07年线性代数å&c

07年线性代数å&c

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Unformatted text preview: 07 年年年年年年年 1 1 1 1 1 1 1 − 1 − 1 , 求 A−1 p329(3) A = 1 − 1 1 − 1 1 − 1 − 1 1 求求 1.(求求 1* 求 )A = A | A| −1 − 4 − 4 − 4 − 4 − 4 − 4 4 4 ∴ * | A |= − 16, A = − 4 4 − 4 4 − 4 4 4 − 4 1/ 4 1/ 4 −1 A= 1/ 4 1/ 4 1/ 4 1/ 4 1/ 4 1 / 4 − 1 / 4 − 1 / 4 − 1 / 4 1 / 4 − 1 / 4 − 1/ 4 − 1/ 4 1/ 4 2 年年年年年年年 1 1 A= 1 1 1 1 −1 −1 1 −1 1 −1 1 − 1 → ...... → − 1 1 1 1 1 1 0 − 2 0 − 2 = B, 0 0 − 2 − 2 0 0 0 2 1 1/ 2 1/ 2 1/ 2 0 − 1 / 2 0 − 1 / 2 −1 −1 ∴A =B = 0 0 − 1 / 2 − 1 / 2 0 1/ 2 0 0 × 年年年年年年年年“年年年年年年年年年年年年年年年年年”年 3 ? 2. 1 1 1 1 1 0 → 0 0 1 1 1 −1 −1 −1 1 1 −1 −1 −1 1 111 1 1 0 1 1/ 2 0 1 1 1/ 2 1 1 0 1/ 2 1 0 → ...... → 0 0 1 0 0 0 1 0 1 0 0 0 → 0 0 1 0 0 0 0 0 1 0 0 0 0 0 − 1/ 2 0 → − 1/ 2 0 0 0 0 − 1 / 2 1 1 0 − 2 − 2 −1 −2 0 −2 −2 1 0 0 0 1 1 − 2 −1 0 −1 111 1 1 0 1 1/ 2 0 0 0 1 1 1/ 2 − 1/ 2 0 0 1 1/ 4 − 1/ 4 1/ 4 1 0 0 1 / 4 1 / 4 − 1 / 4 − 1 / 4 0 1 0 1 / 4 − 1 / 4 1 / 4 − 1 / 4 4 0 0 1 1/ 4 − 1/ 4 − 1/ 4 1/ 4 0 0 0 1/ 4 1/ 4 1/ 4 0 0 0 1 0 0 0 1 0 0 0 1 0 0 − 1/ 2 0 0 0 − 1 / 4 1 / 4 Problem: What is the inverse matrix of block matrix A B H = ? C D 求求求 求 K = ( A − BD −1C ) −1 , N = ( D − CA−1 B ) −1 , K −1 求 H = −1 − D CK − A−1 BN . N 5 年年年年年年年年年年年年 −1 A C A−1 − A−1CB −1 (1) ; = −1 B O B O −1 A O A−1 ( 2) = −1 −1 C B − B CA −1 O A − B −1CA−1 (3) = B C A−1 −1 O ; −1 B B ; O −1 O B C A (4) . = −1 −1 −1 − A CB B O A −1 6 ? ? : AX = B ? ? ? [ A | B] → ...... → [ E | A B] = [ E | X ] −1 1 0 − 2 5 − 1 p340,4(4) A = − 3 4 − 1 , B = − 2 3 , ? ? : AX = B. 2 1 3 1 4 1 0 − 2 5 − 1 1 0 − 2 5 − 1 − 3 4 − 1 − 2 3 → 0 4 − 7 13 0 A= 2 1 3 1 4 0 1 7 − 9 6 7 77 1 0 0 35 1 0 −2 5 − 1 4 → 0 1 7 − 9 6 → 0 1 0 5 49 24 49 0 0 1 − 35 35 0 0 1 − 35 求求求求 [ 求求求求 ] 5.(3) A −1 2 3 −1 = 3 2 − 3 2 − 3 5 6 7 6 13 35 6 5 24 35 p330 5(3),6(1), 1 − 1 3 1 , X = A −1 B = −1 6 1 − −3 6 8 1 3 1 − 6 5 − 6 求求求求求求求求求求求求 (A B) -1 (E A→ B) 3 − 2 1 1 2 1 ..... 2 0 − 0 → 0 1 1 − 0 1 2 −2 3 1 1 1 3 1 − 1 − X =A B = 1− 6 5 −3 − 6 1 01 3 1 0− − 1 6 1 −3 5 − 6 0 1 0 9 1 6 7 7 16 −1 2 6.(1) A = − 7 7 1 1 − 7 7 1 1 20 1 7 5 ∴ X = BA−1 = 7 7 7 7 − 8 15 20 1 7 7 7 7 求求求求求求求求求求求求 3 1 A B = − 2 3 −1 −1 0 1 0 4 2 − 1 4→ − 2 1 1 0 −1 0 − 4 3 1 −2 3 −1 A B → 2 − 1 4→ − 2 1 1 0 −1 0 − 4 E BA−1 0 1 1 3 11 10 0 − 1 6 − 2 9 A E E 求求求 XA = B. → ...... → −1 = B BA X 1 1 1 0 1 1 , B = 1 − 2 1 , 求求 p330,6(2) A = 0 1 − 1 0 0 1 1 1 0 1 ? .A = 0 0 1 − 2 0 1 1 1 1→ 1 − 1 1 1 0 0 1 0 0 0 1 → 1 − 2 3 0 1 − 2 XA = B. 1 0 0 0 1 0 0 0 1 1 − 3 3 0 1 − 2 11 求求 x1 + 2 x2 + 3 x3 = 5 4 x1 + 5 x2 + 6 x3 = 11 7 x + 8 x + 9 x = 17 2 3 1 1 4 求 .A = 7 1 0 0 1 → 0 0 2 3 5 → 5 6 11 8 9 17 5 1 2 3 0 − 3 − 6 − 9 → 0 − 6 − 12 − 18 − 1 − 1 2 3 0 0 x 1 ∴ 求求求 1 2 3 5 0 1 2 3 0 0 0 0 1 − 1 x = c − 2 + 3 .(c求求求求求求 2 x3 1 0 12 求求求求求求 1 2 ? .A = 3 1 ∴ 求求求 2 3 5 1 1 −1 0 −2 x1 + 2 x2 + x3 = 4 2 x + 3 x − x + x = 5 1 2 3 4 3 x1 + 5 x2 + x4 = 9 x1 + x2 − 2 x3 + x4 = 1 1 0 0 4 1 2 1 0 4 0 1 0 − 1 − 3 1 − 3 1 5 → ... → → 0 0 0 − 1 − 3 1 − 3 19 0 0 1 1 0 − 1 − 3 1 − 3 −5 3 0 0 2 −1 0 0 − 2 3 0 0 x1 5 − 2 − 2 x − 3 1 3 2 = c + c + .(c , c ? ? ? ? ? ) x3 1 1 2 0 0 1 2 13 0 1 0 x4 P395(13)n 年年年年年年年年年年年年年年 n 年年年 求 .求 A求 n求求求求求求 ,求 AT = − A 求求求求求求求求 A = AT = − A = (−1) n A A求求 ∴ A ≠ 0,∴ (−1) n = 1, n求求求 . 14 求求求求求求P 330, 25 求求求A求求B求 A + B 求 , 求, 求求求求求求:求求求−1 + B −1 A ( A− 1 + B − 1 ) − 1 求 . A−1 + B −1 = A−1 ( I + AB −1 ) −1 −1 −1 −1 −1 = A ( BB + AB ) = A ( B + A) B , −1 −1 −1 −1 −1 −1 −1 ∴ ( A + B ) = [ A ( A + B) B ] = B( A + B) A. 15 [ 年年年年 ]p330(21) ? .? | A |≠0? , A − 1 1 = A* , | A| AA* = A | E. | | A | ⋅ | A* |=| A |n ,∴| A* |=| A |n −1 求求求求求求求 求 | A |= 0求求 求1 求求 A = O, 求 A* = O,∴ A* |=| A |n −1 ; | (2)求 A ≠ O, 求求求 求求 | A* |= 0.(求求求求求 | A* |≠ 0, 求 A*求求求 求 AA* =| A | I = O, 求求求 求 A ≠ O求求 ( A* ) −1 求求求 A* ) −1 , 求 A = O, . 16 [ 年年年年 ]p332,C(4) (A (4) =A ) ** n −1 = (A−1 A )* = A−1 A (A−1 A ) −1 A n −2 =A A A P333(11) A = 10, 1 0 0 A1 1 * −1 −1 −1 A = AA = = A = 2 2 0 A 10 10 3 4 5 () ( ) 17 1 5 1 − 1 −1 −1 , C = 17 17 p395(14) B , C ⇒ B = −1 2 − 2 3 17 17 B −1 ( AT − E ) − (C − B −1 ) = 0 ⇒ B −1 ( AT − E ) = (C − B −1 ) ⇒ BB −1 ( AT − E ) = B(C − B −1 ) ⇒ ( AT − E ) = BC − E 7 ⇒ AT = BC = 17 − 9 17 5 9 −1 ⇒ A = 2 7 2 7 − 17 ⇒ A = 17 5 − 2 17 17 9 − 17 5 17 3 − 1 2 1 5 2 ? 2.( A ) = C B = 1 1 = 9 7 2 5 18 −1 T −1 −1 年年年年年年 0 求 1.[ AI ] = 0 4 4 0 0 0 → 0 2 0 1 0 0 − 3 0 2 0 0 0 0 1 0 2 0 A−1 , A = 0 0 − 3 4 0 0 0 1 0 0 4 0 0 0 0 1 − 3 0 1 0 → 0 0 − 3 0 1 0 0 0 0 1 0 2 0 1 0 0 1 1 0 0 0 0 1 / 4 0 → 0 1 0 1 / 2 0 0 0 0 0 1 0 − 1 / 3 0 0 1 / 4 0 ∴ A−1 = 1 / 2 0 0 0 − 1/ 3 0 19 求 2.(求求求求求 ) O 求 A= C B 2 , B = 0 O 0 , C = [ 4] − 3 0 1 / 4 0 C −1 O A = −1 = 1 / 2 0 0 B O 0 − 1/ 3 0 −1 0 1 / 4 0 0 − 6 0 求 3. | A |= − 24, A* = − 12 0 0 , A−1 = 1 / 2 0 0 0 8 0 0 − 1/ 3 0 20 0 0 年年年 求 A = ... 0 an a1 0 0 0 0 0 ... ... , ai ≠ 0(i = 1,2,..., n).求 A−1. 0 ... an −1 0 ... 0 0 a2 ... ... a1 O B , C = [ a ]. 求 .求 A = ,B = n C O an − 1 O ∴ A = −1 B −1 − 0 ... 0 an 1 −1 −1 0 C a1 ... 0 . = O −1 0 ... an −1 0 21 求求求 A *求求求求求求求 : (1) AA = A A =| A | I ; * * (2) | A |=| A | * n −1 ; (3)( A* )T = ( AT )* ; (4)( AB ) = B A ; * (5)(kA) = k * * * n −1 * (6)( A ) =| A | ** A; n −2 * −1 A; (7)( A ) 1 = A. | A| 22 23 1 (7) A* = A | A − , 求 A*求求 | A求求 1 ( A* )* = A* | ( A* ) − ,求求 | 2求求 6求 ⇒ 1 ** n− 1 2 (A ) = A | | ⋅ A = A |n − A. | | A| 0 0 2 A* −1 年年年 求 A−1 = 3 1 0, 求求 ). 2 5 2 0 * A −1 1 * −1 求 .( ) = 2( A ) = 2 A, 2 | A| 4 − 2 0 | A−1 |= 2, ( A−1 )* = 0 −10 6 , 1 0 0 1 ∴ A = −1 ( A−1 )* , |A | 24 求求 8 − 4 0 1 −1 * = 2 ⋅ 2 ⋅ 求 A ) = 0 − 20 12 . 2 2 0 0 ( A−1 )* = ( A* ) −1. 年年年年 A 年年年年年年年年年 ? . A* =| A | A−1 , −1 * −1 −1 −1 ∴( A ) =| A | ( A ) 1 = A = ( A* ) −1 | A| 25 A O 求 H = , 求求 O B A, B求 n求求求求求 H *. O O | A || B | I n C O * AC 求 H = , , 求 HH = O BD = | H | I 2 n = O | A || B | I n O D * ∴AC = A || B | I n , BD = A || B | I n . | | ? C = A−1 | A || B |=| B | A* , 求求求 | B | A* O * * D =| A | B ,∴ H = . * | A| B O 26 P392,12(4) 2 A= 0 0 2 2 − 1 , 0 3 1 (1) 年 A 年年年年年年年年年年年 2 年年 A* 年年年年年 年 3 年年 |A*|; (4) 年 A* 年年年年 年 5 年年 (A*)* (6)r(A*) 2−λ ? .(1) | A − λ I |= 1 0 0 2−λ 0 2 − 1 = 0, λ1, 2 = 2 , λ 3 = 3. 3−λ 27 0 1 0 0 λ1, 2 = 2 : A − 2 I = 0 0 2 0 1 2 → 0 0 1 , r = 2, n − r = 1 −1 3 − 2 0 0 0 1 求 x1 = c, 求求求求求 c 0 (c ≠ 0); 0 2− 3 1 2 λ 3 = 3 : A − 3I = 0 2 − 3 − 1 , r = 2, n − r = 1, 0 0 0 ( 2 − 3 ) x1 + x2 + 2 x3 = 0, 求 x3 = c, ( 2 − 3 ) x2 − x3 = 0. 28 ∴ x2 = − ( 2 + 3 )c, x1 = ( 2 + 3 )([2 − ( 2 + 3 )]c, 2( 2 + 3 ) − 2 6 − 5 ? ? ? ? : c − 2− 3 (c ≠ 0). 1 | A| * * * * ( 2) A ? λ = , | A |= 2 3 ,∴λ1, 2 = 6 , λ3 = 2; λ 1 1 * −1 * 2 2 A= A; (3) | A |=| A | = (2 3 ) = 12; (4)( A ) = (2) | A| (5)( A ) = A | A = 2 3 A; | ** (6) r ( A) =3 = n,∴ ( A ) =3. r * 29 23 年年年年年 A*, 年 A. 1 * n−1 ? 1. AA = | A | I .∴ A = | A | ⋅( A ) = | A | * −1 * 求 2.( A ) =| A | ** 年年年 n− 2 2 0 * A = 0 A,∴ A =| A | 9 3 0 2− n * −1 (A ) ; 2− n * n −1 ( A ) =| A | ** 6 , 求 A. 6 6 3 − 9 6 0 2 − 2 ? ? : A= 0 0 1 30 ** (A ) . 31 32 ∴求 λ ≠ 0, λ ≠ 1求求求求求求 λ2 + 3λ − 9 9 3(3 − λ2 ) x1 = , x2 = 2 , x3 = ; 2 2 λ λ λ 求 λ = 1求求求求求求求求 1 4 1 2 1 1 0 1 1 1 0 1 1 0 1 1 → 4 1 2 1 → 0 1 − 2 − 3 6 1 4 3 6 1 4 3 0 0 0 0 求求求求 −1 1 c 2 + − 3 . 1 0 33 ...
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This note was uploaded on 03/29/2012 for the course MATH 120 taught by Professor Folh during the Spring '08 term at Eastern Michigan University.

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