Lecture Notes in Differential Equations - Bruce E. Shapiro
Lecture Notes in Differential Equations - Bruce E. Shapiro Lecture Notes in Differential Equations - Bruce E. Shapiro
122 LESSON 14. REVIEW OF LINEAR ALGEBRA Example 14.2. Let ⎛ A = ⎝ 1 2 3 ⎞ 4 5 6⎠ (14.29) 7 8 9 Then ∣ ∣∣∣ cof a 12 = (−1) 1+2 4 6 7 9∣ = (−1)(36 − 42) = 6 (14.30) Definition 14.26. Let A be a square matrix of order n. The Clasical Adjoint of A, denoted adj A, is the transopose of the matrix that results when every element of A is replaced by its cofactor. Example 14.3. Let ⎛ 1 0 ⎞ 3 A = ⎝4 5 0⎠ (14.31) 0 3 1 The classical adjoint is adj A = Transpose ⎛ ⎞ (1)[(1)(5) − (0)(3)] (−1)[(4)(1) − (0)(0)] (1)[(4)(3) − (5)(0)] ⎝(−1)[(0)(1) − (3)(3)] (1)[(1)(1) − (3)(0)] (−1)[(1)(3) − (0)(0)] ⎠ (1)[(0)(0) − (3)(5)] (−1)[(1)(0) − (3)(4)] (1)[(1)(5) − (0)(4)] ⎛ (14.32) ⎞ ⎛ ⎞ 5 −4 12 5 9 −15 = Transpose ⎝ 9 1 −3⎠ = ⎝−4 1 12 ⎠ (14.33) −15 12 5 12 −3 5 Theorem 14.27. Let A be a non-singular square matrix. Then A −1 = 1 adj A (14.34) det A Example 14.4. Let A be the square matrix defined in equation 14.31. Then det A = 1(5 − 0) − 0 + 3(12 − 0) = 41 (14.35) Hence ⎛ ⎞ A −1 = 1 5 9 −15 ⎝−4 1 12 ⎠ (14.36) 41 12 −3 5
123 In practical terms, computation of the determinant is computationally inefficient, and there are faster ways to calculate the inverse, such as via Gaussian Elimination. In fact, determinants and matrix inverses are very rarely used computationally because there is almost always a better way to solve the problem, where by better we mean the total number of computations as measure by number of required multiplications and additions. Definition 14.28. Let A be a square matrix. Then the eigenvalues of A are the numbers λ and eigenvectors v such that Av = λv (14.37) Definition 14.29. The characteristic equation of a square matrix of order n is the n th order (or possibly lower order) polynomial det(A − λI) = 0 (14.38) Example 14.5. Let A be the square matrix defined in equation 14.31. Then its characteristic equation is 0 = ∣ 1 − λ 0 3 4 5 − λ 0 0 3 1 − λ ∣ (14.39) = (1 − λ)(5 − λ)(1 − λ) − 0 + 3(4)(3) (14.40) = 41 − 11λ + 7λ 2 − λ 3 (14.41) Theorem 14.30. The eigenvalues of a square matrix A are the roots of its characteristic polynomial. Example 14.6. Let A be the square matrix defined in equation 14.31. Then its eigenvalues are the roots of the cubic equation 41 − 11λ + 7λ 2 − λ 3 = 0 (14.42) The only real root of this equation is approximately λ ≈ 6.28761. There are two additional complex roots, λ ≈ 0.356196 − 2.52861i and λ ≈ 0.356196 + 2.52861i. Example 14.7. Let ⎛ 2 −2 ⎞ 3 A = ⎝1 1 1 ⎠ (14.43) 1 3 −1
- Page 79 and 80: 71 From the first of equations (9.5
- Page 81 and 82: 73 Differentiating equations (9.81)
- Page 83 and 84: 75 This has the form Mdt + Ndy = 0
- Page 85 and 86: Lesson 10 Integrating Factors Defin
- Page 87 and 88: 79 Differentiating with respect to
- Page 89 and 90: 81 Proof. In each of the five cases
- Page 91 and 92: 83 as required by equation (10.31).
- Page 93 and 94: 85 Since M y ≠ N t , equation (10
- Page 95 and 96: 87 the revised equation (10.100) is
- Page 97 and 98: 89 Substituting (10.129) into (10.1
- Page 99 and 100: Lesson 11 Method of Successive Appr
- Page 101 and 102: 93 because the integral is zero (th
- Page 103 and 104: 95 Example 11.1. Construct the Pica
- Page 105 and 106: 97 We can then plug this expression
- Page 107 and 108: Lesson 12 Existence of Solutions* I
- Page 109 and 110: 101 • Interchangeability of Limit
- Page 111 and 112: 103 But on the square −1 ≤ t
- Page 113 and 114: 105 Thus lim φ n = φ 0 + lim n→
- Page 115 and 116: 107 because the right hand side doe
- Page 117 and 118: Lesson 13 Uniqueness of Solutions*
- Page 119 and 120: 111 The proof of theorem (13.1) is
- Page 121 and 122: 113 But δ(t) is an absolute value,
- Page 123 and 124: 115 Substituting (13.66) into (13.6
- Page 125 and 126: Lesson 14 Review of Linear Algebra
- Page 127 and 128: 119 Definition 14.10. An m × n (or
- Page 129: 121 Definition 14.19. Matrix Multip
- Page 133 and 134: 125 Simplifying 4x − 2 + 3z = 0 (
- Page 135 and 136: Lesson 15 Linear Operators and Vect
- Page 137 and 138: 129 Example 15.3. By a similar argu
- Page 139 and 140: 131 Therefore ‖y + z‖ 2 ≤ ‖
- Page 141 and 142: 133 Definition 15.5. Two vectors y,
- Page 143 and 144: Lesson 16 Linear Equations With Con
- Page 145 and 146: 137 Hence both r = 1 and r = 3. Thi
- Page 147 and 148: 139 The second order linear initial
- Page 149 and 150: 141 The general solution to is give
- Page 151 and 152: Lesson 17 Some Special Substitution
- Page 153 and 154: 145 Therefore since z = y ′ , Int
- Page 155 and 156: 147 Example 17.5. Solve yy ′′ +
- Page 157 and 158: 149 where I is the identity matrix.
- Page 159 and 160: 151 can be rewritten by solving a =
- Page 161 and 162: Lesson 18 Complex Roots We know for
- Page 163 and 164: 155 Theorem 18.2. Euler’s Formula
- Page 165 and 166: 157 For k = 0, 1, 2, . . . , n −
- Page 167 and 168: 159 and its roots are given by The
- Page 169 and 170: 161 The motivation for equation 18.
- Page 171 and 172: Lesson 19 Method of Undetermined Co
- Page 173 and 174: 165 3. If f(t) = e rt and r is a ro
- Page 175 and 176: 167 Example 19.4. Solve ⎫ y ′
- Page 177 and 178: 169 Adding the two equations gives
- Page 179 and 180: Lesson 20 The Wronskian We have see
122 LESSON 14. REVIEW OF LINEAR ALGEBRA<br />
Example 14.2. Let<br />
⎛<br />
A = ⎝ 1 2 3<br />
⎞<br />
4 5 6⎠ (14.29)<br />
7 8 9<br />
Then<br />
∣ ∣∣∣<br />
cof a 12 = (−1) 1+2 4 6<br />
7 9∣ = (−1)(36 − 42) = 6 (14.30)<br />
Def<strong>in</strong>ition 14.26. Let A be a square matrix of order n. The Clasical<br />
Adjo<strong>in</strong>t of A, denoted adj A, is the transopose of the matrix that results<br />
when every element of A is replaced by its cofactor.<br />
Example 14.3. Let<br />
⎛<br />
1 0<br />
⎞<br />
3<br />
A = ⎝4 5 0⎠ (14.31)<br />
0 3 1<br />
The classical adjo<strong>in</strong>t is<br />
adj A = Transpose<br />
⎛<br />
⎞<br />
(1)[(1)(5) − (0)(3)] (−1)[(4)(1) − (0)(0)] (1)[(4)(3) − (5)(0)]<br />
⎝(−1)[(0)(1) − (3)(3)] (1)[(1)(1) − (3)(0)] (−1)[(1)(3) − (0)(0)] ⎠<br />
(1)[(0)(0) − (3)(5)] (−1)[(1)(0) − (3)(4)] (1)[(1)(5) − (0)(4)]<br />
⎛<br />
(14.32)<br />
⎞ ⎛<br />
⎞<br />
5 −4 12 5 9 −15<br />
= Transpose ⎝ 9 1 −3⎠ = ⎝−4 1 12 ⎠ (14.33)<br />
−15 12 5 12 −3 5<br />
Theorem 14.27. Let A be a non-s<strong>in</strong>gular square matrix. Then<br />
A −1 = 1 adj A (14.34)<br />
det A<br />
Example 14.4. Let A be the square matrix def<strong>in</strong>ed <strong>in</strong> equation 14.31.<br />
Then<br />
det A = 1(5 − 0) − 0 + 3(12 − 0) = 41 (14.35)<br />
Hence<br />
⎛<br />
⎞<br />
A −1 = 1 5 9 −15<br />
⎝−4 1 12 ⎠ (14.36)<br />
41<br />
12 −3 5
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