# EXAM03-solutions.pdf - Version 081 harper(elh2395 EXAM03...

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This preview shows page 1 out of 7 pages. Unformatted text preview: Version 081 – harper (elh2395) – EXAM03 – pencheva – (54150) This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Let A be a 2 × 2 matrix with eigenvalues 2 and 1 and corresponding eigenvectors     −1 1 . , v2 = v1 = 6 −2 Let {xk } be a solution of the difference equation   −1 . xk+1 = Axk , x0 = 14 Compute x1 .   −1 1. x1 = −10   1 correct 2. x1 = 10   1 3. x1 = −10   −10 4. x1 = 1   10 5. x1 = 1   −10 6. x1 = −1 Explanation: To find x1 we must compute Ax0 . Now, express x0 in terms of v1 and v2 . That is, find c1 and c2 such that x0 = c1 v1 + c2 v2 . This is certainly possible because the eigenvectors v1 and v2 are linearly independent (by inspection and also because they correspond to distinct eigenvalues) and hence form a basis for R2 . The row reduction   1 −1 −1 [ v1 v2 x0 ] = −2 6 14   1 0 2 ∼ 0 1 3 1 shows that x0 = 2v1 + 3v2 . Since v1 and v2 are eigenvectors (for the eigenvalues 2 and 1 respectively): x1 = Ax0 = A(2v1 + 3v2 ) = 2Av1 + 3Av2 = 2 · 2v1 + 3 · 1v2       1 −3 4 . = + = 10 18 −8 Consequently, x1 = 002  1 10  . 10.0 points If eigenvectors of an n × n matrix A are a basis for Rn , then A is diagonalizable. True or False? 1. TRUE correct 2. FALSE 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 . 003 10.0 points By diagonalizing the matrix   3 4 , A = −2 −3 compute f (A) for the polynomial f (x) = 2x3 − x2 + 3x − 2 . 1. f (A) =   12 20 correct −10 −18 Version 081 – harper (elh2395) – EXAM03 – pencheva – (54150)  −12 −10 −20 −18  3. f (A) =  −12 10  4. f (A) =  12 10 −20 18  20 18 2. f (A) = 2 while f (−1) = 2x − x + 3x − 2 3 2 x=−1 = −8 . Consequently,    1 2 0 2 1 f (A) = −1 0 −8 −1 −1   12 20 . = −10 −18 Explanation: If A can be diagonalized by   d1 0 −1 P −1 , A = P DP = P 0 d2 004 1 −2  10.0 points A subset H of a vector space V is a subspace of V if the following conditions are satisfied: then i) the zero vector of V is in H, −1 f (A) = P f (D)P   f (d1 ) 0 P −1 . = P 0 f (d2 ) 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. True or False? 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 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 . But 3 − λ det[A − λI] = −2 4 −3 − λ Consequently, the statement is = 8 − (3 − λ)(3 + λ) = λ2 − 1 = 0 , TRUE . i.e., λ1 = 1 and λ2 = −1. Corresponding eigenvectors are     1 2 , , v2 = v1 = −1 −1 005 10.0 points The vectors u1 and u2 shown in x2 so P =  2 −1 1 −1   1 −1 , P −1 =  f (1) 0 0 f (−1) 1 −2  . 2 u2 Thus f (A) =  2 −1 1 −1  1 −1 Now 1 −2  u1 . x1 2 3 2 f (1) = 2x − x + 3x − 2 x=1 = 2, Version 081 – harper (elh2395) – EXAM03 – pencheva – (54150) are eigenvectors corresponding to eigenvalues λ1 = −3 and λ2 = −1 respectively for a 2 × 2 matrix A. 3 x2 4. Which of the following graphs contains the vector A(u1 + u2 )? 2 x2 x1 2 1. 5. −2 correct −2 2 x1 2 x2 2. 6. 2 −2 −2 2 x1 Explanation: As the graph of u1 , u2 shows,     1 2 . u1 = u1 = 2 1 x2 3. 2 2 x1 But then A(u1 + u2 ) = A u1 + A u2 = λ1 u1 + λ2 u2       −7 −1 −6 . = + = −3u1 − u2 = −5 −2 −3 Consequently, A(u1 + u2 ) is contained in the graph Version 081 – harper (elh2395) – EXAM03 – pencheva – (54150) −2 −2 But when −3 −2 1 y = −9 , u1 = 2 , u2 = 2 , −6 1 −2 we see that  y · u1 ku1 k2  −2 u1 = −2 2 , 1 while 006 10.0 points Find the distance from −3 y = −9 −6  y · u2 ku2 k2 3. dist = 11 4. dist = 9 correct 5. dist = 10 Explanation: 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) where y − projW y is in the orthogonal complement W ⊥ . But then dist(y, W ) = ky − projW yk. Now u1 , u2 are non-zero othogonal vectors, so form a basis for W such that     y · u2 y · u1 u1 + u2 . projW y = ku1 k2 ku2 k2 1 u2 = − 2 . −2 projW −2 1 3 y = −2 2 − 2 = −6 , 1 −2 0 and so 1. dist = 7 2. dist = 8  Consequently, 3 to the plane in R spanned by −2 1 u1 = 2 , u2 = 2 . 1 −2 4 y − projW −3 3 2 = −9 − −6 = −3 1 . −6 0 2 Thus dist(y, W ) = 9 . 007 10.0 points Simplify the expression (4u − 3v) · (3u + 4v) − ku − 3vk2 for vectors u, v in Rn . 1. −11kuk2 + 31 u · v + 21kvk2 2. 11kuk2 + 13 u · v + 21kvk2 3. −11kuk2 + 31 u · v − 21kvk2 4. −11kuk2 − 13 u · v − 21kvk2 Version 081 – harper (elh2395) – EXAM03 – pencheva – (54150) 5. 11kuk2 + 13 u · v − 21kvk2 correct 3. c1 = 6. 11kuk2 + 31 u · v + 21kvk2 5 4 3 4. c1 = 0 Explanation: By linearity, 5. c1 = − (4u − 3v) · (3u + 4v) = 4u · (3u + 4v) − 3v · (3u + 4v) 6. c1 = = 12u · u + 16u · v − 9v · u − 12v · v , while 4 3 1 correct 3 Explanation: Since ku − 3vk2 = (u − 3v) · (u − 3v) u1 · u2 = u1 · u3 = u2 · u3 = 0, = u · (u − 3v) − 3v · (u − 3v) = u · u − 3u · v − 3v · u + 9v · v . the vectors u1 , u2 , u3 are mutually orthogonal, hence form a basis for R3 as they are also non-zero. Thus by orthogonality, But y = proju1 y + proju2 y + proju3 y 2 2 v · u = u · v , u · u = kuk , v · v = kvk . = So after expansion the expression becomes so 12kuk2 + 7 u · v − 12kvk2 2 2 − (kuk − 6u · v + 9kvk ) = 11kuk2 + 13 u · v − 21kvk2 . 008 y · u2 y · u3 y · u1 u1 + u2 + u3 , u1 · u1 u2 · u2 u3 · u3 10.0 points c1 = But y · u1 , u1 · u1 c2 = y · u2 , u2 · u2 c3 = y · u3 . u3 · u3 y · u1 (0) + (6) + (0) = . u1 · u1 (9) + (9) + (0) Consequently, Given vectors −3 1 −4 u1 = −3 , u2 = −1 , u3 = 4 , 0 −2 −4 c1 = 009 in R3 , determine c1 so that 1 . 3 10.0 points y = c1 u1 + c2 u2 + c3 u3 If {x1 , x2 , x3 } is a linearly independent set and W = Span{x1 , x2 , x3 }, 0 y = −2 . 4 then any orthogonal set {v1 , v2 , v3 } in W is a basis for W . True or False? when 1. c1 = − 1 3 2. No such value of c1 exists. 1. FALSE correct 2. TRUE Version 081 – harper (elh2395) – EXAM03 – pencheva – (54150) 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 . 010 10.0 points Use the fact that 1 −4 9 −7 A = −1 2 −4 1 5 −6 10 7 1 0 −1 5 ∼ 0 −2 5 −6 0 0 0 0 to determine an orthogonal basis for Col(A). −4 1 1. 2 , −1 −6 5 1 1 2. −1 , −4 5 −1 1 −4 3. −1 , 1 correct 5 1 −4 1 4. 2 , −1 −6 −1 Explanation: The pivot columns of A provide a basis for Col(A). But by row reduction, 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 Thus the pivot columns of A are 1 −4 a1 = −1 , a2 = 2 . 5 −6 6 We apply Gram-Schmidt to produce an orthogonal basis: set u1 = a1 and u 2 = a2 −  a2 · u 1 ku1 k2  u1 −4 1 (−36) −1 = 2 − 27 −6 5 −4 4/3 −8/3 = 2 + −4/3 = 2/3 . −6 20/3 2/3 Consequently, the set of vectors 1 −1 , 5 −4 1 1 is an orthogonal basis for Col(A). 011 10.0 points Let u(t) satisfy du = Au(t), dt   3 . u(0) = 5 Compute u(2) when A is a 2 × 2 matrix with eigenvalues 5 and 2 and corresponding eigenvectors     −1 −1 . , v2 = v1 = −3 1 1. u(2) =  e10 − 2e4 −e10 − 6e4  2. u(2) =  −e10 + 2e4 −e10 − 6e4  3. u(2) =  −e10 + 2e4 e10 + 6e4  4. u(2) =  −e10 − 2e4 e10 − 6e4  Version 081 – harper (elh2395) – EXAM03 – pencheva – (54150) 5. u(2) =  e10 − 2e4 e10 + 6e4 6. u(2) =  e10 + 2e4 −e10 + 6e4 is invertible.  True or False?  1. FALSE correct 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 −1 3 [ v1 v2 u(0) ] = 1 −3 5   1 0 −1 . ∼ 0 1 −2 2. TRUE correct 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× 4−2− 3×4 = 2. Consequently, the statement is This shows that c1 = −1, c2 = −2 and u(0) = −v1 − 2v2 . Since v1 and v2 are eigenvectors corresponding to the eigenvalues 5 and 2 respectively, set u(t) = −e5t v1 − 2e2t v2 . Then u(0) is the given initial value and Au(t) = −e5t Av1 − 2e2t Av2   du(t) = − 5e5t v1 − 2 2e2t v2 = . dt Thus u(t) = −e5t v1 − 2e2t v2 solves the given differential equation. Consequently,  10  e + 2e4 u(2) = . −e10 + 6e4 012 10.0 points The matrix 2 A = 0 1 7 1 3 4 −2 0 2 TRUE . ...
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