# HW 10-solutions.pdf - yick(tay236 HW 10 gilbert(53415 This...

• Homework Help
• 15
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 1 out of 15 pages. Unformatted text preview: yick (tay236) – HW 10 – gilbert – (53415) This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 2 1 −1 4 10.0 points An n × n matrix can be diagonalizable, but not invertible. True or False?  is diagonalizable, i.e., A = P DP −1 with P invertible and D diagonal, which of the following is a choice for D?   −3 0 1. D = 0 2   3 0 2. D = 0 3   3 0 3. D = 0 2   −3 0 4. D = 0 −3 5. A is not diagonalizable correct Explanation: Since det [A − λI] = A is not diagonalizable . 002 If the matrix A = Consequently, 10.0 points  1 1. FALSE 2. TRUE correct Explanation: Consider the 3 × 3 triangular matrix 5 −8 1 A = 0 0 7 0 0 −2 Because A is triangular, its eigenvalues are the entries along the diagonal, i.e., λ = 5, 0, −2. Since these are distinct, A is diagonalizable. On the other hand, one of its eigenvalues is zero, so A is not invertible (or note that det[A] = 0 because det[A] = 5(0)(−2) = 0  2−λ 1 −1 4−λ  is the product of the diagonal values of A, so A is not invertible). Therefore, an n × n matrix A can be diagonalizable, but not invertible. = 1 + (2 − λ)(4 − λ) = 9 − 6λ + λ2 , Consequently, the statement is the eigenvalues of A are the solutions of 9 − 6λ + λ2 = (3 − λ)2 = 0 , TRUE . i.e., λ = 3, 3. On the other hand, when λ = 3,    1 −1 −1 = rref(A − λI) = rref 0 1 1 003 1 0  , so x2 is the only free variable. Thus the eigenspace Nul(A − 3I) has dimension 1. But then, when λ = 3, geo multA (λ) < alg multA (λ) . 10.0 points If the matrix A =  1 1 4 −2  is diagonalizable, i.e., A = P DP −1 with P invertible and D diagonal, which of the following is a choice for P ? yick (tay236) – HW 10 – gilbert – (53415) 1. P =  2. P =  2 −1 −1 2 4 1  4 1  which can be written as x1 + 4x2 = 2x1 , 4. P = −1 1 5. P =  −1 1  4 correct 1  1 4 Consequently, A = P DP −1 with P = Explanation: Since 004 det [A − λI] =  1−λ 1 4 −2 − λ  = −4 − (1 − λ)(2 + λ) = λ2 + λ − 6 , the eigenvalues of A are the solutions of λ2 + λ − 6 = (λ + 3)(λ − 2) = 0 , i.e., λ = −3, 2. Thus A is diagonalizable because the eigenvalues of A are distinct, and A = P DP −1 with which can be written as −1 1 4 1  . 10.0 points 1. FALSE correct 2. TRUE Explanation: If an n×n matrix has n distinct eigenvalues, it is diagonalizable, but the converse does not necessarily have to be true. For example, when 1 A = −3 3 where v1 and v2 are eigenvectors corresponding to λ1 and λ2 respectively. To determine v1 and v2 we solve the equation Ax = λx. Ax = λx, λ = −3:        x x1 + 4x2 x1 1 4 = −3 1 , = x2 x1 − 2x2 x2 1 −2  If an n × n matrix A is diagonalizable, then A has n distinct eigenvalues. True or False? P = [v1 v2 ] x1 + 4x2 = −3x1 , x1 − 2x2 = 2x2 , i.e., x1 = 4x2 . So one choice of v2 is   4 . v2 = 1 3. A is not diagonalizable  2 3 −5 3 then 3 −3 , 1 det[A − λI] = −(λ − 1)(λ + 2)2 = 0. Thus the eigenvalues of A are λ = 1, −2, −2. x1 − 2x2 = −3x2 , i.e., x1 = −x2 . So one choice of v1 is   −1 . v1 = 1 Ax = λx, λ = 2:        x x1 + 4x2 x1 1 4 = 2 1 , = x2 x1 − 2x2 x2 1 −2 Now so 1 rref(A − I) = 0 0 0 −1 1 1 , 0 0 1 Nul(A − I) = s −1 : s in R . 1 yick (tay236) – HW 10 – gilbert – (53415) On the other hand, so 1 1 rref(A + 2I) = 0 0 0 0 1 0, 0 Nul(A + 2I) −1 −1 = s 1 + t 0 : s, t in R . 0 1 But then A is diagonalizable because it has 3 linearly independent eigenvectors −1 −1 1 −1 , 1 , 0 . 0 1 1 Since the eigenvalue λ = −2 is repeated, however, A does not have distinct eigenvalues. Consequently, the statement is FALSE . 005 10.0 points The eigenvalues of the matrix 2 0 −2 A = 1 3 2 0 0 3 are λ = 2, 3, 3. If A is diagonalizable, i.e., A = P DP −1 with P invertible and D diagonal, which of the following is a choice for P ? 1 0 −2 1. P = 1 1 0 0 0 1 2. A is not diagonalizable −1 3. P = 1 0 0 1 0 2 0 1 3 1 0 2 4. P = 1 1 0 0 0 1 −1 0 −2 5. P = 1 1 0 correct 0 0 1 Explanation: We first determine the eigenspaces corresponding to λ = 2, 3: λ = 2 : since 0 0 −2 rref(A − 2I) = rref 1 1 2 0 0 1 1 1 0 = 0 0 1, 0 0 0 there is only free variable and −1 Nul(A − 2I) = s 1 : s in R . 0 Thus Nul(A − 2I) has dimension 1, and −1 v1 = 1 0 is a basis for Nul(A − 2I). λ = 3 : since −1 0 rref(A − 3I) = rref 1 0 0 0 1 0 2 = 0 0 0, 0 0 0 −2 2 0 there are two free variables and −2 0 Nul(A−3I) = s 1 + t 0 : s, t in R . 0 1 Thus Nul(A − 3I) has dimension 2, and −2 0 v2 = 1 , v3 = 0 , 0 1 yick (tay236) – HW 10 – gilbert – (53415) form a basis for Nul(A − 3I). Consequently, A is diagonalizable because v1 , v2 , v3 are linearly independent, and A = P DP −1 with −1 0 −2 P = [v1 v2 v2 ] = 1 1 0 . 0 0 1 006 4 On the other hand, one of its eigenvalues is zero, so A is not invertible (or note that det[A] = 0 because det[A] is the product 5(0)(−2) = 0 of the diagonal values of A, hence not invertible). Therefore, an n × n matrix A can be diagonalizable, but not invertible. Consequently, the statement is 10.0 points FALSE . Every n × n matrix A having linearly independent eigenvectors v1 , v2 , . . . , vn can be diagonalized. 008 By diagonalizing the matrix True or False? 1. TRUE correct A = 2. FALSE Explanation: An n × n matrix A can be diagonalized, i.e. written as A = P DP −1 for some invertible matrix P and diagonal matrix D, if and only if A has linearly independent eigenvectors v1 , v2 , . . . , vn . Consequently, the statement is TRUE . 007 10.0 points 10.0 points If A is a diagonalizable n × n matrix, then A is invertible. True or False? 1. TRUE 2. FALSE correct Explanation: Consider the 3 × 3 triangular matrix 5 −8 1 A = 0 0 7 0 0 −2 Because A is triangular, its eigenvalues are the entries along the diagonal, i.e., λ = 5, 0, −2. Since these are distinct, A is diagonalizable.  3 2  −4 , −3 compute f (A) for the polynomial f (x) = 3x3 − 3x2 + 3x − 2 . 1. f (A) = 2. f (A) = 3. f (A) = 4. f (A) =  13 −24 correct 12 −23   −13 24 −12 23   13 −24 −12 23   −13 24 12 −23  Explanation: If A can be diagonalized by A = P DP −1 = P  d1 0  0 P −1 , d2 then f (A) = P f (D)P −1   f (d1 ) 0 P −1 . = P 0 f (d2 ) yick (tay236) – HW 10 – gilbert – (53415) 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. FALSE 2. TRUE correct Explanation: The eigenspaces But 3 − λ det[A − λI] = 2 Nul(A − λ1 I), −4 −3 − λ Nul(A − λ1 I), Nul(A − λ2 I), Nul(A − λ3 I), i.e., λ1 = 1 and λ2 = −1. Corresponding eigenvectors are     1 2 , , v2 = v1 = 1 1 must be at least four. On the other hand, since A is 4 × 4, the sum of the dimensions of Nul(A − λ1 I), Nul(A − λ2 I), Nul(A − λ3 I), so P = 2 1 1 1  2 1  , P −1 =  1 −1 −1 2  must be at most four. Hence the sum of the dimensions of the three eigenspaces is exactly four. . Thus f (A) = Nul(A − λ2 I) must be at least one-dimensional. Thus the sum of the dimensions of = 8 − (3 − λ)(3 + λ) = λ2 − 1 = 0 ,  5 Consequently, the statement is  1 1 f (1) 0 0 f (−1)  1 −1 −1 2  . TRUE . Now f (1) = 3x3 − 3x2 + 3x − 2 x=1 = 1, while 3 2 f (−1) = 3x − 3x + 3x − 2 x=−1 Consequently,    1 1 0 2 1 f (A) = −1 0 −11 1 1   13 −24 . = 12 −23 009 010 Using the fact that ex = 1 + x + = −11 . −1 2  10.0 points If A is a 4 × 4 matrix having distinct eigenvalues λ1 , λ2 , λ3 such that Nul(A − λ3 I) is two-dimensionsional, then A is diagonalizable. True or False? 10.0 points 1 1 2 x + . . . + xn + . . . , 2! n! compute etA as a matrix-valued function of t when   −8 10 . A = −5 7 " 2t # −3t −3t 2t 2e − e 2(e − e ) 1. etA = e2t − e−3t 2e−3t − e2t 2. etA = " 2e−3t − e2t 2(e2t − e−3t ) e−3t − e2t 2e2t − e−3t correct 3. etA = " 2e2t + e2t 2(e−3t − e2t ) e2t − e−3t 2e−3t + e2t # # yick (tay236) – HW 10 – gilbert – (53415) 4. etA = " 2e−3t + e2t 2(e2t − e−3t ) e−3t − e2t 2e2t + e−3t is a diagonalization of the matrix # 3 A = −6 15 Explanation: If A can be diagonalized by   d1 0 −1 P −1 , A = P DP = P 0 d2 e tD = Pe P −1 = P  etd1 0 0 etd2  P 0 0 1 0 . 5 −4 1. d2 = 1, d3 = −4, then tA 6 −1 0 P = −3 0 . 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 0 1 0 0 = λ2 − λ − 6 = (λ + 3)(λ − 2) = 0 , P =  , P −1 Consequently,   −3t  e 2 1 tA e = 0 1 1 = " 2e−3t − e2t e −3t 011 2t −e = 0 e2t  1 −1  −1 2 1 −1 2e − e −3t . −1 2  # . 2(e2t − e−3t ) 2t  10.0 points Find a matrix P and d2 , d3 so that 3 0 0 P 0 d2 0 P −1 , d1 ≥ d2 ≥ d3 , 0 0 d3 0 3. d2 = 4, d3 = −1, 1 0 P = −3 0 0 so 2 1 1 1 1 P = 0 1 −3 i.e., λ1 = −3 and λ2 = 2. Corresponding eigenvectors are     1 2 , , v2 = v1 = 1 1  1 1 1 1 10 7 − λ 2. d2 = 1, d3 = −4, But −8 − λ det[A − λI] = −5 0 0 1 1 1 0 0 4. d2 = 1, d3 = −4, 1 P = −3 1 0 1 0 1 correct 5. d2 = 4, d3 = −1, 0 0 1 P = 0 1 −3 1 1 0 yick (tay236) – HW 10 – gilbert – (53415) 6. d2 = 4, d3 = −1, 1 0 P = −3 1 0 0 0 1 1 Explanation: The entries 3, d2 , d3 in the diagonal matrix are the respective eigenvalues λ1 , λ2 , λ3 of A. But 3 − λ 0 0 1−λ 0 det[A − λI] = −6 15 5 −4 − λ = −λ3 + 13λ − 12 = −(λ − 3)(λ − 1)(λ + 4) . So λ1 = 3, λ2 = 1, λ3 = −4. Now let u1 , u2 , u3 be eigenvectors of A corresponding to λ1 , λ2 , λ3 respectively. Since the eigenvalues are distinct, To determine u2 we row reduce A − λI with λ2 = 1: 2 0 0 rref(A − I) = rref −6 0 0 15 5 −5 1 0 0 = 0 1 −1 . 0 0 0 Thus 0 u2 = 1 . 1 To determine u3 we row reduce A − λI with λ3 = −4: 7 0 0 rref(A + 4I) = rref −6 5 0 15 5 0 −1 0 0 = 0 −1 0 . 0 0 0 Thus, finally, P = [u1 u2 u3 ] has orthogonal columns. λ1 0 A = P 0 λ2 0 0 is a diagonalization of A. 0 0 P −1 λ3 To determine u1 we row reduce A − λI with λ1 = 3: 0 0 0 rref(A − 3I) = rref −6 −2 0 15 5 −7 −3 −1 0 = 0 0 1. 0 0 0 Thus 1 u1 = −3 . 0 7 0 u3 = 0 . 1 Consequently, d2 = 1, d3 = −4 and 1 P = −3 0 012 0 0 1 0 . 1 1 10.0 points Find a matrix P so that   d1 0 P −1 , P 0 d2 d1 ≥ d2 is a diagonalization of the matrix   0 0 A= −8 −4  1 1. P = −2  0 correct 1 yick (tay236) – HW 10 – gilbert – (53415) 2. P 3. P 4. P 5. P 6. P So, P = [u1 u2 ] and  2 1 = 1 −3   −2 1 = 1 0   −1 −3 = −2 1   0 1 = 1 −2   0 1 = −1 2  A = P DP −1 is a diagonalization of A. Consequently, 0 D= 0 013 det(A − λI) = (0 − λ)(−4 − λ) = λ2 + 4λ = (λ + 0)(λ + 4) = 0.     0 0 λ1 0 . = 0 −4 0 λ2 Now to find the eigenvectors of A, we will solve for the nontrivial solution of the characteristic equation by row reducing the related augmented matrices:   0+0 0 0 [ A − λ1 I 0 ] = −8 −4 + 0 0     −2 −1 0 0 0 0 ∼ = 0 0 0 −8 −4 0   1 , =⇒ u1 = −2 D= while  0+4 0 0 −8 −4 + 4 0     1 0 0 4 0 0 ∼ = 0 0 0 −8 0 0   0 =⇒ u2 = . 1 [ A − λ2 I 0] =   1 0 , P = −2 −4  Explanation: To begin, we must find the eigenvectors and eigenvalues of A. To do this, we will use the characteristic equation, det(A − λI) = 0. That is, we will look for the zeros of the characteristic polynomial. So 8  0 1  . 10.0 points Let A be a 2 × 2 matrix with eigenvalues 4 and 2 and corresponding eigenvectors     2 −2 . , v2 = v1 = 1 −4 Let {xk } be a solution of the difference equation   −4 . xk+1 = Axk , x0 = 1 Compute x1 .   −10 1. x1 = −4   −10 2. x1 = 4  4 3. x1 = −10  10 4. x1 = −4    −4 correct 5. x1 = 10   −4 6. x1 = −10  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 yick (tay236) – HW 10 – gilbert – (53415) 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   −2 2 −4 [ v1 v2 x0 ] = −4 1 1   1 0 −1 ∼ 0 1 −3 shows that x0 = −v1 − 3v2 . Since v1 and v2 are eigenvectors (for the eigenvalues 4 and 2 respectively): x1 = Ax0 = A(−v1 − 3v2 ) = −Av1 − 3Av2 = −4v1 − 3 · 2v2       −4 −12 8 . = + = 10 −6 16 Consequently, x1 = 014  −4 10  . Let A be a 2 × 2 matrix with eigenvalues 5 and 3 and corresponding eigenvectors     4 −3 . , v2 = v1 = −13 12 Determine the solution {xk } of the difference equation   2 . xk+1 = Axk , x0 = −11 1. xk = −2 (5)k v1 + 2 (3)k v2 k 2. xk = −2 (5) v1 − (3) v2 correct 3. xk = −4 (5)k v1 − 2 (3)k v2 4. xk = −4 (5)k v1 − (3)k v2 5. xk = −2 (5)k v1 − 2 (3)k v2 6. xk = −4 (5)k v1 + (3)k v2 Explanation: Since v1 and v2 are eigenvectors corresponding to distinct eigenvalues of A, they form an eigenbasis for R2 . Thus x0 = c1 v1 + c2 v2 To compute c1 and c2 we apply row reduction to the augmented matrix   −3 4 2 [ v1 v2 x0 ] = 12 −13 −11   1 0 −2 . ∼ 0 1 −1 This shows that c1 = −2, c2 = −1 and x0 = −2v1 −v2 . Since v1 and v2 are eigenvectors corresponding to the eigenvalues 5 and 3 respectively, xk = Ak x0 = Ak (−2v1 − v2 ) 10.0 points k 9 = −2Ak v1 − Ak v2 = −2 (5)k v1 − (3)k v2 for k ≥ 0. It follows that Axk = −2 (5)k Av1 − (3)k Av2 = −2 (5)k+1 v1 − (3)k+1 v2 = xk+1 . This shows that {xk } solves the difference equation. Consequently, xk = −2 (5)k v1 − (3)k v2 . 015 10.0 points Let A be a 3 × 3 matrix with eigenvalues 3, 2, and −2 and corresponding eigenvectors 2 −3 1 v1 = 6 , v2 = −12 , v3 = 2 . 4 −15 −3 If {xk } is the solution of the difference equation 0 xk+1 = Axk , x0 = −2 , −4 yick (tay236) – HW 10 – gilbert – (53415) determine x1 . 12 1. x1 = −16 −8 8 2. x1 = −16 −12 −12 3. x1 = 16 8 8 4. x1 = 16 correct −12 12 5. x1 = 16 −8 −8 6. x1 = −16 12 Explanation: To find x1 we must compute Ax0 . First, we express express x0 in terms of v1 , v2 , and v3 : x0 = c1 v1 + c2 v2 + c3 v3 . This is certainly possible as the eigenvectors v1 , v2 , and v3 are linearly independent because the eigenvalues are distinct. Hence they form a basis for R3 . The row reduction 2 −3 1 0 [ v1 v2 v3 x0 ] = 6 −12 2 −2 4 −15 −3 −4 1 0 0 2 ∼ 0 1 0 1 0 0 1 −1 shows that x0 = 2v1 +v2 −v3 . But v1 , v2 and v3 are eigenvectors for the respective eigenvalues 3, 2 and −2, so x1 = Ax0 = A(2v1 + v2 − v3 ) = 2Av1 + Av2 − Av3 = 2 · (3)v1 + (2)v2 − (−2)v2 −6 2 8 12 = 36 + −24 + 4 = 16 . 24 −30 −6 −12 10 Consequently, 8 x1 = 16 . −12 016 10.0 points Let A be a 3 × 3 matrix with eigenvalues 2, −1, and −3 and corresponding eigenvectors 2 v1 = 2 , −6 3 v2 = 0 , −3 −3 v3 = −2 . 8 Determine the solution {xk } of the difference equation xk+1 = Axk , −6 x0 = −4 . 13 1. xk = 3 (2)k v1 − (−1)k v2 − (−3)k v3 2. xk = −3 (2)k v1 + (−1)k v2 − (−3)k v3 3. xk = 3 (2)k v1 − (−1)k v2 + (−3)k v3 4. xk = −3 (2)k v1 + (−1)k v2 + (−3)k v3 5. xk = 3 (2)k v1 + (−1)k v2 + (−3)k v3 6. xk = −3 (2)k v1 − (−1)k v2 − (−3)k v3 correct Explanation: Since v1 , v2 , and v3 are eigenvectors corresponding to distinct eigenvalues of A, they form an eigenbasis for R3 . Thus x0 = c1 v1 + c2 v2 + c3 v3 yick (tay236) – HW 10 – gilbert – (53415) To compute c1 , c2 , and c3 we apply row reduction to the augmented matrix 2 3 −3 −6 [ v1 v2 v3 x0 ] = 2 0 −2 −4 −6 −3 8 13 1 0 0 −3 ∼ 0 1 0 −1 . 0 0 1 −1 This shows that c1 = −3, c2 = −1, c3 = −1 and x0 = −3v1 − v2 − v3 . Since v1 , v2 , and v3 are eigenvectors corresponding to the eigenvalues 2, −1, −3 respectively, xk = Ak x0 = Ak (−3v1 − v2 − v3 ) = −3Ak v1 − Ak v2 − Ak v3 for k ≥ 0. It follows that Axk = −3 (2)k Av1 − (−1)k Av2 − (−3)k Av3 = −3 (2)k+1 v1 − (−1)k+1 v2 − (−3)k+1 v3 = xk+1 . This shows that {xk } solves the difference equation. Consequently, xk = −3 (2)k v1 − (−1)k v2 − (−3)k v3 . 10.0 points Let A be a 3 × 3 matrix with eigenvalues 3, 3 − 1 and corresponding eigenvectors 2 , and 2 −2 −1 −3 v1 = 4 , v2 = 3 , v3 = 7 . 2 −2 −3 Determine the solution {xk } of the difference equation −1 xk+1 = Axk , x0 = 1 . −2 1. xk = − (3)k v1 + 3 k 2. xk = (3) v1 + 3  3 k  1 k 2 v2 + 2 − 2  3 k 2 k − 12 v2 − 2  3 k 2 4. xk = − (3)k v1 − 3 correct 5. xk = (3)k v1 − 3  3 k 2 6. xk = − (3)k v1 + 3 v2 + 2 − 12  3 k 2 v3 v2 + 2 − 12 v2 − 2 − 12  3 k 2 k k v2 − 2 − 12 k v3 v3 k v3 Explanation: Since v1 , v2 , and v3 are eigenvectors corresponding to distinct eigenvalues of A, they form an eigenbasis for R3 . Thus x0 = c1 v1 + c2 v2 + c3 v3 = −3 (2)k v1 − (−1)k v2 − (−3)k v3 017 3. xk = (3)k v1 − 3 11 v3 v3 To compute c1 , c2 , and c3 we apply row reduction to the augmented matrix −2 −1 −3 −1 [ v1 v2 v3 x0 ] = 4 3 7 1 2 −2 −3 −2 1 0 0 −1 ∼ 0 1 0 −3 . 0 0 1 2 This shows that c1 = −1, c2 = −3, c3 = 2 and x0 = −v1 − 3v2 + 2v3 . Since v1 , v2 , and v3 are eigenvectors corresponding to the eigenvalues 3, 32 , − 12 respectively, xk = Ak x0 = Ak (−v1 − 3v2 + 2v3 ) = −Ak v1 − 3Ak v2 + 2Ak v3 k k = − (3)k v1 − 3 23 v2 + 2 − 12 v3 for k ≥ 0. It follows that Axk = − (3)k Av1 − 3 = − (3)k+1 v1 − 3 = xk+1 .   3 k 1 k Av + 2 − Av3 2 2 2   k+1 k+1 3 v2 + 2 − 21 v3 2 This shows that {xk } solves the difference equation. Consequently, xk = − (3)k v1 − 3  3 k 2 v2 + 2 − 12 k v3 . yick (tay236) – HW 10 – gilbert – (53415) 12 Thus 018 10.0 points Find the solution of the differential equation   du 0 = Au(t), u(0) = −8 dt when A is a 2 × 2 matrix with eigenvalues 3 and 2 and corresponding eigenvectors     2 2 . , v2 = v1 = −10 −8 1. u(t) = −8e3t v1 + 8e2t v2 2. u(t) = −4e3t v1 + 4e2t v2 correct u(t) = −4e3t v1 + 4e2t v2 solves the given differential equation. 019 Find the solution of the differential equation   du −1 = Au(t), u(0) = 0 dt when A is the matrix  −13 A= 8 3. u(t) = −4e3t v1 + 8e2t v2 4. u(t) = −8e3t v1 + 4e2t v2 5. u(t) = −4e3t v1 − 8e2t v2 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   2 2 0 [ v1 v2 u(0) ] = −8 −10 −8   1 0 −4 . ∼ 0 1 4 Since v1 and v2 are eigenvectors corresponding to the eigenvalues 3 and 2 respectively, 2t u(t) = −4e v1 + 4e v2 . Then u(0) is the given initial value and Au(t) = −4e3t Av1 + 4e2t Av2   du(t) = −4 3e3t v1 + 4 2e2t v2 = . dt   1. u(t) = 2e−t + 3e−5t 2e−t + 2e−5t 2. u(t) =  −2e−t − 3e−5t 2e−t + 2e−5t  3. u(t) =  −2e−t + 3e−5t 2e−t − 2e−5t  4. u(t) =  2e−t + 3e−5t −2e−t − 2e−5t  5. u(t) =  2e−t − 3e−5t −2e−t + 2e−5t  6. u(t) =  −2e−t − 3e−5t −2e−t − 2e−5t  This shows that c1 = −4, c2 = 4 and u(0) = −4v1 + 4v2 . −12 7  6. u(t) = −8e3t v1 − 4e2t v2 3t 10.0 points correct Explanation: Since −13 − λ det[A − λI] = 8 −12 7 −λ = (−13 − λ)(7 − λ) + 96 = λ2 + 6λ + 5 = (λ + 1)(λ + 5), yick (tay236) – HW 10 – gilbert – (53415) 13 the eigenvalues of A are λ1 = −1, λ2 = −5 and corresponding eigenvectors     3 1 , v2 = v1 = −2 −1 Find the solution of the differential equation   du −8 = Au(t), u(0) = −3 dt form a basis for R2 because λ1 6= λ2 . Thus when A is the matrix   1 1 5 A= 2 0 6 u(0) = c1 v1 + c2 v2 . To compute c1 and c2 we apply row reduction to the augmented matrix   1 3 −1 [ v1 v2 u(0) ] = −1 −2 0   1 0 2 . ∼ 0 1 −1 This shows that c1 = 2, c2 = −1 and Then u(0) is the given initial value and Au(t) = 2e−t Av1 − e−5t Av2   du(t) = 2 −e−t v1 − −5e−5t v2 = . dt Thus u(t) is a solution of the differential equation. But     3 1 −5t −t + (−)e u(t) = 2e −2 −1   2e−t − 3e−5t = . −2e−t + 2e−5t  1  1  1. u(t) = 3e3t − 5e 2 t −3e3t 2. u(t) =  3e3t − 5e 2 t 3e3t 3. u(t) =  3e3t + 5e 2 t 3e3t 4. u(t) =  −3e3t + 5e 2 t 3e3t 5. u(t) =  −3e3t + 5e 2 t −3e3t 6. u(t) =  −3e3t − 5e 2 t −3e3t u(0) = 2v1 − v2 . Since v1 and v2 are eigenvectors corresponding to the eigenvalues −1 and −5 respectively, set u(t) = 2e−t v1 − e−5t v2 . 1  1  1  1  correct Explanation: Since 1 −λ 2 det[A − λI] = 0 5 2 3 − λ = ( 21 − λ)(3 − λ) + 0 3 7 = λ2 − λ + = (λ − 3)(λ− 12 ), 2 2 Consequently, u(t) =  2e−t − 3e−5t −2e−t + 2e−5t  solves the given differential equation. the eigenvalues of A are λ1 = 3, λ2 = corresponding eigenvectors     −1 −1 , v2 = v1 = 0 −1 1 2 and form a basis for R2 because λ1 6= λ2 . Thus 020 10.0 points u(0) = c1 v1 + c2 v2 . yick (tay236) – HW 10 – gilbert – (53415) To compute c1 and c2 we apply row reduction to the augmented matrix   −1 −1 −8 [ v1 v2 u(0) ] = −1 0 −3   1 0 3 . ∼ 0 1...
View Full Document

• • • 