Solution Manual Advanced Modern Engineering Mathematics 4th Edition By Glyn James

Chapter 1. Matrix Analysis 1 Chapter 2. Numerical Solution of Ordinary Differential Equations 86 Chapter 3. Vector Calculus 126 Chapter 4. Functions of a Complex Variable 194 Chapter 5. Laplace Transforms 270 Chapter 6. The z Transform 369 Chapter 7. Fourier Series 413 Chapter 8. The Fourier Transform 489 Chapter 9. Partial Differential Equations 512 Chapter 10. Optimization 573 Chapter 11. Applied Probability and Statistics 639 iii1 Matrix Analysis Exercises 1.3.3 1(a) Yes, as the three vectors are linearly independent and span threedimensional space. 1(b) No, since they are linearly dependent ⎡⎣ 325 ⎤⎦ − 2⎡ ⎣ 1 0 1 ⎤ ⎦ = ⎡ ⎣ 1 2 3⎤ ⎦ 1(c) No, do not span three-dimensional space. Note, they are also linearly dependent. 2 Transformation matrix is A = √12 ⎡ ⎣ 1 1 0 1 0 0 −1 0 √2 ⎤ ⎦ ⎡ ⎣ 1 0 0 0 1 0 0 0 1 ⎤ ⎦ = ⎡ ⎣ √ √1 10 0 1 2 2 − √ √112 2 0 0 ⎤ ⎦ Rotates the (e1,e2) plane through π/4 radians about the e3 axis. 3 By checking axioms (a)–(h) on p. 10 it is readily shown that all cubics ax3 + bx2 + cx + d form a vector space. Note that the space is four dimensional. 3(a) All cubics can be written in the form ax3 + bx2 + cx + d and {1, x, x2, x3} are a linearly independent set spanning four-dimensional space. Thus, it is an appropriate basis. ©c Pearson Education Limited 20112 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 3(b) No, does not span the required four-dimensional space. Thus a general cubic cannot be written as a linear combination of (1 − x), (1 + x), (1 − x3), (1 + x3) as no term in x2 is present. 3(c) Yes as linearly independent set spanning the four-dimensional space a(1 − x) + b(1 + x) + c(x2 − x3) + d(x2 + x3) = (a + b) + (b − a)x + (c + a)x2 + (d − c)x3 ≡ α + βx + γx2 + δx3 3(d) Yes as a linear independent set spanning the four-dimensional space a(x − x2) + b(x + x2) + c(1 − x3) + d(1 + x3) = (a + b) + (b − a)x + (c + d)x2 + (d − c)x3 ≡ α + βx + γx2 + δx3 3(e) No not linearly independent set as (4x3 + 1) = (3x2 + 4x3) − (3x2 + 2x) + (1 + 2x) 4 x + 2x3, 2x − 3x5, x + x3 form a linearly independent set and form a basis for all polynomials of the form α + βx3 + γx5. Thus, S is the space of all odd quadratic polynomials. It has dimension 3. ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 3 Exercises 1.4.3 5(a) Characteristic polynomial is λ3 − p1λ2 − p2λ − p3 with p1 = trace A = 12 B1 = A − 12I = ⎡ ⎣ −4 2 3 9 2 1 −7 − −1 8 ⎤ ⎦ A2 = A B1 = ⎡ ⎣ − −17 18 2 −− −30 7 5 5 −−33 7 ⎤ ⎦ p2 = 12 trace A2 = −40 B2 = A2 + 40I = ⎡ ⎣ − − −5 5 7 −7⎤ ⎦ A3 = A B2 = ⎡ ⎣ 35 0 0 0 35 0 0 0 35 ⎤ ⎦ p3 = 13 trace A3 = 35 Thus, characteristic polynomial is λ3 − 12λ2 + 40λ − 35 Note that B3 = A3 − 35I = 0 confirming check. 5(b) Characteristic polynomial is λ4 − p1λ3 − p2λ2 − p3λ − p4 with p1 = trace A = 4 B1 = A − 4I = ⎡⎢⎣ −2 −1 1 2 0 −3 1 0 −1 1 −3 1 1 1 1 −4 ⎤⎥⎦ A2 = A B1 = ⎡⎢⎣ −3 4 0 −3 −1 −2 −2 1 2 0 −2 −5 −3 −3 −1 3 ⎤⎥⎦ ⇒ p2 = 12 trace A2 = −2 ©c Pearson Education Limited 20114 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition B2 = A2 + 2I = ⎡⎢⎣ −1 4 0 −3 −1 0 −2 1 2 0 0 −5 −3 −3 −1 5 ⎤⎥⎦ A3 = A B2 = ⎡⎢⎣ −5 2 0 −2 1 0 −2 −4 −1 −7 −3 4 0 4 −2 −7 ⎤⎥⎦ ⇒ p3 = 13 trace A3 = −5 B3 = A3 + 5I = ⎡⎢⎣ 0 0 0 −2 1 5 −2 −4 −1 −8 2 4 0 4 −2 −2 ⎤⎥⎦ A4 = A B3 = ⎡⎢⎣ −2 0 0 0 0 −2 0 0 0 0 −2 0 0 0 0 −2 ⎤⎥⎦ ⇒ p4 = 14 trace A4 = −2 Thus, characteristic polynomial is λ4 − 4λ3 + 2λ2 + 5λ + 2 Note that B4 = A4 + 2I = 0 as required by check. 6(a) Eigenvalues given by ¯ ¯ 1−1λ 1−1λ¯ ¯ = λ2 − 2λ = λ(λ − 2) = 0 so eigenvectors are λ1 = 2,λ2 = 0 Eigenvectors given by corresponding solutions of (1 − λi)ei1 + ei2 = 0 ei1 + (1 − λi)ei2 = 0 Taking i = 1,2 gives the eigenvectors as e1 = [1 1]T,e2 = [1 − 1]T (1) 6(b) Eigenvalues given by ¯ ¯ 1−3λ 2−2λ¯ ¯ = λ2 − 3λ − 4 = (λ + 1)(λ − 4) = 0 so eigenvectors are λ1 = 4,λ2 = −1 Eigenvectors given by corresponding solutions of (l − λi)ei1 + 2ei2 = 0 3ei1 + (2 − λi)ei2 = 0 ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 5 Taking i = 1,2 gives the eigenvectors as e1 = [2 3]T,e2 = [1 − 1]T 6(c) Eigenvalues given by ¯¯¯¯¯¯ 1 − λ 0 −4 0 5 − λ 4 −4 4 3 − λ ¯¯¯¯¯¯ = λ3 + 9λ2 + 9λ − 81 = (λ − 9)(λ − 3)(λ + 3) = 0 So the eigenvalues are λ1 = 9,λ2 = 3,λ3 = −3. The eigenvectors are given by the corresponding solutions of (1 − λi)ei1 + 0ei2 − 4ei3 = 0 0ei1 + (5 − λi)ei2 + 4ei3 = 0 −4ei1 + 4ei2 + (3 − λi)ei3 = 0 Taking i = 1,λi = 9 solution is e11 8 = − e12 16 = e13 −16 = β1 ⇒ e1 = [−1 2 2]T Taking i = 2,λi = 3 solution is e21 −16 = − e22 16 = e23 8 = β2 ⇒ e2 = [2 2 − 1]T Taking i = 3,λi = −3 solution is e31 32 = − e32 16 = e33 32 = β3 ⇒ e3 = [2 − 1 2]T 6(d) Eigenvalues given by ¯¯¯¯¯¯ 1 − λ 1 2 0 2 − λ 2 −1 1 3 − λ ¯¯¯¯¯¯ = 0 Adding column 1 to column 2 gives ¯¯¯¯¯¯ 1 − λ 2 − λ 2 0 2 − λ 2 −1 0 3 − λ ¯¯¯¯¯¯ = (2 − λ) ¯¯¯¯¯¯ 1 − λ 1 2 0 1 2 −1 0 3 − λ ¯¯¯¯¯¯ R1−R2(2 − λ) ¯¯¯¯¯¯ 1 − λ 0 0 0 1 2 −1 0 3 − λ ¯¯¯¯¯¯ = (2 − λ)(1 − λ)(3 − λ) ©c Pearson Education Limited 20116 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition so the eigenvalues are λ1 = 3,λ2 = 2,λ3 = 1. Eigenvectors are the corresponding solutions of (A − λiI)ei = 0 When λ = λ1 = 3 we have ⎡⎣ −2 1 2 0 −1 2 −1 1 0 ⎤⎦ ⎡⎣ e11 e12 e13 ⎤⎦ = 0 leading to the solution e11 −2 = − e12 2 = e13 −1 = β1 so the eigenvector corresponding to λ2 = 3 is e1 = β1[2 2 1]T,β1 constant. When λ = λ2 = 2 we have ⎡⎣ −1 1 2 0 0 2 −1 1 1 ⎤⎦ ⎡⎣ e21 e22 e23 ⎤⎦ = 0 leading to the solution e21 −2 = − e22 2 = e23 0 = β3 so the eigenvector corresponding to λ2 = 2 is e2 = β2[1 1 0]T,β2 constant. When λ = λ3 = 1 we have ⎡⎣ 0 1 2 0 1 2 −1 1 2 ⎤⎦ ⎡⎣ e31 e32 e33 ⎤⎦ = 0 leading to the solution e31 0 = − e32 2 = e33 1 = β1 so the eigenvector corresponding to λ3 = 1 is e3 = β3[0 − 2 1]T,β3 constant. 6(e) Eigenvalues given by ¯¯¯¯¯¯ 5 − λ 0 6 0 11 − λ 6 6 6 −2 − λ ¯¯¯¯¯¯ = λ3 − 14λ2 − 23λ − 686 = (λ − 14)(λ − 7)(λ + 7) = 0 so eigenvalues are λ1 = 14,λ2 = 7,λ3 = −7 ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 7 Eigenvectors are given by the corresponding solutions of (5 − λi)ei1 + 0ei2 + 6ei3 = 0 0ei1 + (11 − λi)ei2 + 6ei3 = 0 6ei1 + 6ei2 + (−2 − λi)ei3 = 0 When i = 1,λ1 = 14 solution is e11 12 = −e12 −36 = e13 18 = β1 ⇒ e1 = [2 6 3]T When i = 2,λ2 = 7 solution is e21 −72 = −e22 −36 = e23 −24 = β2 ⇒ e2 = [6 − 3 2]T When i = 3,λ3 = −7 solution is e31 54 = −e32 −36 = e33 −108 = β3 ⇒ e3 = [3 2 − 6]T 6(f) Eigenvalues given by ¯¯¯¯¯¯ 1 − λ −1 0 1 2 − λ 1 −2 1 −1 − λ ¯¯¯¯¯¯ R1+R2 ¯¯¯¯¯¯ −1 − λ 0 −1 − λ 1 2 − λ 1 −2 1 −1 − λ ¯¯¯¯¯¯ = (1 + λ) ¯¯¯¯¯¯ −1 0 0 1 2 − λ 0 −2 1 1 − λ ¯¯¯¯¯¯ = 0, i.e. (1 + λ)(2 − λ)(1 − λ) = 0 so eigenvalues are λ1 = 2,λ2 = 1,λ3 = −1 Eigenvectors are given by the corresponding solutions of (1 − λi)ei1 − ei2 + 0ei3 = 0 ei1 + (2 − λi)ei2 + ei3 = 0 −2ei1 + ei2 − (1 + λi)ei3 = 0 Taking i = 1,2,3 gives the eigenvectors as e1 = [−1 1 1]T,e2 = [1 0 − 1]T,e3 = [1 2 − 7]T ©c Pearson Education Limited 20118 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 6(g) Eigenvalues given by ¯¯¯¯¯¯ 4 − λ 1 1 2 5 − λ 4 −1 −1 −λ ¯¯¯¯¯¯ R1 + (R2 + R3) ¯¯¯¯¯¯ 5 − λ 5 − λ 5 − λ 2 5 − λ 4 −1 −1 −λ ¯¯¯¯¯¯ = (5 − λ) ¯¯¯¯¯¯ 1 0 0 2 3 − λ 2 −1 0 1 − λ ¯¯¯¯¯¯ = (5 − λ)(3 − λ)(1 − λ) = 0 so eigenvalues are λ1 = 5,λ2 = 3,λ3 = 1 Eigenvectors are given by the corresponding solutions of (4 − λi)ei1 + ei2 + ei3 = 0 2ei1 + (5 − λi)ei2 + 4ei3 = 0 −ei1 − ei2 − λiei3 = 0 Taking i = 1,2,3 and solving gives the eigenvectors as e1 = [2 3 − 1]T,e2 = [1 − 1 0]T,e3 = [0 − 1 1]T 6(h) Eigenvalues given by ¯¯¯¯¯¯ 1 − λ −4 −2 0 3 − λ 1 1 2 4 − λ ¯¯¯¯¯¯ R1+2R2 ¯¯¯¯¯¯ 1 − λ 2 − 2λ 0 0 3 − λ 1 1 2 4 − λ ¯¯¯¯¯¯ = (1 − λ) ¯¯¯¯¯¯ 1 0 0 0 3 − λ 1 1 0 4 − λ ¯¯¯¯¯¯ = (1 − λ)(3 − λ)(4 − λ) = 0 so eigenvalues are λ1 = 4,λ2 = 3,λ3 = 1 Eigenvectors are given by the corresponding solutions of (1 − λi)ei1 − 4ei2 − 2ei3 = 0 2ei1 + (3 − λi)ei2 + ei3 = 0 ei1 + 2ei2 + (4 − λi)ei3 = 0 ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 9 Taking i = 1,2,3 and solving gives the eigenvectors as e1 = [2 − 1 − 1]T,e2 = [2 − 1 0]T,e3 = [4 − 1 − 2]T Exercises 1.4.5 7(a) Eigenvalues given by ¯¯¯¯¯¯ 2 − λ 2 1 1 3 − λ 1 1 2 2 − λ ¯¯¯¯¯¯ R1−R2 ¯¯¯¯¯¯ 1 − λ −1 + λ 0 0 3 − λ 1 1 2 2 − λ ¯¯¯¯¯¯ = (1 − λ) ¯¯¯¯¯¯ 1 0 0 1 4 − λ 1 1 3 2 − λ ¯¯¯¯¯¯ = (1 − λ)[λ2 − 6λ + 5] = (1 − λ)(λ − 1)(λ − 5) = 0 so eigenvalues are λ1 = 5,λ2 = λ3 = 1 The eigenvectors are the corresponding solutions of (2 − λi)ei1 + 2ei2 + ei3 = 0 ei1 + (3 − λi)ei2 + ei3 = 0 ei1 + 2ei2 + (2 − λi)ei3 = 0 When i = 1,λ1 = 5 and solution is e11 4 = −e12 −4 = e13 4 = β1 ⇒ e1 = [1 1 1]T When λ2 = λ3 = 1 solution is given by the single equation e21 + 2e22 + e23 = 0 Following the procedure of Example 1.6 we can obtain two linearly independent solutions. A possible pair are e2 = [0 1 2]T , e3 = [1 0 − 1]T ©c Pearson Education Limited 201110 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 7(b) Eigenvalues given by ¯¯¯¯¯¯ −λ −2 −2 −1 1 − λ 2 −1 −1 2 − λ ¯¯¯¯¯¯ = −λ3 + 3λ2 − 4 = −(λ + 1)(λ − 2)2 = 0 so eigenvalues are λ1 = λ2 = 2,λ3 = −1 The eigenvectors are the corresponding solutions of −λiei1 − 2ei2 − 2ei3 = 0 −ei1 + (1 − λi)ei2 + 2ei3 = 0 −ei1 − ei2 + (2 − λi)ei3 = 0 When i = 3,λ3 = −1 corresponding solution is e31 8 = −e32 −1 = e33 3 = β3 ⇒ e3 = [8 1 3]T When λ1 = λ2 = 2 solution is given by −2e21 − 2e22 − 2e23 = 0 (1) −e21 − e22 + 2e23 = 0 (2) −e21 − e22 = 0 (3) From (1) and (2) e23 = 0 and it follows from (3) that e21 = −e22. We deduce that there is only one linearly independent eigenvector corresponding to the repeated eigenvalues λ = 2. A possible eigenvector is e2 = [1 − 1 0]T 7(c) Eigenvalues given by ¯¯¯¯¯¯ 4 − λ 6 6 1 3 − λ 2 −1 −5 −2 − λ ¯¯¯¯¯¯ R1−3R3 ¯¯¯¯¯¯ 1 − λ −3 + 3λ 0 1 3 − λ 2 −1 −5 −2 − λ ¯¯¯¯¯¯ = (1 − λ) ¯¯¯¯¯¯ 1 −3 0 1 3 − λ 2 −1 −5 −2 − λ ¯¯¯¯¯¯ = (1 − λ) ¯¯¯¯¯¯ 1 0 0 1 6 − λ 2 1 −8 −2 − λ ¯¯¯¯¯¯ = (1 − λ)(λ2 + λ + 4) = (1 − λ)(λ − 2)2 = 0 ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 11 so eigenvalues are λ1 = λ2 = 2, λ3 = 1. The eigenvectors are the corresponding solutions of (4 − λi)ei1 + 6ei2 + 6ei3 = 0 ei1 + (3 − λi)ei2 + 2ei3 = 0 −ei1 − 5ei2 − (2 + λi)ei3 = 0 When i = 3,λ3 = 1 corresponding solution is e31 4 = −e32 −1 = e33 −3 = β3 ⇒ e3 = [4 1 − 3]T When λ1 = λ2 = 2 solution is given by 2e21 + 6e22 + 6e23 = 0 e21 + e22 + 2e23 = 0 −e21 − 5e22 − 4e23 = 0 so that e21 6 = −e22 −2 = e23 −4 = β2 leading to only one linearly eigenvector corresponding to the eigenvector λ = 2. A possible eigenvector is e2 = [3 1 − 2]T 7(d) Eigenvalues given by ¯¯¯¯¯¯ 7 − λ −2 −4 3 −λ −2 6 −2 −3 − λ ¯¯¯¯¯¯ R1−2R2 ¯¯¯¯¯¯ 1 − λ −2 + 2λ 0 3 −λ −2 6 −2 −3 − λ ¯¯¯¯¯¯ = (1 − λ) ¯¯¯¯¯¯ 1 −2 0 3 −λ −2 6 −2 −3 − λ ¯¯¯¯¯¯ = (1 − λ) ¯¯¯¯¯¯ 1 0 0 3 6 − λ −2 6 10 −3 − λ ¯¯¯¯¯¯ = (1 − λ)(λ − 2)(λ − 1) = 0 so eigenvalues are λ1 = 2, λ2 = λ3 = 1. The eigenvectors are the corresponding solutions of (7 − λi)ei1 − 2ei2 − 4ei3 = 0 3ei1 − λiei2 − 2ei3 = 0 6ei1 − 2ei2 − (3 + λi)ei3 = 0 ©c Pearson Education Limited 201112 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition When i = 1,λ2 = 2 and solution is e11 6 = −e12 −3 = e13 6 = β1 ⇒ e1 = [2 1 2]T When λ2 = λ3 = 1 the solution is given by the single equation 3e21 − e22 − 2e23 = 0 Following the procedures of Example 1.6 we can obtain two linearly independent solutions. A possible pair are e2 = [0 2 − 1]T ,e3 = [2 0 3]T 8 (A − I) = ⎡ ⎣ −2 3 3 1 2 1 4 −7 −5 ⎤ ⎦ Performing a series of row and column operators this may be reduced to the form ⎡⎣ 0 0 0 0 0 1 1 0 0 ⎤⎦ indicating that (A − I) is of rank 2. Thus, the nullity q = 3 − 2 = 1 confirming that there is only one linearly independent eigenvector associated with the eigenvalue λ = 1. The eigenvector is given by the solution of −4e11 − 7e12 − 5e13 = 0 2e11 + 3e12 + 3e13 = 0 e11 + 2e12 + e13 = 0 giving e11 −3 = −e12 −1 = e13 1 = β1 ⇒ e1 = [−3 1 1]T 9 (A − I) = ⎡ ⎣ − −1 1 1 1 − −1 1 1 1 −1 ⎤ ⎦ ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 13 Performing a series of row and column operators this may be reduced to the form ⎡⎣ 1 0 0 0 0 0 0 0 0 ⎤⎦ indicating that (A−I) is of rank 1. Then, the nullity of q = 3−1 = 2 confirming that there are two linearly independent eigenvectors associated with the eigenvalue λ = 1. The eigenvectors are given by the single equation e11 + e12 − e13 = 0 and two possible linearly independent eigenvectors are e1 = [1 0 1]T and e2 = [0 1 1]T Exercises 1.4.8 10 These are standard results. 11(a) (i) Trace A = 4 + 5 + 0 = 9 = sum eigenvalues; (ii) detA = 15 = 5 × 3 × 1 = product eigenvalues; (iii) A−1 = 1 15 ⎡⎣ 4 −1 −1 −4 1 −14 3 3 18 ⎤⎦ . Eigenvalues given by ¯¯¯¯¯¯ 4 − 15λ −1 −1 −4 1 − 15λ −14 3 3 18 − 15λ ¯¯¯¯¯¯ C3−C2 ¯¯¯¯¯¯ 4 − 15λ −1 0 −4 1 − 15λ −15 + 15λ 3 3 15 − 15λ ¯¯¯¯¯¯ = (15 − 15λ) ¯¯¯¯¯¯ 4 − 15λ −1 0 −4 1 − 15λ −1 3 3 1 ¯¯¯¯¯¯ = (15 − 15λ)(15λ − 5)(15λ − 3) = 0 confirming eigenvalues as 1, 1 3, 1 5 . ©c Pearson Education Limited 201114 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition (iv) AT = ⎡ ⎣ 4 2 1 5 1 4 0 − −1 1 ⎤ ⎦ having eigenvalues given by ¯¯¯¯¯¯ 4 − λ 2 −1 1 5 − λ −1 1 4 −λ ¯¯¯¯¯¯ = (λ − 5)(λ − 3)(λ − 1) = 0 that is, eigenvalue as for A. 11(b) (i) 2A = ⎡ ⎣ −8 2 2 4 10 8 2 −2 0 ⎤ ⎦ having eigenvalues given by ¯¯¯¯¯¯ 8 − λ 2 2 4 10 − λ 8 −2 −2 −λ ¯¯¯¯¯¯ C1−C2 ¯¯¯¯¯¯ 6 − λ 2 2 −6 + λ 10 − λ 8 0 −2 −λ ¯¯¯¯¯¯ = (6 − λ) ¯¯¯¯¯¯ 1 2 2 −1 10 − λ 8 0 −2 −λ ¯¯¯¯¯¯ = (6 − λ) ¯¯¯¯¯¯ 1 2 2 0 12 − λ 10 0 −2 −λ ¯¯¯¯¯¯ = (6 − λ)(λ − 10)(λ − 2) = 0 Thus eigenvalues are 2 times those of A; namely 6, 10 and 2. (ii) A + 2I = ⎡⎣ 6 1 1 2 7 4 −1 −1 2 ⎤⎦ having eigenvalues given by ¯¯¯¯¯¯ 6 − λ 1 1 2 7 − λ 4 −1 −1 2 − λ ¯¯¯¯¯¯ = −λ3 + 15λ2 − 71λ + 105 = −(λ − 7)(λ − 5)(λ − 3) = 0 confirming the eigenvalues as 5 + 2,3 + 2,1 + 2. Likewise for A − 2I ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 15 (iii) A2 = ⎡ ⎣ − 22 6 −6 −5 ⎤ ⎦ having eigenvalues given by ¯¯¯¯¯¯ 17 − λ 8 8 14 23 − λ 22 − λ −6 −6 −5 − λ ¯¯¯¯¯¯ R1 + (R2) + R3) ¯¯¯¯¯¯ 25 − λ 25 − λ 25 − λ 14 23 − λ 22 −6 −6 −5 − λ ¯¯¯¯¯¯ = (25 − λ) ¯¯¯¯¯¯ 1 0 0 14 9 − λ 8 −6 0 1 − λ ¯¯¯¯¯¯ = (25 − λ)(9 − λ)(1 − λ) = 0 that is, eigenvalues A2 are 25, 9, 1 which are those of A squared. 12 Eigenvalues of A given by ¯¯¯¯¯¯ −3 − λ −3 −3 −3 1 − λ −1 −3 −1 1 − λ ¯¯¯¯¯¯ R3+R2 ¯¯¯¯¯¯ −3 − λ −3 −3 −3 1 − λ −1 0 −2 + λ 2 − λ ¯¯¯¯¯¯ = (λ − 2) ¯¯¯¯¯¯ −3 − λ −3 −3 −3 1 − λ −1 0 1 −1 ¯¯¯¯¯¯ C3+C2(λ − 2) ¯¯¯¯¯¯ −3 − λ −3 −6 −3 (1 − λ) −λ 0 1 0 ¯¯¯¯¯¯ = −(λ − 2)(λ + 6)(λ − 3) = 0 so eigenvalues are λ1 = −6,λ2 = 3,λ3 = 2 Eigenvectors are given by corresponding solutions of (−3 − λi)ei1 − 3ei2 − 3ei3 = 0 −3ei1 + (1 − λi)ei2 − ei3 = 0 −3ei1 − ei2 + (1 − λi)ei3 = 0 Taking i = 1,2,3 gives the eigenvectors as e1 = [2 1 1]T, e2 = [−1 1 1]T, e3 = [0 1 − 1]T It is readily shown that eT 1 e2 = eT 1 e3 = eT 2 e3 = 0 so that the eigenvectors are mutually orthogonal. ©c Pearson Education Limited 201116 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 13 Let the eigenvector be e = [a b c]T then since the three vectors are mutually orthogonal a + b − 2c = 0 a + b − c = 0 giving c = 0 and a = −b so an eigenvector corresponding to λ = 2 is e = [1 −1 0]T . Exercises 1.5.3 14 Taking x(0) = [1 1 1]T iterations may then be tabulated as follows: Iteration k 0 1 2 3 4 1 0.9 0.874 0.869 0.868 x(k) 1 1 1 1 1 1 0.5 0.494 0.493 0.492 9 7.6 7.484 7.461 7.457 A x(k) 10 8.7 8.61 8.592 8.589 5 4.3 4.242 4.231 4.228 λ ≃ 10 8.7 8.61 8.592 8.589 Thus, estimate of dominant eigenvalue is λ ≃ 8.59 and corresponding eigenvector x ≃ [0.869 1 0.493]Tor x ≃ [0.61 0.71 0.35]T in normalised form. 15(a) Taking x(0) = [1 1 1]T iterations may then be tabulated as follows: Iteration k 0 1 2 3 4 5 6 1 0.75 0.667 0.636 0.625 0.620 0.619 x(k) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 2.5 2.334 2.272 2.250 2.240 A x(k) 4 3.75 3.667 3.636 3.625 3.620 4 3.75 3.667 3.636 3.625 3.620 λ ≃ 4 3.75 3.667 3.636 3.625 3.620 Thus, correct to two decimal places dominant eigenvalue is 3.62 having corresponding eigenvectors [0.62 1 1]T . ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 17 15(b) Taking x(0) = [1 1 1]T iterations may be tabulated as follows: Iteration k 0 1 2 3 4 5 1 0.364 0.277 0.257 0.252 0.251 x(k) 1 0.545 0.506 0.501 0.493 0.499 1 1 1 1 1 1 4 2.092 1.831 1.771 1.756 A x(k) 6 3.818 3.566 3.561 3.49 11 7.546 7.12 7.03 6.994 λ ≃ 11 7.546 7.12 7.03 6.994 Thus, correct to two decimal places dominant eigenvalue is 7 having corresponding eigenvector [0.25 0.5 1]T . 15(c) Taking x(0) = [1 1 1 1]T iterations may then be tabulated as follows: Iteration k 0 1 2 3 4 5 6 1 1 1 1 1 1 1 x(k) 1 0 −0.5 −0.6 −0.615 −0.618 − 0.618 1 1 −0.5 −0.6 −0.615 −0.618 −0.618 1 1 1 1 1 1 1 1 2 2.5 2.6 2.615 2.618 A x(k) 0 −1 −1.5 −1.6 −1.615 −1.618 0 −1 −1.5 −1.6 −1.615 −1.618 1 2 2.5 2.6 2.615 2.618 λ ≃ 1 2 2.5 2.6 2.615 2.618 Thus, correct to two decimal places dominant eigenvalue is 2.62 having corresponding eigenvector [1 − 0.62 − 0.62 1]T . 16 The eigenvalue λ1 corresponding to the dominant eigenvector e1 = [1 1 2]T is such that A e1 = λ1e1 so ⎡⎣ 3 1 1 1 3 1 1 1 5 ⎤⎦ ⎡⎣ 112 ⎤⎦ = λ1 ⎡ ⎣ 1 1 2 ⎤ ⎦ so λ1 = 6. ©c Pearson Education Limited 201118 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Then A1 = A − 6ˆe1ˆeT 1 where ˆe1 = £√16 √16 √26¤T so A1 = ⎡ ⎣ 3 1 1 1 3 1 1 1 5 ⎤ ⎦ − ⎡ ⎣ 1 1 2 1 1 2 2 2 4 ⎤ ⎦ = ⎡ ⎣ −2 0 0 2 1 −1 1 − −1 1 ⎤ ⎦ Applying the power method with x(0) = [1 1 1]T y(1) = A1x(0) = ⎡ ⎣ −1 11 ⎤ ⎦ = x(1) y(2) = A1x(1) = ⎡ ⎣ −3 33 ⎤ ⎦ = 3 ⎡ ⎣ −1 11 ⎤ ⎦ Clearly, λ2 = 3 and ˆe2 = √13[1 1 − 1]T . Repeating the process A2 = A1 − λ2ˆe2ˆeT 2 = ⎡ ⎣ −2 0 0 2 1 −1 1 − −1 1⎤ ⎦ − ⎡ ⎣ −1 1 1 1 1 −1 1 − −1 1⎤ ⎦ = ⎡ ⎣ −1 0 0 0 1 1 0 −1 0 ⎤ ⎦ Taking x(0) = [1 − 1 0]T the power method applied to A2 gives y(1) = A2x(0) = ⎡ ⎣ −2 02 ⎤ ⎦ = 2 ⎡ ⎣ −1 01⎤ ⎦ and clearly, λ3 = 2 with ˆe3 = √12[1 − 1 0]T . 17 The three Gerschgorin circles are | λ − 5 |= 2, | λ |= 2, | λ + 5 |= 2 ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 19 which are three non-intersecting circles. Since the given matrix A is symmetric its three eigenvalues are real and it follows from Theorem 1.2 that 3 < λ1 < 7 , −2 < λ2 < 2 , −7 < λ3 < 7 (i.e., an eigenvalue lies within each of the three circles). 18 The characteristic equation of the matrix A is ¯¯¯¯¯¯ 10 − λ −1 0 −1 2 − λ 2 0 2 3 − λ ¯¯¯¯¯¯ = 0 that is (10 − λ)[(2 − λ)(3 − λ) − 4] − (3 − λ) = 0 or f(λ) = λ3 − 15λ2 + 51λ − 17 = 0 Taking λ0 = 10 as the starting value the Newton–Raphson iterative process produces the following table: i λi f(λi) f′(λi) − f(λi) f′(λi) 0 10 7 −51.00 0.13725 1 10.13725 −0.28490 −55.1740 −0.00516 2 10.13209 −0.00041 −55.0149 −0.000007 Thus to three decimal places the largest eigenvalue is λ = 10.132 Using Properties 1.1 and 1.2 of section 1.4.6 we have 3X i =1 λi = trace A = 15 and 3Y i =1 λi =| A |= 17 ©c Pearson Education Limited 201120 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Thus, λ2 + λ3 = 15 − 10.132 = 4.868 λ2λ3 = 1.67785 so λ2(4.868 − λ2) = 1.67785 λ2 2 − 4.868λ2 + 1.67785 = 0 λ2 = 2.434 ± 2.0607 that is λ2 = 4.491 and λ3 = 0.373 19(a) If e1,e2,...,en are the corresponding eigenvectors to λ1,λ2,...,λn then (KI−A)ei = (K−λi)ei so that A and (KI−A) have the same eigenvectors and eigenvalues differ by K. Taking x(o) = nP i=1 αrei then x(p) = (KI − A)x(p−1) = (KI − A)2x(p−2) = ... = nX r =1 αr(K − λr)per Now K − λn > K − λn−1 > ... > K − λ1 and x(p) = αn(K − λn)pen + nX r =1 αr(K − λr)per = (K − λn)p[αnen + n−1 X r =1 αr£KK −− λλnr ¤per] → (K − λn)pαnen = Ken as p → ∞ Also x (p+1) i x (p) i → (K − λn)p+1 (K − λn)p αnen αnen = K − λn Hence, we can find λn 19(b) Since A is a symmetric matrix its eigenvalues are real. By Gerschgorin’s theorem the eigenvalues lie in the union of the intervals | λ − 2 |≤ 1, | λ − 2 |≤ 2, | λ − 2 |≤ 1 i.e. | λ − 2 |≤ 2 or 0 ≤ λ ≤ 4. ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 21 Taking K = 4 in (a) KI − A = 4I − A = ⎡ ⎣ 2 1 0 1 2 1 0 1 2 ⎤ ⎦ Taking x(0) = [1 1 1]T iterations using the power method are tabulated as follows: Iteration k 0 1 2 3 4 5 1 0.75 0.714 0.708 0.707 0.707 x(k) 1 1 1 1 1 1 1 0.75 0.714 0.708 0.707 0.707 3 2.5 2.428 2.416 2.414 A x(k) 4 3.5 3.428 3.416 3.414 3 2.5 2.428 2.416 2.414 λ ≃ 4 3.5 3.428 3.416 3.414 Thus λ3 = 4 − 3.41 = 0.59 correct to two decimal places. Exercises 1.6.3 20 Eigenvalues given by Δ = ¯¯¯¯¯¯ −1 − λ 6 −12 0 −13 − λ 30 0 −9 20 − λ ¯¯¯¯¯¯ = 0 Now Δ = (−1 − λ) ¯¯¯¯ −13 − λ 30 −9 20 − λ ¯¯¯¯ = (−1 − λ)(λ2 − 7λ + 10) = (−1 − λ)(λ − 5)(λ − 2) so Δ = 0 gives λ1 = 5,λ2 = 2,λ3 = −1 Corresponding eigenvectors are given by the solutions of (A − λiI)ei = 0 When λ = λ1 = 5 we have ⎡⎣ −6 6 −12 0 −18 30 0 −9 15 ⎤⎦ ⎡⎣ e11 e12 e13 ⎤⎦ = 0 ©c Pearson Education Limited 201122 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition leading to the solution e11 −36 = −e12 −180 = e13 108 = β1 so the eigenvector corresponding to λ1 = 5 is e1 = β1[1 − 5 − 3]T When λ = λ2 = 2, we have ⎡⎣ −3 6 −12 0 −15 30 0 −9 18 ⎤⎦ ⎡⎣ e21 e22 e23 ⎤⎦ = 0 leading to the solution e21 0 = −e22 −90 = e23 45 = β2 so the eigenvector corresponding to λ2 = 2 is e2 = β2[0 2 1]T When λ = λ3 = −1, we have ⎡⎣ 0 6 −12 0 −12 30 0 −9 21 ⎤⎦ ⎡⎣ e31 e32 e33 ⎤⎦ = 0 leading to the solution e31 18 = −e32 0 = e33 0 = β3 so the eigenvector corresponding to λ3 = −1 is e3 = β3[1 0 0]T A modal matrix M and spectral matrix Λ are M = ⎡ ⎣ − −1 0 1 5 2 0 3 1 0 ⎤ ⎦ Λ = ⎡ ⎣ 5 0 0 0 2 0 0 0 −1 ⎤ ⎦ M−1 = ⎡ ⎣ 0 1 0 3 1 −1 2 − −2 5 ⎤ ⎦ and matrix multiplication confirms M−1A M = Λ 21 From Example 1.9 the eigenvalues and corresponding normalised eigenvectors of A are λ1 = 6,λ2 = 3,λ3 = 1 ˆe1 = √15[1 2 0]T,ˆe2 = [0 0 1]T,ˆe3 = √15[−2 1 0]T, Mˆ = √15 ⎡ ⎣ 1 0 2 0 1 0 √5 0 −2⎤ ⎦ ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 23 Mˆ T A M = 1 5 ⎡⎣ 1 2 0 0 0 √5 −2 1 0 ⎤⎦ ⎡⎣ 2 2 0 2 5 0 0 0 3 ⎤⎦ ⎡⎣ 1 0 −2 2 0 1 0 √5 0 ⎤⎦ = 15 ⎡⎣ 6 12 0 0 0 3√5 −2 1 0 ⎤⎦ ⎡⎣ 1 0 −2 2 0 1 0 √5 0 ⎤⎦ = 15 ⎡⎣ 30 0 0 0 15 0 0 0 5 ⎤⎦ = ⎡⎣ 6 0 0 0 3 0 0 0 1 ⎤⎦ = Λ 22 The eigenvalues of A are given by ¯¯¯¯¯¯ 5 − λ 10 8 10 2 − λ −2 8 −2 11 − λ ¯¯¯¯¯¯ = −(λ3−18λ2−81λ+1458) = −(λ−9)(λ+9)(λ−18) = 0 so eigenvalues are λ1 = 18,λ2 = 9,λ3 = −9 The eigenvectors are given by the corresponding solutions of (5 − λi)ei1 + 10ei2 + 8ei3 = 0 10ei1 + (2 − λi)ei2 − 2ei3 = 0 8ei1 − 2ei2 + (11 − λi)ei3 = 0 Taking i = 1,2,3 and solving gives the eigenvectors as e1 = [2 1 2]T, e2 = [1 2 − 2]T,e3 = [−2 2 1]T Corresponding normalised eigenvectors are eˆ1 = 1 3 [2 1 2]T, eˆ2 = 1 3 [1 2 − 2]T, eˆ3 = 1 3 [−2 2 1]T Mˆ = 1 3 ⎡⎣ 2 1 −2 1 2 2 2 −2 1 ⎤⎦ , Mˆ T = 1 3 ⎡⎣ 2 1 2 1 2 −2 −2 2 1 ⎤⎦ ©c Pearson Education Limited 201124 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition Mˆ T A M = 1 9 ⎡⎣ 2 1 2 1 2 −2 −2 2 1 ⎤⎦ ⎡⎣ 5 10 8 10 2 −2 8 −2 11 ⎤⎦ ⎡⎣ 2 1 −2 1 2 2 2 −2 1 ⎤⎦ = 19 ⎡⎣ 36 18 36 9 18 −18 18 −18 −9 ⎤⎦ ⎡⎣ 2 1 −2 1 2 2 2 −2 1 ⎤⎦ = ⎡⎣ 4 2 4 1 2 −2 2 −2 −1 ⎤⎦ ⎡⎣ 2 1 −2 1 2 2 2 −2 1 ⎤⎦ = ⎡⎣ 18 0 0 0 9 0 0 0 −9 ⎤⎦ = Λ 23 A = ⎡ ⎣ −1 1 0 1 1 2 1 − −2 1⎤ ⎦ Eigenvalues given by 0 = ¯¯¯¯¯¯ 1 − λ 1 −2 −1 2 − λ 1 0 1 −1 − λ ¯¯¯¯¯¯ = −(λ3 −2λ2 −λ+2) = −(λ−1)(λ−2)(λ+1) = 0 so eigenvalues are λ1 = 2,λ2 = 1,λ3 = −1. The eigenvectors are given by the corresponding solutions of (1 − λi)ei1 + ei2 − 2ei3 = 0 −ei1 + (2 − λi)ei2 + ei3 = 0 0ei1 + ei2 − (1 + λi)ei3 = 0 Taking i = 1,2,3 and solving gives the eigenvectors as e1 = [1 3 1]T,e2 = [3 2 1]T,e3 = [1 0 1]T M = ⎡ ⎣ 1 3 1 3 2 0 1 1 1 ⎤ ⎦ , Λ = ⎡ ⎣ 2 0 0 0 1 0 0 0 −1 ⎤ ⎦ M−1 = −1 6 ⎡⎣ 2 −2 −2 −3 0 −3 1 2 −7 ⎤⎦ ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 25 Matrix multiplication then confirms M−1 A M = Λ and A = M Λ M−1 24 Eigenvalues given by ¯¯¯¯¯¯ 3 − λ −2 4 −2 −2 − λ 6 4 6 −1 − λ ¯¯¯¯¯¯ = −λ3 + 63λ − 162 = −(λ + 9)(λ − 6)(λ − 3) = 0 so the eigenvalues are λ1 = −9,λ2 = 6,λ3 = 3. The eigenvectors are the corresponding solutions of (3 − λi)ei1 − 2ei2 + 44ei3 = 0 −2ei1 − (2 + λi)ei2 + 6ei3 = 0 4ei1 + 6ei2 − (1 + λi)ei3 = 0 Taking i = 1,2,3 and solving gives the eigenvectors as e1 = [1 2 − 2]T,e2 = [2 1 2]T,e3 = [−2 2 1]T Since eT 1 e2 = eT 1 e3 = eT 2 e3 = 0 the eigenvectors are orthogonal L = [eˆ1 eˆ2 eˆ3] = 1 3 ⎡⎣ 1 2 −2 2 1 2 −2 2 1 ⎤⎦ L A L ˆ = 1 9 ⎡⎣ 1 2 −2 2 1 2 −2 2 1 ⎤⎦ ⎡⎣ 3 −2 4 −2 −2 6 4 6 −1 ⎤⎦ ⎡⎣ 1 2 −2 2 1 2 −2 2 1 ⎤⎦ = 19 ⎡⎣ −9 −18 18 12 6 12 −6 6 3 ⎤⎦ ⎡⎣ 1 2 −2 2 1 2 −2 2 1 ⎤⎦ = 19 ⎡⎣ −81 0 0 0 54 0 0 0 27 ⎤⎦ = ⎡⎣ −9 0 0 0 6 0 0 0 3 ⎤⎦ = Λ 25 Since the matrix A is symmetric the eigenvectors e1 = [1 2 0]T, e2 = [−2 1 0]T, e3 = [e31 e32 e33]T ©c Pearson Education Limited 201126 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition are orthogonal. Hence, eT 1 e3 = e31 + 2e32 = 0 and eT 2 e3 = −2e31 + e32 = 0 Thus, e31 = e32 = 0 and e33 arbitrary so a possible eigenvector is e3 = [0 0 1]T . Using A = M Λ ˆ Mˆ T where Λ = ⎡ ⎣ 6 0 0 0 1 0 0 0 3⎤ ⎦ gives A = ⎡ ⎣ √ √0 0 1 1 25 5 −√√1255 0 0 ⎤ ⎦ ⎡ ⎣ 6 0 0 0 1 0 0 0 3 ⎤ ⎦ ⎡ ⎣ −√0 0 1 1√525 √ √215 5 0 0 ⎤ ⎦ = ⎡⎣ 2 2 0 2 5 0 0 0 3 ⎤⎦ 26 A − I = ⎡ ⎣ −2 3 3 1 2 1 4 −7 −5 ⎤ ⎦ ∼ ⎡ ⎣ 0 0 0 0 1 0 0 −1 0⎤ ⎦ is of rank 2 Nullity (A − I) = 3 − 2 = 1 so there is only one linearly independent vector corresponding to the eigenvalue 1. The corresponding eigenvector e1 is given by the solution of (A − I)e1 = 0 or −4e11 − 7e12 − 5e13 = 0 2e11 + 3e12 + 3e13 = 0 e11 + 2e12 + 212 = 0 that is, e1 = [−3 1 1]T . To obtain the generalised eigenvector e∗ 1 we solve (A − I)e∗ 1 = e1 or ⎡⎣ −4 −7 −5 2 3 3 1 2 1 ⎤⎦ ⎡⎣ e∗ 11 e∗ 12 e∗ 13 ⎤⎦ = ⎡⎣ −3 11 ⎤⎦ giving e∗ 1 = [−1 1 0]T . To obtain the second generalised eigenvector e∗∗ 1 we solve (A − I)e∗∗ 1 = e∗ 1 or ⎡⎣ −4 −7 −5 2 3 3 1 2 1 ⎤⎦ ⎡⎣ e∗∗ 11 e∗∗ 12 e∗∗ 13 ⎤⎦ = ⎡⎣ −1 10 ⎤⎦ ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 27 giving e∗∗ 1 = [2 − 1 0]T . M = [e1 e∗ 1 e∗∗ 1 ] = ⎡⎣ −3 −1 2 1 1 −1 1 0 0 ⎤⎦ detM = −1 and M−1 = − ⎡ ⎣ − −0 0 1 1 − −2 1 −− −11 2 ⎤ ⎦ = ⎡ ⎣ 0 0 1 1 2 1 1 1 2 ⎤ ⎦ Matrix multiplication then confirms M−1 A M = ⎡ ⎣ 1 1 0 0 1 1 0 0 1 ⎤ ⎦ 27 Eigenvalues are given by | A − λI |= 0 that is, λ4 − 4λ3 − 12λ2 + 32λ + 64 = (λ + 2)2(λ − 4)2 = 0 so the eigenvalues are −2, −2, 4 and 4 as required. Corresponding to the repeated eigenvalue λ1,λ2 = −2 (A + 2I) = ⎡⎢⎣ 3 0 0 −3 0 3 −3 0 −0.5 −3 3 0.5 −3 0 0 3 ⎤⎥⎦ ∼ ⎡⎢⎣ 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 ⎤⎥⎦ is of rank 2 Thus, nullity (A+2I) is 4−2 = 2 so there are two linearly independent eigenvectors corresponding to λ = −2. Corresponding to the repeated eigenvalues λ3,λ4 = 4 (A − 4I) = ⎡⎢⎣ −3 0 0 −3 0 −3 −3 0 −0.5 −3 −3 0.5 −3 0 0 −3 ⎤⎥⎦ ∼ ⎡⎢⎣ 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 ⎤⎥⎦ is of rank 3 Thus, nullity (A − 4I) is 4 − 3 = 1 so there is only one linearly independent eigenvector corresponding to λ = 4. ©c Pearson Education Limited 201128 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition When λ = λ1 = λ2 = −2 the eigenvalues are given by the solution of (A+2I)e = 0 giving e1 = [0 1 1 0]T,e2 = [1 0 0 1]T as two linearly independent solutions. When λ = λ3 = λ4 = 8 the eigenvectors are given by the solution of (A − 4I)e = 0 giving the unique solution e3 = [0 1 −1 0]T . The generalised eigenvector e∗ 3 is obtained by solving (A − 4I)e∗ 3 = e3 giving e∗ 3 = (6 − 1 0 − 6]T . The Jordan canonical form is J = ⎡⎢⎢⎢⎢⎢⎢⎢⎣ −2 0 0 0 0 −2 0 0 0 0 4 1 0 0 0 4 ⎤⎥⎥⎥⎥⎥⎥⎥⎦ Exercises 1.6.5 28 The quadratic form may be written in the form V = xTAx where x = [ x1 x2 x3 ]T and A = ⎡ ⎣ 2 2 1 2 5 2 1 2 2 ⎤ ⎦ The eigenvalues of A are given by ¯¯¯¯¯¯ 2 − λ 2 1 2 5 − λ 2 1 2 2 − λ ¯¯¯¯¯¯ = 0 ⇒ (2 − λ)(λ2 − 7λ + 6) + 4(λ − 1) + (λ − 1) = 0 ⇒ (λ − 1)(λ2 − 8λ + 7) = 0 ⇒ (λ − 1)2(λ − 7) = 0 giving the eigenvalues as λ1 = 7,λ2 = λ3 = 1 Normalized eigenvector corresponding to λ1 = 7 is ˆe1 = £ √16 √26 √16 ¤T ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 29 and two orthogonal linearly independent eigenvectors corresponding to λ − 1 are ˆe2 = £ √12 0 − √12 ¤T ˆe3 = £ − √13 √13 − √13 ¤T Note that ˆe2 and ˆe3 are automatically orthogonal to ˆe1. The normalized orthogonal modal matrix Mˆ and spectral matrix Λ are Mˆ = ⎡⎢⎣ 1√ 6 1√ 2 − 1√ 3 2√ 6 0 √13 1√ 6 − 1√ 2 − 1√ 3 ⎤⎥⎦ ,Λ = ⎡⎣ 7 0 0 0 1 0 0 0 1 ⎤⎦ such that Mˆ TAMˆ = Λ. Under the orthogonal transformation x = My ˆ the quadratic form V reduces to V = yTMˆ TAMy ˆ = yTΛy = [ y1 y2 y3 ] ⎡ ⎣ 7 0 0 0 1 0 0 0 1 ⎤ ⎦ ⎡ ⎣ y y y1 2 3 ⎤ ⎦ = 7y2 1 + y2 2 + y2 3 29(a) The matrix of the quadratic form is A = ⎡ ⎣ −1 21 2 − −1 2 1 7 −1 ⎤ ⎦ and its leading principal minors are 1, ¯¯¯¯ 1 −1 −1 2 ¯¯¯¯ = 1,detA = 2 Thus, by Sylvester’s condition (a) the quadratic form is positive definite. 29(b) Matrix A = ⎡⎣ 1 −1 2 −1 2 −1 2 −1 5 ⎤⎦ and its leading principal minors are 1, ¯¯¯¯ 1 −1 −1 2 ¯¯¯¯ = 1,detA = 0 Thus, by Sylvester’s condition (c) the quadratic form is positive semidefinite. ©c Pearson Education Limited 201130 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 29(c) Matrix A = ⎡ ⎣ −1 21 2 − −1 2 1 4 −1 ⎤ ⎦ and its leading principal minors are 1, ¯¯¯¯ 1 −1 −1 2 ¯¯¯¯ = 1,detA = −1. Thus, none of Sylvester’s conditions are satisfied and the quadratic form is indefinite. 30(a) The matrix of the quadratic form is A = · −ab c −b ¸ and its leading principal minors are a and ac − b2. By Sylvester’s condition (a) in the text the quadratic form is positive definite if and only if a > 0 and ac − b2 > 0 that is, a > 0 and ac > b2 30(b) The matrix of the quadratic form is A = ⎡ ⎣ −201 −a b b1 0 3 ⎤ ⎦ having principal minors 2,2a − 1 and detA = 6a − 2b2 − 3. Thus, by Sylvester’s condition (a) in the text the quadratic form is positive definite if and only if 2a − 1 > 0 and 6a − 2b2 − 3 > 0 or 2a > 1 and 2b2 < 6a − 3 31 The eigenvalues of the matrix A are given by 0 = ¯¯¯¯¯¯ 2 − λ 1 −1 1 2 − λ 1 −1 1 2 − λ ¯¯¯¯¯¯ R1+R3 ¯¯¯¯¯¯ 3 − λ 3 − λ 0 1 2 − λ 1 −1 1 2 − λ ¯¯¯¯¯¯ = (3 − λ) ¯¯¯¯¯¯ 1 1 0 1 2 − λ 1 −1 1 2 − λ ¯¯¯¯¯¯ = (3 − λ) ¯¯¯¯¯¯ 1 0 0 1 1 − λ 1 −1 2 2 − λ ¯¯¯¯¯¯ = (3 − λ)(λ2 − 3λ) ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 31 so the eigenvalues are 3,3,0 indicating that the matrix is positive semidefinite. The principal minors of A are 2, ¯¯¯¯ 2 1 1 2 ¯¯¯¯ = 3, detA = 0 confirming, by Sylvester’s condition (a), that the matrix is positive semidefinite. 32 The matrix of the quadratic form is A = ⎡ ⎣ K1 1 −K1 1 1 1 −1 ⎤ ⎦ having principal minors K, ¯¯¯¯ K 1 1 K ¯¯¯¯ = K2 − 1 and detA = K2 − K − 3 Thus, by Sylvester’s condition (a) the quadratic form is positive definite if and only if K2 − 1 = (K − 1)(K + 1) > 0 and K2 − K − 3 = (K − 2)(K + 1) > 0 i.e. K > 2 or K < −1. If K = 2 then detA = 0 and the quadratic form is positive semidefinite. 33 Principal minors of the matrix are 3 + a, ¯¯¯¯ 3 + a 1 1 a ¯¯¯¯ = a2 + 3a − 1,detA = a3 + 3a2 − 6a − 8 Thus, by Sylvester’s condition (a) the quadratic form is positive definite if and only if 3 + a > 0,a2 + 3a − 1 > 0 and a3 + 3a2 − 6a − 8 > 0 or (a + 1)(a + 4)(a − 2) > 0 3 + a > 0 ⇒ a > −3 a2 + 3a − 1 > 0 ⇒ a < −3.3 or a > 0.3 (a + 1)(a + 4)(a − 2) > 0 ⇒ a > 2 or − 4 < a < −1 Thus, minimum value of a for which the quadratic form is positive definite is a = 2. ©c Pearson Education Limited 201132 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 34 A = ⎡ ⎣ −1 2 22 −λ3 − −λ2 3 ⎤ ⎦ Principal minors are 1, ¯¯¯¯ 1 2 2 λ ¯¯¯¯ = λ − 4, detA = λ2 − 8λ + 15 = 0 Thus, by Sylvester’s condition (a) the quadratic form is positive definite if and only if λ − 4 > 0 ⇒ λ > 4 and (λ − 5)(λ − 3) > 0 ⇒ λ < 3 or λ > 5 Thus, it is positive definite if and only if λ > 5. Exercises 1.7.1 35 The characteristic equation of A is ¯¯¯¯ 5 − λ 6 2 3 − λ ¯¯¯¯ = λ2 − 8λ + 3 = 0 Now A2 = ·5 6 2 3 ¸ · 5 6 2 3¸ = · ¸ so A2 − 8A + 3I = · ¸ − · ¸ + ·3 0 0 3 ¸ = · 0 0 0 0 ¸ so that A satisfies its own characteristic equation. 36 The characteristic equation of A is ¯¯¯¯ 1 − λ 2 1 1 − λ ¯¯¯¯ = λ2 − 2λ − 1 = 0 ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 33 By Cayley–Hamilton theorem A2 − 2A − I = 0 36(a) Follows that A2 = 2A + I = ·2 4 2 2 ¸ + · 1 0 0 1 ¸ = ·3 4 2 3 ¸ 36(b) A3 = 2A2 + A = · 6 8 4 6 ¸ + ·1 2 1 1 ¸ = ·7 10 5 7 ¸ 36(c) A4 = 2A3 + A2 = · ¸ + ·3 4 2 3 ¸ = · ¸ 37(a) The characteristic equation of A is ¯¯¯¯ 2 − λ 1 1 2 − λ ¯¯¯¯ = 0 that is, λ2 − 4λ + 3 = 0 Thus, by the Cayley–Hamilton theorem A2 − 4A + 3I = 0 I = 1 3 [4A − A2] so that A−1 = 1 3 [4I − A] = 13 ½·4 0 0 4 ¸ − ·2 1 1 2 ¸¾ = 1 3 · −21 2 −1 ¸ 37(b) The characteristic equation of A is ¯¯¯¯¯¯ 1 − λ 1 2 3 1 − λ 1 2 3 1 − λ ¯¯¯¯¯¯ = 0 that is, λ3 − 3λ2 − 7λ − 11 = 0 ©c Pearson Education Limited 201134 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition A2 = ⎡ ⎣ 1 1 2 3 1 1 2 3 1 ⎤ ⎦ ⎡ ⎣ 1 1 2 3 1 1 2 3 1 ⎤ ⎦ = ⎡ ⎣ 13 8 8 8 8 5 8 7 8⎤ ⎦ Using (1.44) A−1 = 1 11 (A2 − 3A − 7I) = 1 11 ⎡⎣ −2 5 −1 −1 −3 5 7 −1 −2 ⎤⎦ 38 A2 = ⎡ ⎣ 2 3 1 3 1 2 1 2 3 ⎤ ⎦ ⎡ ⎣ 2 3 1 3 1 2 1 2 3 ⎤ ⎦ = ⎡ ⎣ ⎤ ⎦ The characteristic equation of A is λ2 − 6λ2 − 3λ + 18 = 0 so by the Cayley–Hamilton theorem A3 = 6A2 + 3A − 18I giving A4 = 6(6A2 + 3A − 18I) + 3A2 − 18A = 39A2 − 108I A5 = 39(6A2 + 3A − 18I) + 108A = 234A2 + 9A − 702I A6 = 234(6A2 + 3A − 18I) + 9A2 − 702A = 1413A2 − 4212I A7 = 1413(6A2 + 3A − 18I) + 4212A = 8478A2 + 27A − 25434I Thus, A7 − 3A6 + A4 + 3A3 − 2A2 + 3I = 4294A2 + 36A − 12957I = ⎡⎣ ⎤⎦ ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 35 39(a) Eigenvalues A are λ = 1 (repeated). Thus, eAt = α0I + α1A with et = α0 + α1 tet = α1 ¾ ⇒ α1 = tet, α0 = (1 − t)et so eAt = (1 − t)etI + tetA = ·te ett e0t ¸ 39(b) Eigenvalues A are λ = 1 and λ = 2. Thus, eAt = α0I + α1A with et = α0 + α1 e2t = α0 + 2α1 ¾ ⇒ α0 = 2et − e2t, α1 = e2t − et so eAt = (2et − e2t)I + (e2t − et)A = ·e2te−t et e02t ¸ 40 Eigenvalues of A are λ1 = π, λ2 = π 2 ,λ3 = π 2 . Thus, sinA = α0A + α1A + α2A2 with sin π = 0 = α0 + α1π + α2π2 sin π2 = 1 = α0 + α1 π2 + α2 π2 4 cos π2 = 0 = α1 + πα2 Solving gives α0 = 0, α1 = 4 π , α2 = − 4 2π so that sinA = 4 π A − 4 π2A2 = ⎡⎣ 0 0 0 0 1 0 0 0 1 ⎤⎦ 41(a) dA dt = · dt ddt d((5 t2 −+ 1) t) dt d dt (dt2(2−t −t + 3) 3) ¸ = ·−2t1 2t 2− 1 ¸ ©c Pearson Education Limited 201136 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 41(b) Z12 Adt = · RR1212((5 t2 −+ 1) t)dt dt R12R(12t2(2−t −t + 3) 3)dtdt ¸ = ⎡ ⎣ 10 3 7 2 23 06 ⎤ ⎦ 42 A2 = · t2 5 0 + 1 t − 1 ¸ · t2 5 0 + 1 t − 1 ¸ = · t4 + 25tt22 + 5 + 5 5 t − 4 t3 − tt2−+5t − 1 ¸ d dt(A2) = · 4t3 + 4 10tt + 5 3t2 −52t + 1 ¸ 2AdA dt = · 4t320+ 4 t t 2t20+ 1 ¸ Thus, d dt(A2) ∕= 2Addt A . Exercises 1.8.4 43(a) row rank A = ⎡ ⎣ 1 2 3 4 3 4 7 10 2 1 5 7 ⎤ ⎦ row2 row3 − −→3row1 2row1 ⎡ ⎣ 1 2 3 4 00 −−23 −−21 −−21 ⎤ ⎦ − 12 row2 → ⎡⎣ 1 2 4 4 0 1 1 1 0 −3 −1 −1 ⎤⎦ row3 + 3row2 → ⎡⎣ 1 2 3 4 0 1 1 1 0 0 2 2 ⎤⎦ echelon form, row rank 3 column rank A col2 − 2col1 → col3 − 3col1 col4 − 4col1 ⎡⎣ 1 0 0 0 3 −2 −2 2 2 −3 2 0 ⎤⎦ col3 − col2 → col4 − col2 ⎡⎣ 1 0 0 0 0 −2 0 0 2 −3 2 2 ⎤⎦ col4 − col3 → ⎡⎣ 1 0 0 0 3 −2 0 0 2 −3 2 0 ⎤⎦ echelon form,column rank3 Thus row rank(A) = column rank(A) = 3 ©c Pearson Education Limited 2011Glyn James, Advanced Modern Engineering Mathematics, 4th Edition 37 (b) A is of full rank since rank(A)=min(m,n)=min(3,4)= 3 44(a) AAT = ·4 11 14 8 7 −2 ¸ ⎡ ⎣ 11 7 14 4 8 −2 ⎤ ⎦ = · ¸ = 9· 37 9 9 13 ¸ The eigenvalues λi of AAT are given by the solutions of the equations ¯¯¯ AAT − λI¯ ¯ ¯ = ¯ ¯ ¯ ¯ − λ 8− λ ¯ ¯ ¯ ¯ = 0 ⇒ λ2 − 450λ + 32400 = 0 ⇒ (λ − 360)(λ − 90) = 0 giving the eigenvalues as λ1 = 360,λ2 = 90. Solving the equations. (AAT − λiI)ui = 0 gives the corresponding eigenvectors as u1 = [ 3 1 ]T ,u2 = [ 1 −2 ]T with the corresponding normalized eigenvectors being uˆ1 = £ √310 √110 ¤T ,uˆ2 = £ √110 − √310 ¤T leading to the orthogonal matrix Uˆ = " √ √3 110 10 −√√110 310 # ATA = ⎡ ⎣ 11 7 14 4 8 −2 ⎤ ⎦ ·4 11 14 8 7 −2 ¸ = ⎡ ⎣ ⎤ ⎦ Solving ¯ ¯ATA − μI¯ ¯ = ¯¯¯¯¯¯ 80 − μ 100 40 100 170 − μ 140 − μ ¯¯¯¯¯¯ = 0 gives the eigenvalues μ1 = 360,μ2 = 90,μ3 = 0 with corresponding normalized eigenvectors vˆ1 = [ 1 3 2 3 23 ]T ,vˆ2 = [ − 32 − 1 3 23 ]T ,vˆ3 = [ 2 3 − 2 3 13 ]T ©c Pearson Education Limited 201138 Glyn James, Advanced Modern Engineering Mathematics, 4th Edition leading to the orthogonal matrix Vˆ = ⎡ ⎣ 1 32 3 23 − −232 3 1 3 −231 32 3 ⎤ ⎦ The singular values of A are σ1 = √360 = 6√10 and σ2 = √90 = 3√10 giving Σ = · 6√0 3 10 0 0 √10 0 ¸ Thus, the SVD form of A is A = UΣ ˆ Vˆ T = " √ √3 110 10 −√√110 310 # ·6√0 3 10 0 0 √10 0¸ ⎡ ⎣ −1 32 32 3 − − 23 13 ⎤ ⎦ (Direct multiplication confirms A = · 4 11 14 8 7 −2 ¸) (b) Using (1.55) the pseudo inverse of A is A† = VΣ ˆ ∗Uˆ, Σ∗ = ⎡ ⎣ 6√0 0 0 110 3√0210 ⎤ ⎦ ⇒ ⎡ ⎣ 1 32 3 23 − −232 3 1 3 −2 3 1 32 3 ⎤ ⎦ ⎡ ⎣ 6√0 0 0 110 3√0110 ⎤ ⎦ " √ √3 110 10 −√√110 310 # ⇒ A† = 180 1 ⎡ ⎣ −10 4 8 1 13 −10 ⎤ ⎦ AA† = 1 180 · 4 11 14 8 7 −2 ¸ ⎡ ⎣ −10 4 8 1 13 −10 ⎤ ⎦ = 180 1 · ¸ = I (c) Rank(A) = 2 so A is of full rank. Since number of rows is less than the number of columns A† may be determined using (1.58b) as A† = AT(AAT)−1 = ⎡ ⎣ 11 7 14 4 8 −2⎤ ⎦ · ¸−1 = 180 1 ⎡ ⎣ −10 4 8 1 13 −10 ⎤ ⎦ which confirms with the value determined in (b).