 · u2 u3 · u3 v1 and v2 are linearly independent (by inso spection and also because they correspond to y · u1 y · u2 y · u3 distinct eigenvalues) and hence form a basis c1 = , c2 = , c3 = . for R2 . The row reduction u1 · u1 u2 · u2 u3 · u3   But −2 −2 0 [ v1 v2 x0 ] = −2 −5 −6 y · u1 (0) + (4) + (12)   = . 1 0 −2 u1 · u1 (0) + (16) + (16) ∼ 0 1 2 Consequently, shows that x0 = −2v1 + 2v2 . Since v1 and v2 are eigenvectors (for the eigenvalues 4 and 2 1 c1 = . respectively): 2 009 10.0 points Let A be a 2 × 2 matrix with eigenvalues 4 and 2 and corresponding eigenvectors x1 = Ax0 = A(−2v1 + 2v2 ) = −2Av1 + 2Av2 = −2 · 4v1 + 2 · 2v2       8 −8 16 . = + = −4 −20 16 Version 101 – EXAM03 – gilbert – (54620) 6 This shows that c1 = −1, c2 = 4 and Consequently, x1 = 010  8 −4  u(0) = −v1 + 4v2 . . Since v1 and v2 are eigenvectors corresponding to the eigenvalues 4 and 2 respectively, set u(t) = −e4t v1 + 4e2t v2 . 10.0 points Let u(t) satisfy du = Au(t), dt u(0) =   −13 . 14 Compute u(2) when A is a 2 × 2 matrix with eigenvalues 4 and 2 and corresponding eigenvectors     −3 1 . , v2 = v1 = 3 −2   −e8 + 12e4 1. u(2) = −2e8 + 12e4 2. u(2) =  −e8 + 12e4 2e8 − 12e4 3. u(2) =  e8 − 12e4 2e8 − 12e4 4. u(2) =  e − 12e −2e8 + 12e4  5. u(2) =  e8 + 12e4 −2e8 − 12e4  6. u(2) =  −e8 − 12e4 2e8 + 12e4  Then u(0) is the given initial value and Au(t) = −e4t Av1 + 4e2t Av2   du(t) = − 4e4t v1 + 4 2e2t v2 = . dt Thus u(t) = −e4t v1 + 4e2t v2 solves the given differential equation. Consequently, u(2) =  011 8 4  correct  −e8 − 12e4 2e8 + 12e4  . 10.0 points If {x1 , x2 , x3 } is a linearly independent set and W = Span{x1 , x2 , x3 }, then any orthogonal set {v1 , v2 , v3 } in W is a basis for W . True or False? 1. TRUE Explanation: Since v1 and v2 are eigenvectors corresponding to distinct eigenvalues of A, they form an eigenbasis for R2 . Thus u(0) = c1 v1 + c2 v2 To compute c1 and c2 we apply row reduction to the augmented matrix   1 −3 −13 [ v1 v2 u(0) ] = −2 3 14   1 0 −1 . ∼ 0 1 4 2. FALSE correct Explanation: The three orthogonal vectors must be nonzero to be a basis for a three-dimensional subspace. Consequently, the statement is incomplete and thus FALSE . 012 10.0 points Version 101 – EXAM03 – gilbert – (54620) By diagonalizing the matrix   3 4 , A = −2 −3 compute f (A) for the polynomial 3 2 f (x) = 4x + x − 3x + 1 .   −5 2 −4 1 2. f (A) =  5 −2 3. f (A) =  5 2  4 correct −1  4. f (A) =  −5 −2 1. f (A) = 4 1 −4 −1 so P =  2 −1 1 −1  −1 =  1 −1 , P  f (1) 0 0 f (−1) 1 −2  Explanation: If A can be diagonalized by   d1 0 −1 P −1 , A = P DP = P 0 d2 then f (A) = P f (D)P −1   f (d1 ) 0 P −1 . = P 0 f (d2 ) Now A can be diagonalized if we can find an eigenbasis of R2 of eigenvectors v1 , v2 of A corresponding to eigenvalues λ1 , λ2 , for then:   λ1 0 P −1 , P = [v1 v2 ] . A = P 0 λ2 But 4 −3 − λ = 8 − (3 − λ)(3 + λ) = λ2 − 1 = 0 , i.e., λ1 = 1 and λ2 = −1. Corresponding eigenvectors are     1 2 , , v2 = v1 = −1 −1 . Thus f (A) =  2 −1 1 −1  1 −1 1 −2 Now  3 − λ det[A − λI] = −2 7 f (1) = 4x3 + x2 − 3x + 1 x=1 = 3, while f (−1) = 4x3 + x2 − 3x + 1 x=−1 Consequently,    1 3 0 2 1 f (A) = −1 0 1 −1 −1   5 4 . = −2 −1 = 1. 1 −2   . ...
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