# HW07.pdf - janecka(cdj2326 u2013 HW 07 u2013 gilbert...

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This preview shows page 1 out of 12 pages. Unformatted text preview: janecka (cdj2326) – HW 07 – gilbert – (53435) This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consequently, A is not diagonalizable . 002 If the matrix 1 10.0 points If the matrix A =  2 1 −1 4  A = 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 2 3. A is not diagonalizable correct 4. D =  3 0 0 3  5. D =  −3 0 0 −3  det [A − λI] = 2−λ 1 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 ?   −1 1 1. P = 1 4   2 4 2. P = −1 1   −1 4 3. P = 2 1   −1 4 correct 4. P = 1 1 5. A is not diagonalizable Explanation: Since   −1 4−λ Explanation: Since  det [A − λI] = 2 = 1 + (2 − λ)(4 − λ) = 9 − 6λ + λ , 4 −2 − λ  the eigenvalues of A are the solutions of 9 − 6λ + λ2 = (3 − λ)2 = 0 , λ2 + λ − 6 = (λ + 3)(λ − 2) = 0 , i.e., λ = 3, 3. 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) . 1−λ 1 = −4 − (1 − λ)(2 + λ) = λ2 + λ − 6 , the eigenvalues of A are the solutions of On the other hand, when λ = 3,    1 −1 −1 = rref(A − λI) = rref 0 1 1  i.e., λ = −3, 2. Thus A is diagonalizable because the eigenvalues of A are distinct, and A = P DP −1 with P = [v1 v2 ] where v1 and v2 are eigenvectors corresponding to λ1 and λ2 respectively. To determine v1 and v2 we solve the equation Ax = λx. janecka (cdj2326) – HW 07 – gilbert – (53435) Ax = λx, λ = −3:        x x1 + 4x2 x1 1 4 = −3 1 , = x2 x1 − 2x2 x2 1 −2 which can be written as x1 + 4x2 = −3x1 , x1 − 2x2 = −3x2 , Ax = λx, λ = 2:        x x1 + 4x2 x1 1 4 = 2 1 , = x2 x1 − 2x2 x2 1 −2 which can be written as x1 − 2x2 = 2x2 , i.e., x1 = 4x2 . So one choice of v2 is   4 . v2 = 1 Consequently, A = P DP −1 with   −1 4 P = . 1 1 003 1 0 −2 3. P = 1 1 0 0 0 1 4. A is not diagonalizable i.e., x1 = −x2 . So one choice of v1 is   −1 . v1 = 1 x1 + 4x2 = 2x1 , 2 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 1 0 2 2. P = 1 1 0 0 0 1 −1 5. P = 1 0 0 −2 1 0 correct 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 janecka (cdj2326) – HW 07 – gilbert – (53435) Thus Nul(A − 3I) has dimension 2, and 0 −2 v2 = 1 , v3 = 0 , 0 1 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 1 004 10.0 points  −12 2. f (A) = 3   15 −12 correct 3. f (A) = 6 −3   −15 12 4. f (A) = −6 3  then f (A) = P f (D)P −1   f (d1 ) 0 P −1 . = P 0 f (d2 ) True or False? 2. TRUE correct 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 . 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 3 − λ det[A − λI] = 2 Consequently, the statement is TRUE . 005 10.0 points By diagonalizing the matrix   3 −4 , A = 2 −3 compute f (A) for the polynomial f (x) = 4x3 + 3x2 − x + 3 . 1. f (A) =  −15 6 12 −3  15 −6 Explanation: If A can be diagonalized by   d1 0 −1 P −1 , A = P DP = P 0 d2 Every n × n matrix A having linearly independent eigenvectors v1 , v2 , . . . , vn can be diagonalized. 1. FALSE 3 −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 so P =  2 1  2 1 1 1 1 1  , P −1 =  1 −1 −1 2  . Thus f (A) =  f (1) 0 0 f (−1)  1 −1 −1 2  . janecka (cdj2326) – HW 07 – gilbert – (53435) Now f (1) = 4x3 + 3x2 − x + 3 x=1 = 9, while f (−1) = 4x3 + 3x2 − x + 3 = 3. x=−1 Consequently,    1 9 0 2 1 f (A) = −1 0 3 1 1   15 −12 . = 6 −3 006 −1 2 = λ2 − λ − 3 = (λ − 1)(λ + 3) = 0 , 10.0 points 1 1 2 x + . . . + xn + . . . , 2! n! 2. etA = 2et − e−3t 2(e−3t − et ) et − e−3t 2e−3t − et # cor- i.e., λ1 = 1 and λ2 = −3. Corresponding eigenvectors are     1 2 , , v2 = v1 = 1 1 so P =  2 1 3. etA = tA 4. e = 2e−3t + e−3t 2(et − e−3t ) e−3t − et 2et + e−3t " 2et + e−3t 2(e−3t − et ) et − e−3t 2e−3t + et e = Pe P −1 = P  etd1 0 0 etd2  # P −1 , P −1 = " = 0 e−3t  1 −1  1 −1 . −1 2  2(e−3t − et ) et − e−3t 2e−3t − et 10.0 points d1 ≥ d2 is a diagonalization of the matrix   2 3 A= −4 −5   1 1  4 −3 −3 1. P = −4 .  −1 2 2et − e−3t Find a matrix P so that   d1 0 P −1 , P 0 d2 then tD  007 # Explanation: If A can be diagonalized by   d1 0 −1 P −1 , A = P DP = P 0 d2 tA 1 1 Consequently,   t 2 1 e tA e = 0 1 1 rect " −8 −7 − λ 5 − λ det[A − λI] = 4  compute etA as a matrix-valued function of t when   5 −8 . A = 4 −7 # " −3t 2e − et 2(et − e−3t ) tA 1. e = e−3t − et 2et − e−3t " 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 Using the fact that ex = 1 + x + 4 −1 2. P = 1  # . janecka (cdj2326) – HW 07 – gilbert – (53435) 1 3. P = −1  1 4. P = 1 So, P = [u1 u2 ] and  −3 correct 4  4 3 is a diagonalization of A. Consequently, 1 −1  3 4 −1 6. P = −1 A = P DP −1   −3 5. P = 4 −1 D= 0  008 = λ2 + 3λ + 2 = (λ + 1)(λ + 2) = 0. λ1 D= 0 0 λ2   −1 = 0  0 . −2 while  2+2 3 0 −4 −5 + 2 0     4 3 0 4 3 0 ∼ = 0 0 0 −4 −3 0   −3 . =⇒ u2 = 4 [ A − λ2 I 0] =  . 10.0 points The eigenvalues of an n × n matrix A are the entries on the main diagonal of A. True or False? 1. FALSE correct 2. TRUE Consequently, the statement is 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:   2+1 3 0 [ A − λ1 I 0 ] = −4 −5 + 1 0     −1 −1 0 3 3 0 ∼ = 0 0 0 −4 −4 0   1 , =⇒ u1 = −1  −3 4 Explanation: The eigenvalues of a triangular n × n matrix A are the entries on the main diagonal of A. This is not necessarily true for matrices that are not triangular. det(A − λI) = (2 − λ)(−5 − λ) + 12    1 0 , P = −1 −2   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 5 FALSE . 009 When 10.0 points  3 A= 0 6 1  find matrices D and P in a diagonalization of A given that λ1 > λ2 .     1 −3 −1 0 , P = 1. D = 0 −1 0 −3  −1 2. D = 0  3 3. D = 0   −1 0 , P = 0 −3   −1 0 , P = 0 1   1 3 0 , P = 4. D = 0 0 1  3 −1   3 correct −1 −3 −1  janecka (cdj2326) – HW 07 – gilbert – (53435)   0 3 0 , P = 5. D = −1 0 1  6. D =  −1 0  −1 3  0 0 , P = −1 −3  −1 3  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. det(A − λI) = (3 − λ)(1 − λ) = λ2 − 4λ + 3 = (λ − 3)(λ − 1) = 0. So  λ1 D= 0 0 λ2   3 = 0  0 . 1 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:   3−3 6 0 [ A − λ1 I 0 ] = 0 1−3 0     0 1 0 0 6 0 ∼ = 0 0 0 0 −2 0   −1 , =⇒ u1 = 0 while [ A − λ2 I 0]  2 6 0 0 = =⇒  3−1 6 0 = 0 1−1 0    −1 −3 0 0 ∼ 0 0 0 0   3 . u2 = −1  So, P = [u1 u2 ] and A = P DP −1 is a diagonalization of A. 6 Consequently,    −1 3 0 , P = D= 0 0 1 010 3 −1  . 10.0 points For n × n matrix A is said to be diagonalizable when A = P DP −1 for some matrix D and invertible matrix P . True or False? 1. TRUE 2. FALSE correct Explanation: An n × n matrix A is said to be diagonalizable when A can be factored as A = P DP −1 with D a DIAGONAL matrix and P an invertible matrix. Consequently, the statement is FALSE . 011 10.0 points If an n × n matrix A is diagonalizable, then A has n distinct eigenvalues. True or False? 1. TRUE 2. FALSE correct 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 then 3 −5 3 3 −3 , 1 det[A − λI] = −(λ − 1)(λ + 2)2 = 0. janecka (cdj2326) – HW 07 – gilbert – (53435) 7 Thus the eigenvalues of A are λ = 1, −2, −2. 1. FALSE Now so 1 rref(A − I) = 0 0 0 −1 1 1 , 0 0 1 Nul(A − I) = s −1 : s in R . 1 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 . 1 0 1 Since the eigenvalue λ = −2 is repeated, however, A does not have distinct eigenvalues. Consequently, the statement is FALSE . 2. TRUE correct Explanation: The eigenspace corresponding to λ3 must be at least one-dimensional because all eigenspaces must have a dimension at least one. But if this eigenspace is one-dimensional, then the sum of the dimensions of the three eigenspaces is six, which is less than seven, in which case A will not be diagonalizable. On the other hand, if the eigenspace corresponding to λ3 is two-dimensional, then the sum of the dimensions of the three eigenspaces is seven, in which case A will be diagonalizable. Consequently, the statement is TRUE . 013 10.0 points Find the solution of the differential equation   du 4 = Au(t), u(0) = −15 dt when A is a 2 × 2 matrix with eigenvalues 3 and 2 and corresponding eigenvectors     2 −2 . , v2 = v1 = −3 6 1. u(t) = −6e3t v1 − 2e2t v2 012 10.0 points 2. u(t) = −3e3t v1 − e2t v2 correct If A is a 7 × 7 matrix having eigenvalues λ1 , λ2 , and λ3 such that 3. u(t) = −3e3t v1 + 2e2t v2 (i) the eigenspace corresponding to λ1 is two-dimensional, 4. u(t) = −6e3t v1 − e2t v2 (ii) the eigenspace corresponding to λ2 is three-dimensional, 5. u(t) = −3e3t v1 − 2e2t v2 then A need not be diagonalizable. 6. u(t) = −6e3t v1 + e2t v2 True or False? Explanation: janecka (cdj2326) – HW 07 – gilbert – (53435) 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 4 [ v1 v2 u(0) ] = 6 −3 −15   1 0 −3 . ∼ 0 1 −1 This shows that c1 = −3, c2 = −1 and u(0) = −3v1 − v2 . Since v1 and v2 are eigenvectors corresponding to the eigenvalues 3 and 2 respectively, u(t) = −3e3t v1 − e2t v2 . Then u(0) is the given initial value and Au(t) = −3e3t Av1 − e2t Av2   du(t) = −3 3e3t v1 − 2e2t v2 = . dt Thus u(t) = −3e3t v1 − e2t v2 solves the given differential equation. 014 10.0 points Find the solution of the differential equation   du −2 = Au(t), u(0) = 6 dt when A is the matrix  5 A= −9 1. u(t) =  6 −10 −6e−t − 8e−4t 6e−t + 12e−4t   8 2. u(t) =  6e−t + 8e−4t −6e−t − 12e−4t  3. u(t) =  −6e−t − 8e−4t −6e−t − 12e−4t  4. u(t) =  6e−t + 8e−4t 6e−t + 12e−4t 5. u(t) =  6e−t − 8e−4t −6e−t + 12e−4t 6. u(t) =  −6e−t + 8e−4t 6e−t − 12e−4t   correct  Explanation: Since 5 − λ det[A − λI] = −9 6 −10 − λ = (5 − λ)(−10 − λ) + 54 = λ2 + 5λ + 4 = (λ + 1)(λ + 4), the eigenvalues of A are λ1 = −1, λ2 = −4 and corresponding eigenvectors     −2 −3 , v2 = v1 = 3 3 form a basis for R2 because λ1 6= λ2 . Thus u(0) = c1 v1 + c2 v2 . To compute c1 and c2 we apply row reduction to the augmented matrix   −3 −2 −2 [ v1 v2 u(0) ] = 3 3 6   1 0 −2 . ∼ 0 1 4 This shows that c1 = −2, c2 = 4 and u(0) = −2v1 + 4v2 . Since v1 and v2 are eigenvectors corresponding to the eigenvalues −1 and −4 respectively, set u(t) = −2e−t v1 + 4e−4t v2 . janecka (cdj2326) – HW 07 – gilbert – (53435) Then u(0) is the given initial value and 6. u(t) = Au(t) = −2e−t Av1 + 4e−4t Av2   du(t) = −2 −e−t v1 + 4 −4e−4t v2 = . dt Thus u(t) is a solution of the differential equation. But     −4t −2 −t −3 + (+4)e u(t) = −2e 3 3   6e−t − 8e−4t = . −6e−t + 12e−4t Consequently, u(t) =  6e−t − 8e−4t −6e−t + 12e−4t  solves the given differential equation. 015 10.0 points Find the solution of the differential equation   du −16 = Au(t), u(0) = −14 dt when A is the matrix  1 A= −1 12e3t + 4e2t 12e3t + 2e2t 2 4 1. u(t) = 2. u(t) =  12e − 4e 12e3t − 2e2t 3. u(t) =  −12e3t + 4e2t 12e3t − 2e2t 4. u(t) =   −12e3t + 4e2t −12e3t + 2e2t 5. u(t) =  12e3t − 4e2t −12e3t + 2e2t 2t −12e3t − 4e2t −12e3t − 2e2t  correct Explanation: Since 1 − λ det[A − λI] = −1 2 4 −λ = (1 − λ)(4 − λ) + 2 = λ2 − 5λ + 6 = (λ − 3)(λ−2), the eigenvalues of A are λ1 = 3, λ2 = 2 and corresponding eigenvectors     −2 −4 , v2 = v1 = −1 −4 form a basis for R2 because λ1 6= λ2 . Thus u(0) = c1 v1 + c2 v2 . To compute c1 and c2 we apply row reduction to the augmented matrix   −4 −2 −16 [ v1 v2 u(0) ] = −4 −1 −14   1 0 3 . ∼ 0 1 2 This shows that c1 = 3, c2 = 2 and u(0) = 3v1 + 2v2 .   3t  9  Since v1 and v2 are eigenvectors corresponding to the eigenvalues 3 and 2 respectively, set u(t) = 3e3t v1 + 2e2t v2 . Then u(0) is the given initial value and  Au(t) = 3e3t Av1 + 2e2t Av2     du(t) = 3 3e3t v1 + 2 2e2t v2 = . dt Thusu(t) solves the differential equation. But     2t −2 3t −4 + 2e u(t) = 3e −1 −4   −12e3t − 4e2t = −12e3t − 2e2t janecka (cdj2326) – HW 07 – gilbert – (53435) 10 This shows that c1 = −2, c2 = 1 and Consequently, u(t) =  −12e3t − 4e2t −12e3t − 2e2t  solves the given differential equation. 016 u(0) = −2v1 + v2 . Since v1 and v2 are eigenvectors corresponding to the eigenvalues −1 and −2 respectively, set 10.0 points u(t) = −2e−t v1 + e−2t v2 . Let u(t) satisfy du = Au(t), dt u(0) =   9 . −12 Compute u(2) when A is a 2 × 2 matrix with eigenvalues −1 and −2 and corresponding eigenvectors     3 −3 . , v2 = v1 = −6 3   −6e−2 − 3e−4 1. u(2) = 6e−2 + 6e−4 2. u(2) =  6e−2 − 3e−4 6e−2 − 6e−4 3. u(2) =  −6e−2 + 3e−4 −6e−2 + 6e−4  4. u(2) =  6e + 3e −6e−2 − 6e−4  5. u(2) =  −6e−2 + 3e−4 6e−2 − 6e−4  6. u(2) =  6e−2 − 3e−4 −6e−2 + 6e−4  −2  −4 Then u(0) is the given initial value and Au(t) = −2e−t Av1 + e−2t Av2   du(t) = −2 −e−t v1 + −2e−2t v2 = . dt Thus u(t) = −2e−t v1 + e−2t v2 solves the given differential equation. Consequently, u(2) = 017 correct To compute c1 and c2 we apply row reduction to the augmented matrix   −3 3 9 [ v1 v2 u(0) ] = 3 −6 −12   1 0 −2 . ∼ 0 1 1 6e−2 + 3e−4 −6e−2 − 6e−4  . 10.0 points Find the solution of the differential equation   du 2 = Au(t), u(0) = −5 dt when A is the matrix 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   1 6 A= 2 −6 3 1. u(t) =  e2t + e 2 t 3 −2e2t − 3e 2 t 2. u(t) =  −e2t + e 23t 2e2t − 3e 2 t 3. u(t) =  e2t − e 2 t3 2e2t − 3e 2 t 3 3   1 1   correct janecka (cdj2326) – HW 07 – gilbert – (53435) 3 4. u(t) =  e2t − e 2 t 3 −2e2t + 3e 2 t 5. u(t) =  −e2t − e 23t 2e2t + 3e 2 t 6. u(t) =  −e2t + e 2 t3 −2e2t + 3e 2 t 3 This shows that u(t) solves the differential equation. But     3 −1 2t −1 t − e2 u(t) = −e 2 3   3 e2t + e 2 t 3 . = −2e2t − 3e 2 t   3  Consequently, Explanation: Since u(t) = 3 −λ det[A − λI] = −3 1 2 1 2 −λ the eigenvalues of A are λ1 = 2, λ2 = corresponding eigenvectors     −1 −1 , v2 = v1 = 3 2  3 e2t + e 2 t 3 −2e2t − 3e 2 t  solves the given differential equation. = (3 − λ)( 12 − λ) + 6 7 = λ2 − λ + 3 = (λ − 2)(λ− 32 ), 2 3 2 018 and form a basis for R2 because λ1 6= λ2 . Thus u(0) = c1 v1 + c2 v2 . To compute c1 and c2 we apply row reduction to the augmented matrix   −1 −1 2 [ v1 v2 u(0) ] = 2 3 −5   1 0 −1 . ∼ 0 1 −1 This shows that c1 = −1, c2 = −1 and u(0) = −v1 − v2 . Since v1 and v2 are eigenvectors corresponding to the eigenvalues 2 and 23 respectively, set 3 u(t) = −e2t v1 − e 2 t v2 . Then u(0) is the given initial value and 3 11 Au(t) = −e2t Av1 − e 2 t Av2  3   du(t) 2t . = − 2e v1 − 32 e 2 t v2 = dt 10.0 points Let A be a 2 × 2 matrix with eigenvalues −1 and −3 and corresponding eigenvectors     1 4 . , v2 = v1 = 1 −4 Let {xk } be a solution of the difference equation   6 . xk+1 = Axk , x0 = −2 Compute x1 .   10 1. x1 = 2   2 2. x1 = −10   −10 3. x1 = 2   −2 4. x1 = −10   2 5. x1 = 10   −10 correct 6. x1 = −2 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 janecka (cdj2326) – HW 07 – gilbert – (53435) 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   4 1 6 [ v1 v2 x0 ] = −4 1 −2   1 0 1 ∼ 0 1 2 shows that x0 = v1 + 2v2 . Since v1 and v2 are eigenvectors (for the eigenvalues −1 and −3 respectively): x1 = Ax0 = A(v1 + 2v2 ) = Av1 + 2Av2 = −1v1 + 2 · −3v2       −10 −6 −4 . = + = −2 −6 4 Consequently, x1 = 019  −10 −2  . 10.0 points Let A be a 3 × 3 matrix with eigenvalues 3, −1, and −2 and corresponding eigenvectors 1 2 3 v1 = 1 , v2 = 4 , v3 = 4 . 2 0 2 If {xk } is the solution of the difference equation 3 xk+1 = Axk , x0 = 3 , 2 determine x1 . 14 1. x1 = −15 13 13 2. x1 = 15 14 −14 3. x1 = −15 −13 12 −13 4. x1 = 15 −14 −13 5. x1 = −15 correct −14 14 6. x1 = 15 13 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 1 2 3 3 [ v1 v2 v3 x0 ] = 1 4 4 3 2 0 2 2 1 0 0 −1 ∼ 0 1 0 −1 0 0 1 2 shows that x0 = −v1 − v2 + 2v3 . But v1 , v2 and v3 are eigenvectors for the respective eigenvalues 3, −1 and −2, so x1 = Ax0 = A(−v1 − v2 + 2v3 ) = −Av1 − Av2 + 2Av3 = −(3)v1 − (−1)v2 + 2 · (−2)v2 −3 2 −12 −13 = −3 + 4 + −16 = −15 . −6 0 −8 −14 Consequently, −13 x1 = −15 . −14 ...
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