CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS

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CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS EXERCISE 01 Page 545 1. Use matrices to solve: 3x + 4y x + 5y + 7 3x + 4y x + 5y 7 Hence, 3 4 x 0 5 y 7 The inverse of 3 4 5 is: 1 5 4 1 5 4 15 8 3 7 3 Thus, x 1 5 4 0 1 8 y 7 3 7 7 1 4 3 x 4 and y 3. Use matrices to solve: p + 5q + 14.6 3.1p + 1.7 q +.06 p + 5q 14.6 3.1p + 1.7 q.06 Hence, 5 p 14.6 3.1 1.7 q.06 The inverse of 5 3.1 1.7 is: 1 1.7 5 1 1.7 5 3.4 15.5 3.1 1.1 3.1 Thus, p 1 1.7 5 14.6 1 14.5 q 1.1 3.1.06 1.1 41.14 1. 3.4 p 1. and q 3.4 816

3. Use matrices to solve: x + y + 3z 5 x 3y z 3 3x + 4y + 5z 3 Since x + y + 3z 5 x 3y z 3 3x + 4y + 5z 3, 1 3 x 5 3 1 y 3 3 4 5 z 3 Matrix of cofactors is: 11 7 1 14 10 7 7 7 and the transpose of cofactors is: 11 7 7 14 7 1 10 7 1 3 3 1 3 4 5 1( 11) (7) + 3( 1) 8 The inverse of 1 3 3 1 3 4 5 is: 11 7 1 7 14 7 8 1 10 7 Thus, x 11 7 5 8 1 1 1 y 7 14 7 3 8 1 8 8 z 1 10 7 3 56 x 1, y 1 and z 4. Use matrices to solve: 3a + 4b 3c a + b + c 15 7a 5b + 4c 6 Since 3a + 4b 3c a + b + c 15 7a 5b + 4c 6, 3 4 3 a b 15 7 5 4 c 6 817

Matrix of cofactors is: 18 4 1 33 43 14 0 14 and the transpose of cofactors is: 18 1 14 33 0 4 43 14 3 4 3 7 5 4 3(18) 4( ) + 3( 4) 154 The inverse of 3 4 3 7 5 4 is: 18 1 14 1 33 0 154 4 43 14 Thus, a 18 1 14 385.5 1 1 b 33 0 15 539 3.5 154 154 c 4 43 14 610016.5 a.5, b 3.5 and c 6.5 5. Use matrices to solve: p + q + 3r + 7.8 p + 5q r 1.4 5p q + 7r 3.5 Since p + q + 3r 7.8 p + 5q r 1.4 5p q + 7r 3.5 1 3 p 7.8 5 1 q 1.4 5 1 7 r 3.5 Matrix of cofactors is: 34 19 7 17 8 11 17 7 1 and the transpose of cofactors is: 34 17 17 19 8 7 7 11 1 1 3 5 1 5 1 7 1(34) (19) + 3( 7) 85 The inverse of 1 3 5 1 5 1 7 is: 34 17 17 1 19 8 7 85 7 11 1 818

Thus, p 34 17 17 7.8 348.5 4.1 1 1 q 19 8 7 1.4 161.5 1.9 85 85 r 7 11 1 3.5 9.5.7 p 4.1, q 1.9 and r.7 6. In two closed loops of an electrical circuit, the currents flowing are given by the simultaneous equations I 1 + I + 4 5I 1 + 3I 1 Use matrices to solve for I 1 and I I 1 + I 4 5I 1 + 3I 1 Hence, 1 I1 4 5 3 I 1 The inverse of 1 5 3 is: 1 3 1 3 3 10 5 1 7 5 1 Thus, I1 1 3 4 1 14 I 7 5 1 1 7 1 3 I 1 and I 3 7. The relationship between the displacement s, velocity v and acceleration a of a piston is given by the equations s + v + a 4 3s v + 4a 5 3s + v a 4 Use matrices to determine the values of s, v and a. Since s + v + a 4 3s v + 4a 5 819

3s + v a 4 1 s 4 3 1 4 v 5 3 1 a 4 Matrix of cofactors is: 7 15 9 6 7 4 10 7 and the transpose of cofactors is: 7 6 10 15 7 9 4 7 1 3 1 4 3 1 1( 7) ( 15) + (9) 41 The inverse of 1 3 1 4 3 1 is: 7 6 10 1 15 7 41 9 4 7 Thus, s 7 6 10 4 8 1 1 v 15 7 5 13 3 41 41 a 9 4 7 4 164 4 s, v 3 and a 4 8. In a mechanical system, acceleration x, velocity x and distance x are related by the simultaneous equations: 3.4 x+ 7.0 x 13.x 11.39 6.0 x+ 4.0 x+ 3.5x 4.98.7 x+ 6.0 x+ 7.1x 15.91 Use matrices to find the values of x, x and x. Since 3.4 x+ 7.0 x 13.x 11.39 6.0 x+ 4.0 x+ 3.5x 4.98.7 x+ 6.0 x+ 7.1x 15.91, x 3.4 7.0 13. 11.39 6.0 4.0 3.5 x 4.98.7 6.0 7.1 x 15.91 80

Matrix of cofactors is: 7.4 5.05 46.8 18.9 59.78 1.5 77.3 67.3 55.6 and the transpose of cofactors is: 7.4 18.9 77.3 5.05 59.78 67.3 46.8 1.5 55.6 3.4 7.0 13. 6.0 4.0 3.5 3.4(7.4) 7.0 ( 5.05) + ( 13.)( 46.8) 1007.7.7 6.0 7.1 The inverse of 3.4 7.0 13. 6.0 4.0 3.5.7 6.0 7.1 is: 7.4 18.9 77. 1 5.05 59.78 67.3 1007.7 46.8 1.5 55.6 Thus, x 7.4 18.9 77.3 11.39 503.635 0.5 1 1 x 5.05 59.78 67.3 4.98 775.5979 0.77 1007.7 1007.7 x 46.8 1.5 55.6 15.91 1410.178 1.4 x.5, x.77 and x 1.4 81

EXERCISE 0 Page 548 1. Use determinants to solve the simultaneous equations: 3x 5y 17.6 7y x Since 3x 5y + 17.6 x x + 7y x y 1 5 17.6 3 17.6 3 5 7 7 x y 1 13. 30.8 11 13. 11 1. and y 30.8 11.8. Use determinants to solve the simultaneous equations:.3m 4.4n 6.84 8.5n 6.7m 1.3 Since.3m 4.4n 6.84 6.7m + 8.5n 1.3 m n 1 4.4 6.84.3 6.84.3 4.4 8.5 1.3 6.7 1.3 6.7 8.5 m 63.55 9.93 m n 1 63.55 48.657 9.93 6.4 and n 48.657 9.93 4.9 3. Use determinants to solve the simultaneous equations: 3x + 4y + z 10 x 3y + 5z + 9 x + y z 6 Since 3x + 4y + z 10 x 3y + 5z + 9 x + y z 6 8

x y z 1 4 1 10 3 1 10 3 4 10 3 4 1 3 5 9 5 9 3 9 3 5 1 6 1 1 6 1 6 1 1 x y z 1 4( 1) 1(0) 10( 7) 3( 1) 1( 1) 10( 7) 3(0) 4( 1) 10(7) 3( 7) 4( 7) + 1(7) x y z 1 14 8 14 14 Hence, x 14 14 8 1, y 14 and z 14 14 1 4. Use determinants to solve the simultaneous equations: 1.p.3q 3.1r + 10.1 4.7p + 3.8q 5.3r 1.5 3.7p 8.3q + 7.4r + 8.1 Since 1.p.3q 3.1r + 10.1 4.7p + 3.8q 5.3r 1.5 3.7p 8.3q + 7.4r + 8.1 p q r 1.3 3.1 10.1 1. 3.1 10.1 1..3 10.1 1..3 3.1 3.8 5.3 1.5 4.7 5.3 1.5 4.7 3.8 1.5 4.7 3.8 5.3 8.3 7.4 8.1 3.7 7.4 8.1 3.7 8.3 8.1 3.7 8.3 7.4 p q.3(10.17) + 3.1( 71.67) + 10.1( 15.87) 1.(10.17) + 3.1(11.6) + 10.1(54.39) r 1 1.( 71.67) +.3(11.6) + 10.1( 53.07) 1.( 15.87) +.3(54.39) 3.1( 53.07) p q r 1 405.855 117.565 135.85 70.57 Hence, p 405.855 70.57 1.5, q 117.565 70.57 4.5 and r 135.85 70.57.5 83

5. Use determinants to solve the simultaneous equations: x y z 1 + 3 5 0 x y z 19 + 4 3 40 x + y z 59 60 Since x y z 1 + + 3 5 0 (1) x y z 19 + 4 3 40 () x + y z 59 60 (3) Multiplying each term in (1) by 60 gives: 30x 0y + 4z + 3 Multiplying each term in () by 10 gives: 30x + 80y 60z 57 Multiplying each term in (3) by 60 gives: 60x + 60y 60z 59 x y z 1 0 4 3 30 4 3 30 0 3 30 0 4 80 60 57 30 60 57 30 80 57 30 80 60 60 60 59 60 60 59 60 60 59 60 60 60 x y 0(10) 4( 1300) + 3( 100) 30(10) 4(1650) + 3(1800) z 1 30( 1300) + 0(1650) + 3( 3000) 30( 100) + 0(1800) + 4( 3000) x y z 1 5 00 30 600 15 000 7 000 Hence, x 5 00 7 000 7 30 600, y 0 7 000 17 40 and z 15 000 7 000 5 4 6. In a system of forces, the relationship between two forces F 1 and F is given by 5F 1 + 3F + 6 3F 1 + 5F + 18 Use determinants to solve for F 1 and F 84

Since 5F 1 + 3F + 6 3F 1 + 5F + 18 F1 F 1 3 6 5 6 5 3 5 18 3 18 3 5 F1 F 1 4 7 16 F 1 4 16 1.5 and F 7 4.5 16 7. Applying mesh-current analysis to an a.c. circuit results in the following equations: (5 j4) I 1 ( j4) I 100 0 (4 + j3 j4) I ( j4) I 1 Solve the equations for I 1 and I (5 j4) I 1 ( j4) I 100 0 ( j4) I 1 + (4 + j3 j4) I + 0 (5 j4) I 1 + j4 I 100 j4 I 1 + (4 j) I + 0 Hence, I1 I 1 j4 100 (5 j4) 100 (5 j4) j4 (4 j) 0 j4 0 j4 (4 j) I1 I 1 100(4 j) j400 (5 j4)(4 j) j4 ( ) Thus, I 1 I1 I 1 400 j100 j400 3 j1 400 j100 41.31 14.04 3 j1 38.75 33.7 10.77 19.3 A and I j400 400 90 3 j1 38.75 33.7 10.45 56.73 A 85

8. Kirchhoff s laws are used to determine the current equations in an electrical network and show that i 1 + 8i + 3i 3 31 3i 1 i + i 3 5 i 1 3i + i 3 6 Use determinants to solve for i 1, i and i 3 Since i 1 + 8i + 3i 3 + 31 3i 1 i + i 3 + 5 i 1 3i + i 3 6 i i i 8 3 31 1 3 31 1 8 31 1 8 3 1 5 3 1 5 3 5 3 1 3 6 6 3 6 3 i1 i i3 1 8( 16) 3(7) + 31( 1) 1( 16) 3( 8) + 31(4) 1(7) 8( 8) + 31( 5) 1( 1) 8(4) + 3( 5) 1 3 1 i i i 40 19 96 48 1 3 1 Hence, i 1 40 48 5, i 19 48 4 and i 3 96 48 9. The forces in three members of a framework are F 1, F and F 3. They are related by the simultaneous equations shown below. 1.4 F 1 +.8 F +.8 F 3 5.6 4. F 1 1.4 F + 5.6 F 3 35.0 4. F 1 +.8 F 1.4 F 3 5.6 Find the values of F 1, F and F 3 using determinants. 1.4 F 1 +.8 F +.8 F 3 5.6 4. F 1 1.4 F + 5.6 F 3 35.0 4. F 1 +.8 F 1.4 F 3 + 5.6 86

Hence, F1 F F3 1.8.8 5.6 1.4.8 5.6 1.4.8 5.6 1.4.8.8 1.4 5.6 35.0 4. 5.6 35.0 4. 1.4 35.0 4. 1.4 5.6.8 1.4 5.6 4. 1.4 5.6 4..8 5.6 4..8 1.4 F1 F.8( 17.64).8(90.16) 5.6( 13.7) 1.4( 17.64).8(170.5) 5.6( 9.4) F3 1 1.4(90.16).8(170.5) 5.6(17.64) 1.4( 13.7).8( 9.4) +.8(17.64) F1 F F3 1 5.008 337.51 450.016 11.504 Thus, F 1 5.008 11.504 F 337.51 11.504 3 and F 3 450.016 11.504 4 10. Mesh-current analysis produces the following three equations: 0 0 (5 + 3 j4) I (3 j4) I 1 10 90 (3 j4 + ) I (3 j4) I I 1 3 15 0 10 90 (1 + ) I I 3 Solve the equations for the loop currents I 1, I and I 3 Rearranging gives: (8 j4) I 1 (3 j4) I + 0 I 3 0 Hence, (3 j4) I 1 + (5 j4) I I 3 j10 0 I 1 I + 14 I 3 + (15 + j10) I1 I I3 (3 j4) 0 0 (8 j4) 0 0 (8 j4) (3 j4) 0 (5 j4) j10 (3 j4) j10 (3 j4) (5 j4) j10 14 (15 + j10) 0 14 (15 + j10) 0 (15 + j10) 1 (8 j4) (3 j4) 0 (3 j4) (5 j4) 0 14 I1 I (3 j4) (15 + j10) + j140 0 14(5 j4) 4 (8 j4) (15 + j10) + j140 0 14(3 j4) [ ] [ ] [ ] [ ] 87

I3 (8 j4) (5 j4)(15 + j10) j0) + (3 j4) (3 j4)(15 + j10) 0 (3 j4) [ ] [ ] [ ] 1 (8 j4) 14(5 j4) 4 + (3 j4) 14(3 j4) [ ] [ ] I1 I (3 j4) 30 + j10 0 66 j56 (8 j4) 30 + j10 0 4 + j56) [ ] [ ] [ ] [ ] I3 (8 j4) 115 j30 + (3 j4) 85 + j30 40(3 j4) [ ] [ ] 1 (8 j4) 66 j56 + (3 j4) 4 + j56 [ ] [ ] I1 I ( 90 + j360 + j10 + 480) 130 + j110 40 + j960 + j10 + 480 + 840 j110 Hence, I 1 I3 90 j40 j460 10 55 + j430 + 10 10 + j160 1 58 j448 j64 4 16 + j168 + j168 + 4 I1 I I3 1 ( 1710 + j640) (1080 j40) (545 j110) (40 j376) ( 1710 + j640) 185.84 0.5 (40 j376) 550.44 43.09 3.317.57 A I I 3 (1080 j40) 1080.74.1 (40 j376) 550.44 43.09 1.963 40.97 A (545 j110) 555.99 11.41 555.99 191.41 (40 j376) 550.44 43.09 550.44 43.09 1.010 148.3 A 88

EXERCISE 03 Page 550 1. Repeat Problems 3, 4, 5, 7 and 8 of Exercise 01 on page 545, using Cramer s rule.. Repeat Problems 3, 4, 8 and 9 of Exercise 0 on page 548, using Cramer s rule. Just a sample of these questions are detailed below. The remainder are left to the reader. 1. Q(3), Exercise 01 Use Cramers rule to solve: x + y + 3z 5 x 3y z 3 3x + 4y + 5z 3 x 5 3 3 3 1 3 4 5 5( 11) (18) + 3(1) 8 1 3 1( 11) (7) + 3( 1) 8 3 1 3 4 5 1 y z 1 5 3 3 1 3 3 5 1(18) 5(7) + 3(15) 8 1 8 8 8 1 5 3 3 3 4 3 1( 1) (15) + 5( 1) 56 8 8 8 1. Q(7), Exercise 01 Use Cramer s rule to solve: s + v + a 4 3s v + 4a 5 3s + v a 4 89

s 4 5 1 4 4 1 4( 7) ( 9) + (46) 8 1 1( 7) ( 15) + (9) 41 3 1 4 3 1 v a 1 4 3 5 4 3 4 1 1( 9) 4( 15) + ( 87) 13 3 41 41 41 1 4 3 1 5 3 4 1( 46) ( 87) + 4(9) 164 4 41 41 41. Q(8), Exercise 0 Use Cramer s rule to solve: i1+ 8i + 3i3 31 3i1 i + i3 5 i 3i + i 6 1 3 i 1 31 8 3 5 1 6 3 31( 1) 8( 16) + 3(7) 40 1 8 3 1( 1) 8(4) + 3( 5) 48 3 1 3 5 i i 3 1 31 3 3 5 1 6 1( 16) + 31(4) + 3(8) 19 4 48 48 48 1 8 31 3 5 3 6 1( 7) 8(8) 31( 5) 96 48 48 48 830

EXERCISE 04 Page 551 1. In a mass-spring-damper system, the acceleration x m/s, velocity x m/s and displacement x m are related by the following simultaneous equations: 6. x+ 7.9 x + 1.6 x 18.0 7.5 x+ 4.8 x + 4.8 x 6.39 13.0 x+ 3.5 x 13.0 x 17.4 By using Gaussian elimination, determine the acceleration, velocity and displacement for the system, correct to decimal places. 6. x+ 7.9 x+ 1.6x 18.0 7.5 x+ 4.8 x+ 4.8x 6.39 (1) () 13.0 x+ 3.5 x 13.0x 17.4 () 7.5 6. (1) gives: 0 4.7565 x 10.44 x 15.384 ( ) (3) 13.0 6. (1) gives: 0 13.065 x 39.419 x 55.14 (3 ) 13.065 (3 ) ( ) gives: 0 + 0 10.737 x 1.886 4.7565 1.886 x 1.0 10.737 From (3 ), 13.065 x 39.419(1.) 55.14 (3) 13.065 x 55.14 + 39.419(1.) and x 55.14 + 39.419(1.) 13.065.60 From (1), 6. x+ 7.9(0.60) + 1.6(1.) 18.0 and x 6. x 18.0 4.74 15. 1.86 1.86 6. 0.30 831

. The tensions, T1, T and T 3 in a simple framework are given by the equations: 5T + 5T + 5T 7.0 1 3 T + T + 4T.4 1 3 4T + T 4.0 1 Determine T 1, T and T 3 using Gaussian elimination. 5T + 5T + 5T 7.0 (1) 1 3 T + T + 4T.4 () 1 3 4T + T + 0T 4.0 (3) 1 3 () 1 5 (1) gives: 0 + T + 3T3 1.0 ( ) (3) 4 5 (1) gives: 0 T 4T3 1.6 (3 ) (3 ) 1 ( ) gives: T 3.4 T 3. In (3 ) T 4(0.) 1.6 and T.4 In (1) 5T 1 + 5(0.4) + 5(0.) 7.0 T 1.6 + 0.8 0.8 5T 1 7.0.0 1.0 4.0 and T 1.8 3. Repeat Problems 3, 4, 5, 7 and 8 of Exercise 01 on page 545, using the Gaussian elimination method. These problems are left to the reader using the same method as in the above examples. 4. Repeat Problems 3, 4, 8 and 9 of Exercise 0 on page 548, using the Gaussian elimination method. These problems are left to the reader using the same method as in the above examples. 83

EXERCISE 05 Page 556 1. Determine the (a) eigenvalues (b) eigenvectors for the matrix: 4 1 1 (a) The eigenvalue is determined by solving the characteristic equation A λi 4 1 0 λ 1 1 0 1 4 λ 0 1 1 0 λ λ 4 1 1 λ (Given a square matrix, we can get used to going straight to this characteristic equation) Hence, ( λ)( 1 λ) ( 4)( 1) λ + λ + λ 4 and λ λ 6 (λ 3)(λ + ) λ 3 λ 3 or λ + λ (Instead of factorizing, the quadratic formula could be used; even electronic calculators can solve quadratic equations) Hence, the eigenvalues of the matrix 4 1 1 are 3 and (b) From (a) the eigenvalues of 4 1 1 are λ 1 3 and λ Using the equation (A λi)x for λ 1 3 3 4 x1 1 1 3 x 0 1 4 x1 1 4 x 0 833

x 4x 1 x1 4x Hence, whatever value x is, the value of x 1 will be 4 times greater. Hence the simplest 4 eigenvector is: x1 1 Using the equation (A λi)x for λ and 4 x 0 1 1 x 0 1 4 4 x1 1 1 x 0 4x 4x 1 x + 1 x 0 From either of these two equations, x1 x or x x1 Hence, whatever value x 1 is, the value of x will be the same. Hence the simplest eigenvector 1 is: x 1 4 Summarizing, x1 1 is an eigenvector corresponding to λ 1 3 1 and x 1 is an eigenvector corresponding to λ. Determine the (a) eigenvalues (b) eigenvectors for the matrix: 3 6 1 4 (a) The eigenvalue is determined by solving the characteristic equation A λi 3 λ 6 1 4 λ Hence, (3 λ)(4 λ) (6)(1) 834

1 3λ 4λ + λ 6 and λ 7λ + 6 (λ 1)(λ 6) λ 1 λ 1 or λ 6 λ 6 Hence, the eigenvalues of the matrix 3 6 1 4 are 1 and 6 (b) From (a) the eigenvalues of 3 6 1 4 are λ 1 1 and λ 6 Using the equation (A λi)x for λ 1 1 3 1 6 x1 1 4 1 x 0 6 x1 1 3 x 0 x + 6x 1 and x1+ 3x x1 3x Hence, whatever value x is, the value of x 1 will be 3 times greater. Hence the simplest 3 eigenvector is: x1 1 Using the equation (A λi)x for λ 6 and 3 6 6 x1 1 4 6 x 0 3 6 x1 1 x 0 3x + 6x 1 x x 1 835

From either of these two equations, x1 x Hence, whatever value x is, the value of x 1 will be two times greater. Hence the simplest eigenvector is: x 1 3 Summarizing, x1 1 and x 1 is an eigenvector corresponding to λ 1 1 is an eigenvector corresponding to λ 6 3. Determine the (a) eigenvalues (b) eigenvectors for the matrix: 3 1 0 (a) The eigenvalue is determined by solving the characteristic equation A λi 3 λ 1 0 λ Hence, (3 λ)( λ) (1)( ) 3λ + λ + and λ 3λ + (λ 1)(λ ) λ 1 λ 1 or λ λ Hence, the eigenvalues of the matrix 3 1 0 are 1 and (b) From (a) the eigenvalues of 3 1 0 are λ 1 1 and λ Using the equation (A λi)x for λ 1 1 3 1 1 x1 0 1 x 0 836

1 x1 1 x 0 x + x 1 and x1 x x x1 Hence, whatever value x 1 is, the value of x will be times greater. Hence the simplest 1 eigenvector is: x1 Using the equation (A λi)x for λ and 3 1 x1 0 x 0 1 1 x1 x 0 x + 1 x 0 x x 1 From either of these two equations, x1 x or x x1 Hence, whatever value x 1 is, the value of x will be 1 times greater. Hence the simplest 1 eigenvector is: x 1 1 Summarizing, x1 1 and x 1 is an eigenvector corresponding to λ 1 1 is an eigenvector corresponding to λ 4. Determine the (a) eigenvalues (b) eigenvectors for the matrix: 1 1 1 4 4 1 1 5 837

(a) The eigenvalue is determined by solving the characteristic equation A λi 1 λ 1 1 4 λ 4 1 1 5 λ Hence, using the top row: ( 1 λ)[ ( λ)(5 λ) (4)(1) ] 1[ 4(5 λ) (4)( 1) ] + 1[ ( 4)(1) ( 1)( λ) ( 1 λ)[ 10 λ 5λ + λ 4 ] + 1[ 0 + 4λ + 4] + 1[ 4 + λ ] ( 1 λ)[ λ 7λ + 6 ] 16 + 4λ λ and λ + 7λ 6 λ 3 + 7 λ 6λ 16 + 4λ λ λ 3 + 6 λ Using the factor theorem or an electronic calculator, λ, λ 6 or λ + 4λ 4 Hence, the eigenvalues of the matrix 1 1 1 4 4 1 1 5 are, 6 and 1 1 1 (b) From (a), the eigenvalues of 4 4 1 1 5 Using the equation (A λi)x for λ 1 From the second equation, 1 1 1 x1 4 4 x 1 1 5 x 3 1 1 x 0 4 0 4 x 1 1 1 3 x 3x1 x + x3 4x1+ 4x3 x + x + 3x 1 3 x x 1 3 are λ 1, λ 6 and λ 3 838

Substituting in the first equation, 3x x + x x x 1 1 1 Hence,when x 1, x and x 1 1 3 1 Hence the simplest eigenvector corresponding to λ 1 is: x1 1 Using the equation (A λi)x for λ 6 1 6 1 1 x1 4 6 4 x 0 1 1 5 6 x 7 1 1 x1 4 4 4 x 1 1 1 x 7x1 x + x3 4x1 4x + 4x3 x1+ x x3 From the third equation, x1 x x3 Substituting in the first equation, 7( x x ) x + x 8x + 8x x x Hence, x1 x x3 Hence, if x 1, x3 1 and x1 3 3 3 3 0 Hence the simplest eigenvector corresponding to λ 6 is: x 1 1 Using the equation (A λi)x for λ 3 1 1 1 x1 4 4 x 1 1 5 x 1 1 1 x1 4 4 4 x 1 1 7 x 839

From the second equation, Substituting in the first equation, x1 x + x3 4x1+ 4x + 4x3 x + x + 7x 1 3 x1 x + x3 ( x + x ) x + x x and x x 3 3 3 1 Hence, if x1 1, x 1 and x3 1 Hence the simplest eigenvector corresponding to λ is: x3 1 0 5. Determine the (a) eigenvalues (b) eigenvectors for the matrix: 1 1 0 1 1 0 1 1 (a) The eigenvalue is determined by solving the characteristic equation A λi 1 λ 1 0 1 λ 1 0 1 1 λ Hence, using the top row: (1 λ)[ ( λ)(1 λ) ( 1)( 1) ] 1[ 1(1 λ) ( 1)(0) ] + 0 (1 λ)[ λ λ + λ 1 ] + 1[ 1 + λ ] (1 λ)[ λ 3λ + 1 ] 1 + λ and λ 3λ + 1 λ 3 + 3 λ λ 1 + λ λ 3 + 4 λ 3λ λ 3 4 λ + 3λ λ( λ 4λ + 3) and λ(λ 1)(λ 3) Hence, λ, λ 1 or λ 3 840

Hence, the eigenvalues of the matrix 1 1 0 1 1 0 1 1 are 0, 1 and 3 1 1 0 (b) From (a), the eigenvalues of 1 1 0 1 1 Using the equation (A λi)x for λ 1 From the first equation, From the third equation, 1 0 1 0 x1 1 0 1 x 0 1 1 0 x 1 1 0 x1 1 1 x 0 1 1 x x1 x x1+ x x3 x + x 3 0 x x x 1 x 3 are λ 1, λ 1 and λ 3 3 Hence,when x1 1, x 1 and x3 1 1 Hence the simplest eigenvector corresponding to λ 1 is: x1 1 1 Using the equation (A λi)x for λ 1 1 1 1 0 x1 1 1 1 x 0 1 1 1 x 0 1 0 x1 1 1 1 x 0 1 0 x x x + x x 1 3 0 841

From the second equation, x x x since x 1 3 Hence, if x1 1, x3 1 and x 1 Hence the simplest eigenvector corresponding to λ 1 is: x 1 Using the equation (A λi)x for λ 3 From the first equation, 1 3 1 0 x1 1 3 1 x 0 1 1 3 x 1 0 x1 1 1 1 x 0 1 x x1 x x1 x x3 x x 3 Substituting in the second equation, x x 1 x ( x ) x x x 1 1 3 1 3 Hence, if x1 1, x and x3 1 1 Hence the simplest eigenvector corresponding to λ 3 3 is: x3 1 6. Determine the (a) eigenvalues (b) eigenvectors for the matrix: 1 3 1 1 (a) The eigenvalue is determined by solving the characteristic equation A λi λ 1 3 λ 1 1 λ 84

Hence, using the top row: ( λ)[ (3 λ)( λ) (1)() ] [ 1( λ) (1)(1) ] + [ (1)() 1(3 λ) ] ( λ)[ 6 3λ λ + λ ] [ λ 1 ] [ 3 + λ] ( λ)[ λ 5λ + 4 ] + λ + λ and λ 10λ + 8 λ 3 + 5 λ 4λ λ 3 + 7 λ 14λ + 8 λ 3 7 λ + 14λ 8 Hence, by calculator, λ 1, λ or λ 4 Hence, the eigenvalues of the matrix 1 3 1 1 are 1, and 4 (b) From (a), the eigenvalues of 1 3 1 1 Using the equation (A λi)x for λ 1 1 1 x1 1 3 1 1 x 1 1 x 1 x1 1 1 x 1 1 x x1+ x x3 x + x + x 1 3 Subtracting one equation from the other gives: and from the first equation, x 3 x x 1 are λ 1 1, λ and λ 3 4 Hence,when x 1, x and x 1 3 Hence the simplest eigenvector corresponding to λ 1 1 is: x1 1 0 843

Using the equation (A λi)x for λ From the first equation, x1 1 3 1 x 1 x 0 x1 1 1 1 x 1 0 x x x3 x1+ x + x3 x + x 1 x x 3 From the third equation, x x ie.. x x 1 1 3 Hence, if x 1, x3 1 and x1 Hence the simplest eigenvector corresponding to λ is: x 1 1 Using the equation (A λi)x for λ 4 4 x1 1 3 4 1 x 1 4 x x1 1 1 1 x 1 x x1+ x x3 x1 x + x3 x + x x 1 3 Subtracting the third equation from the first equation gives: From the second equation, x 1 x x since x 1 3 844

Hence, if x 1, x3 1 and x1 0 Hence the simplest eigenvector corresponding to λ 3 4 is: x3 1 1 7. Determine the (a) eigenvalues (b) eigenvectors for the matrix: 1 1 0 1 1 3 (a) The eigenvalue is determined by solving the characteristic equation A λi 1 λ 1 0 λ 1 1 3 λ Hence, using the top row: (1 λ)[ ( λ)(3 λ) (1)() ] 1[ 0 ( 1)() ] + [ 0 ( 1)( λ) (1 λ)[ 6 λ 3λ + λ ] + [ λ ] (1 λ)[ λ 5λ + 4 ] + 4 λ and λ 5λ + 4 λ 3 + 5 λ 4λ + λ λ 3 + 6 λ 11λ + 6 λ 3 6 λ + 11λ 6 Hence, using a calculator, λ 1, λ or λ 3 Hence, the eigenvalues of the matrix 1 1 0 1 1 3 are 1, and 3 1 1 (b) From (a), the eigenvalues of 0 1 1 3 Using the equation (A λi)x for λ 1 1 are λ 1 1, λ and λ 3 3 1 1 1 x1 0 1 x 1 1 3 1 x 845

From the first equation, From the third equation, 0 1 x1 0 1 x 1 1 x x + x3 x + x + x 1 3 x x 3 x + ( x ) + x x 1 3 3 1 Hence,when x 1, x and x 3 3 1 0 Hence the simplest eigenvector corresponding to λ 1 1 is: x1 1 Using the equation (A λi)x for λ 1 1 x1 0 x 1 1 3 x 1 1 x1 0 0 x 1 1 1 x x1+ x + x3 x 3 x1+ x + x3 From the second equation, x 3 From the first equation, x x 1 Hence, if x1 1, x 1 and x3 1 Hence the simplest eigenvector corresponding to λ is: x 1 0 Using the equation (A λi)x for λ 3 3 846

From the third equation, From the second equation, 1 3 1 x1 0 3 x 1 1 3 3 x 1 x1 0 1 x 1 1 0 x x1+ x + x3 x + x3 x + x x x 1 0 x 1 x 3 Hence, if x1 1, x 1 and x3.5 or if x1, x and x3 1 Hence the simplest eigenvector corresponding to λ 3 3 is: x3 1 847

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