Sect3 6
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Sect3 6 Sect3 6 Document Transcript

  • SECTION 3.6 DETERMINANTS 0 0 3 4 0 1. 4 0 0 = + (3) = 3 ⋅ 4 ⋅ 5 = 60 0 5 0 5 0 2 1 0 2 1 1 1 2. 1 2 1 = + (2) − (1) = 2(4 − 1) − (2 − 0) = 4 1 2 0 2 0 1 2 1 0 0 0 0 5 0 2 0 5 0 6 8 3. = + (1) 6 9 8 = − (5) = − 5(42 − 0) = − 210 3 6 9 8 0 7 0 10 7 4 0 10 7 5 11 8 7 5 11 8 3 −2 6 23 5 8 4. = − (−3) 3 −2 6 = 3(−4) = − 12(30 − 24) = − 72 0 0 0 −3 3 6 0 4 0 0 4 0 17 0 0 1 0 0 2 0 0 0 2 0 0 0 0 0 3 0 0 0 3 0 3 0 5. 0 0 0 3 0 = +1 = + 2 0 0 4 = 2( +5) = 2 ⋅ 5 ⋅ 3 ⋅ 4 = 120 0 0 0 4 0 4 0 0 0 0 4 5 0 0 0 5 0 0 0 5 0 0 0 3 0 11 −5 0 3 0 −5 0 −2 4 13 6 5 3 0 0 −2 4 6 5 4 5 6. 0 0 5 0 0 = +5 = 5(−2) −2 4 5 = − 10(+3) = 60 7 6 17 7 6 7 7 6 −9 17 7 7 6 7 0 0 2 0 0 0 8 2 0 1 1 1 1 1 1 R 2− 2 R1 7. 2 2 2 = 0 0 0 = 0 3 3 3 3 3 3
  • 2 3 4 2 3 4 R 2 + R1 2 3 8. −2 −3 1 = 0 0 5 = −5 = − 5(4 − 9) = 25 3 2 3 2 7 3 2 7 3 −2 5 3 −2 5 R 3− 2 R1 3 5 9. 0 5 17 = 0 5 17 = + 2 = 5(6 − 0) = 30 0 2 6 −4 12 0 0 2 −3 6 5 0 0 −7 R1+ 3 R 2 1 −2 10. 1 −2 −4 = 1 −2 −4 = + ( −7) = − 7( −5 + 4) = 7 2 −5 2 −5 12 2 −5 12 1 2 3 4 1 2 3 4 5 6 7 0 5 6 7 R 4− 2 R1 0 5 6 7 8 9 11. = = +1 0 8 9 = + 5 = 5 ⋅ 8 = 40 0 0 8 9 0 0 8 9 0 1 0 0 1 2 4 6 9 0 0 0 1 2 0 0 −3 2 0 0 −3 1 11 12 0 1 11 12 R 4 + 2 R1 0 1 11 12 5 13 12. = = + 2 0 5 13 = + 2 = 2 ⋅ 5 = 10 0 0 5 13 0 0 5 13 0 1 0 0 1 −4 0 0 7 0 0 0 1 −4 4 −1 −4 4 −1 0 20 11 R 2+ R3 R1+ 4 R 3 20 11 13. −1 −2 2 = 0 2 5 = 0 2 5 = +1 = 100 − 22 = 78 2 5 1 4 3 1 4 3 1 4 3 4 2 −2 1 1 3 R 2−3 R1 1 1 3 R1− R 2 R 3+ 5 R1 −2 −14 14. 3 1 −5 = 3 1 −5 = 0 −2 −14 = + 1 = − 22 1 18 −5 −4 3 −5 −4 3 0 1 18 −2 5 4 0 13 14 0 13 14 R1+ 2 R 3 R 2−5 R 3 13 14 15. 5 3 1 = 5 3 1 = 0 −17 −24 = + 1 = − 74 −17 −24 1 4 5 1 4 5 1 4 5
  • 2 4 −2 10 0 −4 10 0 −4 R1− 2 R 3 R 2+ 2 R 3 10 −4 16. −5 −4 −1 = −5 −4 −1 = −13 0 1 = − 2 = 84 −13 1 −4 2 1 −4 2 1 −4 2 1 2 3 3 1 2 3 3 1 4 3 −3 R 2+ R1 4 3 −3 0 4 3 −3 R 3− R1 0 4 3 −3 R 3+ R1 17. = = 2 −4 −4 −4 = 2 0 −1 −7 = 8 2 −1 −1 −3 0 −4 −4 −4 −4 −3 2 0 0 −1 0 −4 −3 2 0 −4 −3 2 1 4 4 1 1 4 4 1 1 −2 2 R 2+ 9 R1 1 −2 2 0 1 −2 2 R 3−3 R1 0 1 −2 2 R 3− R1 18. = = 1 −9 −11 1 = 0 −29 19 = 135 3 3 1 4 0 −9 −11 1 1 −3 −2 0 −1 −4 0 1 −3 −2 0 1 −3 −2 1 0 0 3 1 0 0 3 1 −2 0 1 0 0 0 1 −2 0 R 3+ 2 R1 0 1 −2 0 C 2+ 2 C1 19. = = 1 3 −2 9 = 3 4 9 = 39 −2 3 −2 3 0 3 −2 9 −3 3 3 −3 −3 3 0 −3 3 3 0 −3 3 3 1 2 1 −1 1 2 1 −1 R 2− 2 R1 −3 1 5 R 2+ 2 R1 −3 1 5 2 1 3 3 R 4+ R1 0 −3 1 5 R 3+ R1 20. = = 1 1 −2 3 = −5 0 13 = 79 0 1 −2 3 0 1 −2 3 6 −1 3 3 0 8 −1 4 −2 4 0 6 −1 3 3 4 1 2 4 1 3 2 21. ∆ = = 1; x = = 10, y = = −7 5 7 ∆1 7 ∆5 1 5 8 1 3 8 1 5 3 22. ∆ = = 1; x = = − 1, y = = 1 8 13 ∆ 5 13 ∆8 5 17 7 1 6 7 1 17 6 23. ∆ = = 1; x = = 2, y = = −4 12 5 ∆4 5 ∆ 12 4 11 15 1 10 15 1 11 10 24. ∆ = = 1; x = = 5, y = = −3 8 11 ∆ 7 11 ∆ 8 7
  • 5 6 1 12 6 1 5 12 25. ∆ = = 2; x = = 6, y = = −3 3 4 ∆ 6 4 ∆3 6 6 7 1 3 7 1 1 6 3 26. ∆ = = − 2; x = = , y = = 0 8 9 ∆4 9 2 ∆8 4 5 2 −2 1 2 −2 1 1 27. ∆ = 1 5 −3 = 96; x1 = −2 5 −3 = , ∆ 3 5 −3 5 2 −3 5 5 1 −2 5 2 1 1 2 1 1 x2 = 1 −2 −3 = − , x3 = 1 5 −2 = − ∆ 3 ∆ 3 5 2 5 5 −3 2 5 4 −2 4 4 −2 1 4 28. ∆ = 2 0 3 = 35; x1 = 2 0 3 = , ∆ 7 2 −1 1 1 −1 1 5 4 −2 5 4 4 1 3 1 2 x2 = 2 2 3 = , x3 = 2 0 2 = ∆ 7 ∆ 7 2 1 1 2 −1 1 3 −1 −5 3 −1 − 5 1 29. ∆ = 4 −4 −3 = 23; x1 = −4 −4 −3 = 2, ∆ 1 0 −5 2 0 −5 3 3 −5 3 −1 3 1 1 x2 = 4 −4 −3 = 3, x3 = 4 −4 −4 = 0 ∆ ∆ 1 2 −5 1 0 2 1 4 2 3 4 2 1 1 30. ∆ = 4 2 1 = 56; x1 = 1 2 1 = − , ∆ 7 2 −2 −5 −3 −2 −5 1 3 2 1 4 3 1 9 1 2 x2 = 4 1 1 = , x3 = 4 2 1 = ∆ 14 ∆ 7 2 −3 −5 2 −2 −3
  • 2 0 −5 −3 0 −5 1 8 31. ∆ = 4 −5 3 = 14; x1 = 3 −5 3 = − , ∆ 7 −2 1 1 1 1 1 2 −3 −5 2 0 −3 1 10 1 1 x2 = 4 3 3 = − , x3 = 4 −5 3 = ∆ 7 ∆ 7 −2 1 1 −2 1 1 3 4 −3 5 4 −3 1 7 32. ∆ = 3 −2 4 = 6; x1 = 7 −2 4 = − , ∆ 3 3 2 −1 3 2 −1 3 5 −3 3 4 5 1 1 x2 = 3 7 4 = 9, x3 = 3 −2 7 = 8 ∆ ∆ 3 3 −1 3 2 3 4 4 4 = 16 15 13  −1 1 33. det A = − 4, A  4  28 25 23    −2 −3 12  1  34. det A = 35, A −1 = 9 −4 −19 35   13 2 −8     −15 25 −26  1  35. det A = 35, A −1 =  10 −5 8  35  15 −25 19     5 20 −17  1  36. det A = 23, A −1 = 10 17 −11 23    1 4 −8     11 −14 −15 −1 1   37. det A = 29, A =  −17 19 10  29  18 −15 −14  
  •  −6 10 2 = 15 −21 −6  −1 1 38. det A = 6, A  6 12 −18 −6     −21 −1 −13 1  39. det A = 37, A −1 =  4 9 6   37  −6 5 −9     9 12 −13 −1 1   40. det A = 107, A =  11 −21 −4  107  −15 −20 −14     a1  41. If A =   and B = [b1 b 2 ] in terms of the two row vectors of A and the two column a 2   a1b1 a1b 2  vectors of B, then AB =   , so a 2b1 a 2b 2  a b a 2b1  b1  T T ( AB ) =  1 1 =  T  a1 a1  = BT AT , T T a1b 2 a 2b 2   b 2    because the rows of A are the columns of AT and the columns of B are the rows of BT.  a b   x   ax + by ax by 42. det AB = det      = = +   c d  y   cx + dy cx + dy cx + dy x x y y x y = ac + ad + bc + bd = ad + bc x y x y y x x x x = ad − bc = (ad − bc) = (det A)(det B) y y y
  • 43. We expand the left-hand determinant along its first column: ka11 a12 a13 ka21 a22 a23 ka31 a32 a33 = ka11 ( a12 a23 − a22 a13 ) − ka21 ( a12 a33 − a32 a13 ) + ka31 ( a12 a23 − a22 a13 ) = k  a11 ( a12 a23 − a22 a13 ) − a21 ( a12 a33 − a32 a13 ) + a31 ( a12 a23 − a22 a13 )   a11 a12 a13 = k a21 a22 a23 a31 a32 a33 44. We expand the left-hand determinant along its third row: a21 a22 a23 a11 a12 a13 a31 a32 a33 = a31 ( a22 a13 − a23a12 ) − a32 ( a21a13 − a23 a11 ) + a33 ( a21a13 − a22 a11 ) = −  a31 ( a23a12 − a22 a13 ) − a32 ( a23a11 − a21a13 ) + a33 ( a22 a11 − a21a13 )   a11 a12 a13 = k a21 a22 a23 a31 a32 a33 45. We expand the left-hand determinant along its third column: a1 b1 c1 + d1 a2 b2 c2 + d 2 a3 b3 c3 + d3 = ( c1 + d1 )( a2b3 − a3b2 ) − ( c2 + d 2 )( a1b3 − a3b1 ) + (c3 + d3 )( a1b2 − a2b1 ) = c1 ( a2b3 − a3b2 ) − c2 ( a1b3 − a3b1 ) + c3 ( a1b2 − a2b1 ) + d1 ( a2b3 − a3b2 ) − d 2 ( a1b3 − a3b1 ) + d3 ( a1b2 − a2b1 ) a1 b1 c1 a1 b1 d1 = a2 b2 c2 + a2 b2 d2 a3 b3 c3 a3 b3 d3
  • 46. We expand the left-hand determinant along its first column: a1 + kb1 b1 c1 a2 + kb2 b2 c2 a3 + kb3 b3 c3 = ( a1 + kb1 )(b2c3 − b3c2 ) − ( a2 + kb2 )(b1c3 − c3b3 ) + ( a3 + kb3 )(b1c2 − b2 c1 ) =  a1 (b2 c3 − b3c2 ) − a2 (b1c3 − c3b3 ) + a3 (b1c2 − b2c1 )   + k b1 (b2 c3 − b3c2 ) − b2 (b1c3 − c3b3 ) + b3 (b1c2 − b2 c1 )   a1 b1 c1 b1 b1 d1 a1 b1 c1 = a2 b2 c2 + k b2 b2 d 2 = a2 b2 c2 a3 b3 c3 b3 b3 d3 a3 b3 c3 47. We illustrate these properties with 2 × 2 matrices A =  aij  and B = bij  .     T a a21  a a12  (A ) T T =  11 =  11 = A a22  a22  (a)  a12   a22  T  ca ca12   ca ca21   a11 a21  (cA ) =  11 =  11  = c a  = cA T T (b)  ca21 ca22  ca12 ca22   12 a22  a + b a12 + b12   a11 + b11 a21 + b21  T (A + B) =  11 11  = a + b T (c)   a21 + b21 a22 + b22   12 12 a22 + b22  a a21  b11 b21  =  11  + b b  = A + B T T  a12 a22   12 22  48. The ijth element of ( AB)T is the jith element of AB, and hence is the product of the jth row of A and the ith column of B. The ijth element of BT AT is the product of the ith row of BT and the jth column of AT . Because transposition of a matrix changes the ith row to the ith column and vice versa, it follows that the ijth element of BT AT is the product of the jth row of A and the ith column of B. Thus the matrices ( AB)T and BT AT have the same ijth elements, and hence are equal matrices. a1 b1 c1 a1 a2 a3 49. If we write A = a2 b2 c2 and A = b1 T b2 b3 , then expansion of A along its a3 b3 c3 c1 c2 c3 first row and of AT along its first column both give the result a1 (b2 c3 − b3c2 ) + b1 ( a2 c3 − a3c2 ) + c1 ( a2b3 − a3b2 ).
  • If A 2 = A then A = A , so A − A = A ( A − 1) = 0, and hence it follows 2 2 50. immediately that either A = 0 or A = 1. If A n = 0 then A = 0, so it follows immediately that A = 0. n 51. −1 If AT = A −1 then A = AT = A −1 = A . Hence A 2 52. = 1, so it follows that A = ±1. −1 53. If A = P −1BP then A = P −1BP = P −1 B P = P B P = B. 54. If A and B are invertible, then A ≠ 0 and B ≠ 0. Hence AB = A B ≠ 0, so it follows that AB is invertible. Conversely, AB invertible implies that AB = A B ≠ 0, so it follows that both A ≠ 0 and B ≠ 0, and therefore that both A and B are invertible. 55. If either AB = I or BA = I is given, then it follows from Problem 54 that A and B are both invertible because their product (one way or the other) is invertible. Hence A–1 exists. So if (for instance) it is AB = I that is given, then multiplication by A–1 on the right yields B = A −1. 56. The matrix A–1 in part (a) and the solution vector x in part (b) have only integer entries because the only division involved in their calculation — using the adjoint formula for the inverse matrix or Cramer's rule for the solution vector — is by the determinant A = 1. a d f bc −cd de − bf  If A =  0 b e  then A −1 = 1  57.   0 ac −ae  . abc   0 0  b 0  0 ab   58. The coefficient determinant of the linear system c cos B + b cos C = a c cos A + a cos C = b b cos A + a cos B = c in the unknowns {cos A, cos B, cos C} is
  • 0 c b c 0 a = 2abc. b a 0 Hence Cramer's rule gives a c b 1 ab 2 − a 3 + ac 2 b2 − a 2 + c2 cos A = b 0 a = = , 2abc 2abc 2bc c a 0 whence a 2 = b 2 + c 2 − 2bc cos A. 59. These are almost immediate computations. 60. (a) In the 4 × 4 case, expansion along the first row gives 2 1 0 0 2 1 0 1 1 0 2 1 0 1 2 1 0 2 1 = 21 2 1 −0 2 1 = 21 2 1− , 0 1 2 1 1 2 0 1 2 0 1 2 0 1 2 0 0 1 2 so B4 = 2 B3 − B2 = 2(4) − (3) = 5. The general recursion formula Bn = 2 Bn −1 − Bn − 2 results in the same way upon expansion along the first row. (b) If we assume inductively that Bn −1 = (n − 1) + 1 = n and Bn − 2 = (n − 2) + 1 = n − 1, then the recursion formula of part (a) yields Bn = 2 Bn −1 − Bn − 2 = 2(n ) − (n − 1) = n + 1. 61. Subtraction of the first row from both the second and the third row gives 1 a a2 1 a a2 1 b b 2 = 0 b − a b 2 − a 2 = (b − a)(c 2 − a 2 ) − (c − a )(b 2 − a 2 ) 1 c c2 0 c − a c2 − a2 = (b − a)(c − a)(c + a) − (c − a)(b − a)(b + a) = (b − a)(c − a) [(c + a) − (b + a)] = (b − a)(c − a)(c − b).
  • 62. Expansion of the 4 × 4 determinant defining P ( y ) along its 4th row yields 1 x1 x12 P( y ) = y 1 x2 3 x2 +  = y 3V ( x1 , x2 , x3 ) + lower-degree terms in y. 2 2 1 x3 x3 Because it is clear from the determinant definition of P ( y ) that P( x1 ) = P( x2 ) = P( x3 ) = 0, the three roots of the cubic polynomial P( y ) are x1 , x2 , x3 . The factor theorem therefore says that P( y ) = k ( y − x1 )( y − x2 )( y − x3 ) for some constant k, and the calculation above implies that k = V ( x1 , x2 , x3 ) = ( x3 − x1 )( x3 − x2 )( x2 − x1 ). Finally we see that V ( x1 , x2 , x3 , x4 ) = P( x4 ) = V ( x1 , x2 , x3 ) ⋅ ( x4 − x1 )( x4 − x2 )( x4 − x1 ) = ( x4 − x1 )( x4 − x2 )( x4 − x1 )( x3 − x1 )( x3 − x2 )( x2 − x1 ), which is the desired formula for V ( x1 , x2 , x3 , x4 ) . 63. The same argument as in Problem 62 yields P( y ) = V ( x1 , x2 ,  , xn −1 ) ⋅ ( y − x1 )( y − x2 ) ⋅  ⋅ ( y − xn −1 ). Therefore V ( x1 , x2 ,  , xn ) = ( xn − x1 )( xn − x2 ) ⋅ ⋅ ( xn − xn −1 )V ( x1 , x2 ,  , xn−1 ) n −1 = ( xn − x1 )( xn − x2 ) ⋅ ⋅ ( xn − xn −1 ) ∏ ( xi − x j ) i> j n = ∏ ( x − x ). i> j i j 64. (a) V(1, 2, 3, 4) = (4 – 1)(4 – 2)(4 – 3)(3 – 1)(3 – 2)(2 – 1) = 12 (b) V(–1, 2,–2, 3) = 3 − ( −1) [3 − 2] 3 − ( −2 ) ( −2 ) − ( −1) ( −2 ) − 2   2 − ( −1) = 240       