Mathematical Methods for Physics and Engineering - Matematica.NET

8.20 HINTS AND ANSWERS8.5 Use the ( property of the determinant ) of a matrix product.0 − tan(θ/2)8.7 (d) S =.tan(θ/2) 0(e) Note that (I + K)(I − K) =I − K 2 =(I − K)(I + K).8.9 (b) 32iA.8.11 a = b cos γ + c cos β, and cyclic permutations; a 2 = b 2 + c 2 − 2bc cos α, and cyclicpermutations.8.13 (a) 2 −1/2 (0 0 1 1) T , 6 −1/2 (2 0 − 1 1) T ,39 −1/2 (−1 6 − 1 1) T , 13 −1/2 (2 1 2 − 2) T .(b) 5 −1/2 (1 2 0 0) T , (345) −1/2 (14 − 7 10 0) T ,(18 285) −1/2 (−56 28 98 69) T .8.15 C does not commute with the others; A, B and D have (1 − 2) T and (2 1) T ascommon eigenvectors.8.17 For A : (1 0 − 1) T , (1 α 1 1) T , (1 α 2 1) T .For B : (1 1 1) T , (β 1 γ 1 − β 1 − γ 1 ) T , (β 2 γ 2 − β 2 − γ 2 ) T .The α i , β i and γ i are arbitrary.Simultaneous and orthogonal: (1 0 − 1) T , (1 1 1) T , (1 − 2 1) T .8.19 α j =(v · e j∗ )/(λ j − µ), where λ j is the eigenvalue corresponding to e j .(a) x = (2 1 3) T .(b) Since µ is equal to one of A’s eigenvalues λ j , the equation only has a solutionif v · e j∗ = 0; (i) no solution; (ii) x =(1 1 3/2) T .8.21 U = (10) −1/2 (1, 3i;3i, 1), Λ = (1, 0; 0, 11).8.23 J =(2y 2 −4y +4)/(y 2 +2), with stationary values at y = ± √ 2 and correspondingeigenvalues 2 ∓ √ 2. From the trace property of A, the third eigenvalue equals 2.8.25 Ellipse; θ = π/4, a = √ 22; θ =3π/4, b = √ 10.8.27 The direction of the eigenvector having the unrepeated eigenvalue is(1, 1, −1)/ √ 3.8.29 (a) A = SA ′ S † ,whereS is the matrix whose columns are the eigenvectors of thematrix A to be constructed, and A ′ =diag(λ, µ, ν).(b) A =(λ +2µ +3ν, 2λ − 2µ, λ +2µ − 3ν; 2λ − 2µ, 4λ +2µ, 2λ − 2µ;λ +2µ − 3ν, 2λ − 2µ, λ +2µ +3ν).1(c) (1, 5, −2; 5, 4, 5; −2, 5, 1).38.31 The null space is spanned by (2 0 1 0) T and (1 − 2 0 1) T .8.33 x =3,y =1,z =2.8.35 First show that A is singular. η =1,x =1+2z, y = −3z; η =2,x =2z,y =1− 3z.8.37 L =(1, 0, 0; 1 , 1, 0; 2 , 3, 1),3 3U =(3, 6, 9; 0, −2, 2; 0, 0, 4).(i) x =(−1 1 2) T . (ii) x =(−3 2 2) T .8.39 A is not positive definite, as L 33 is calculated to be √ −6.B = LL T , where the non-zero elements of L areL 11 = √ 5,L 31 = √ 3/5, L 22 = √ 3,L 33 = √ 12/5.8.41⎛( )2 1A † A =, U = √ 1 ⎝1 26√ √ ⎞−1 3√ 22 0 2−1 − √ ⎠√, V = 1 (1 1√3 2 2 1 −18.43 The singular values are 12 √ 6, 0, 18 √ 3 and the calculated best solution is x =1.71,y = −1.94,z = −1.71. The null space is the line x = z,y = 0 and the generalSVD solution is x =1.71 + λ, y = −1.94, z= −1.71 + λ.).315