lenear-bengurion-industry

lenear-bengurion-industry - 0 αβ εφ χ δ ϕ γ ι...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 0 αβ εφ χ δ ϕ γ η ι κλ µν οπ ϖθ ϑρ σςτυ ωξψζ www.GooL.co.il © ¯ – ¯ 1 ¯ , , . . . ¯ ¯ ¯ , – ¯ ¯ ¯ ¯ . . ¯, www.GooL.co.il ¯ ¯, ¯ , , ¯ ¯ . . ¯ www.GooL.co.il/linearit.html : ¯ ¯ , . www.GooL.co.il © ¯ – ¯ 2 ¯ 3 ........................ ¯ 1 10 ........................................................................ 2 16 ................................................................... 3 22 ............................................................. 4 30 ............................... ¯, , 31 ..................................... 34 ( ) ............................................. www.GooL.co.il © 5 6 7 ¯ – ¯ ¯ 3 1 – : ¯ ¯ (1) x + y = 3 (4 2 x + y = 3 (3 x − 4 y = −7 (2 x + 10 y = 11 (1 2x + y = 4 x− y =0 x − y = −1 : 2x − 2 y = 0 ¯ (2) x = 3 (4 2 x + y + z = 3 (3 x − 4 y + z = −7 (2 x + 10 y = 11 (1 2x + y = 4 x−z =0 x − y = −1 x+ y+ z =5 z+t =8 2x − 2 = 0 x+ y =3 ¯ .( (3) ) 3 −4 8 1 (3 2 −3 6 0 −1 4 −5 1 4 1 0 2 (2 −1 2 1 0 0 3 1 −1 3 5 −1 0 (1 2 1 4 2 5 0 −2 6 R1 → R1 +3 R3 , R2 → R2 +3 R3 R2 →4 R2 , R2 → R2 + R1 R1 ↔ R2 , R1 →2 R1 R1 →5 R1 −8 R2 R2 ↔ R3 , R3 → R3 −3R2 R3 → R3 + R1 , R1 ↔ R3 ¯ (4) : 1 −2 4 6 −3 9 (1 → 4 1 1 4 1 1 1 (2 1 0 −4 1 1 0 −4 4 2 → 0 2 17 −3 1 1 0 1 0 4 0 1 0 4 1 0 1 0 (3 0 1 → 4 2 4 4 4 4 www.GooL.co.il © ¯ – ¯ 4 . , . :( 1,3,5,7 1 2 −1 3 (3 1 3 1 5 3 8 4 17 3 6 3 −6 5 ( 2 2 4 1 −2 3 1 2 −1 2 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 1 0 1 0 (6 0 0 1 * 1 1+ i ( 9 1+ i 2i 2 + i 1 + 3i F= , F= ) . 1 2 −3 −2 4 1 (1 2 5 −8 −1 6 4 1 4 −7 5 2 8 1 3 5 (5 1 2 1 6 2 5 3 1 −1 −2 2 1 −2 3 5 −4 −1 11 4 0 11 −5 2 −5 3 1 3 −1 . ,9 .( , 5 (4 3 1 2 1 3 5 (7 1 2 1 6 2 5 3 −1 1 2 −2 −1 3 5 −4 −1 −2 3 −2 −5 1 −1 3 −2 2 3 (8 4 −3 4 2 3 −1 −2 9 3 0 2 1 5 3 2 1 5 −6 6 3 1 1 2 1 2 1 (5) ¯) * ¯ (6) 8 x − 4 y = 10 −6 x + 3 y = 1 (3 4 x + 8 y = 20 3 x + 6 y = 14 (2 2x + 3y = 8 5 x − 4 y = −3 (1 x + 2 y + 3z = 3 4 x + 6 y + 16 z = 8 3 x + 2 y + 17 z = 1 (6 x + 2 y + 3 z = −11 2 x + 3 y − z = −5 3x + y − z = 2 (5 2 x1 − x2 − 3 x3 = 5 3 x1 − 2 x2 + 2 x3 = 5 10 x1 − 6 x2 − 2 x3 = 32 (4 3x − 2 y = 1 −9 x + 6 y = −3 (9 4x − 7 y = 0 8 x − 14 y = 2 (8 x + 3y = 2 2 x + y = −1 (7 6x − 4 y = 2 x + 2 y + 2 z = 2 (12 3x − 2 y − z = 5 2 x − 5 y + 3 z = −4 2 x + 8 y + 12 z = 0 −16 x + 28 y = 4 x − y = −2 x1 + 5 x2 + 4 x3 − 13 x4 = 3 (11 3x1 − x2 + 2 x3 + 5 x4 = 2 2 x1 + 2 x2 + 3 x3 − 4 x4 = 0 www.GooL.co.il © 2 x + 5 y − 8 z + 6t = 5 6 x + 8 y − 10 z + 4t = 8 ¯ – x + 2 y − 3 z + 2t = 2 (10 ¯ 5 : ¯ ( . )k ¯ ¯ . . . (7) . . x + 2ky + z = 0 (3 x + ky + z = 1 (2 3 x + y + kz = 2 x + y + kz = 1 5 x − 7 y + (k + 3) z = k + 1 x + 9ky + 5 z = −2 kx + y + z = 1 3 x − y + (k + 3) z = 3 x + ky + 3 z = 2 (6 kx − y + z = 4 3 x + y + (2 + k ) z = 0 x − y + z =1 2 kx − y = 1 (5 2 x − y + z = 0 (4 (k − 2) x + ky = −2 (k 2 − 1) z = 9 : x + 2y − z = 0 5 x + (1 − k ) y + k 2 z = 1 ¯ ( . 3 x + 4 y − z = 2 (3 kx − 2 y + z = −1 x + 8 y − 3z = k 2 x + 6 y − 2 z = 0.5k + 1 )k ¯ . . ¯ . . 2 x − 3 y + z = 1 (2 4 x + ( k − 5k ) y + 2 z = k ¯ x + y − z + t = 1 (3 ax + y + z + t = b 3 x + 2 y + at = 1 + a ( ¯ (8) . 2 x + ky = 3 (1 ( k + 3) x + 2 y = k 2 + 5 6 x + 3ky = 7k 2 + 2 2 : (1 2 )b .. a . 2 x + 4 y + az = −1 (2 ¯ . (9) . x + 2 y − 4 z = b (1 x + 2 y + 4 z = −4 x + 2 y − 4z = 0 x + 2 y + 6 z = −2b 7 x − 10 y + 16 z = 7 2 x − ay + 3 z = 1 : ¯ (10) x + az = 1 y + 2z = 2 bx + cy + dz = 3 . ¯ . ¯ ¯ a , b, c, d ¯ b, c, d a¯ .F . . ¯ (11) z1 + iz2 + (1 − i ) z3 = 1 + 4i (2 x1 + 2 x2 + 3x3 = 1 (1 iz1 + z2 + (1 + i) z3 = 2 + i 2 x1 + 4 x2 + 4 x3 = 2 (−1 + 3i ) z1 + (3 − i ) z2 + (2 + 4i) z3 = 5 − i 3x1 + x3 = 0 F= ,F= www.GooL.co.il ¯ © F= – ¯ 5 6 x + y − z = 1 2 2 3 x − 7 y + ( k + 1) z = k − 1 : ¯ 4 x − 6 y + (k + 2) z = 4 . (12) ¯ . . . .3 . . .2 . .1 : ¯ k¯ . ¯ . z=0 . ¯ k¯ ¯ . z=0 ¯ . k¯ . (1, 2, 3) . . . k . ?¯ ¯ .¯ " (1, 0, 0) , k . (13) . . ( ¯ ¯ . . ¯ . x1 ) (14) ¯ ( www.GooL.co.il © ) . . x4 . x4 = 0 :( ¯ ¯ – ¯ )3 * 7 1 – . 1 0 0 0 3 (4 2 1 0 0 4 0 0 1 1 8 (4 (2 (2 2 1 1 3 (3 1 −4 1 −7 1 −1 0 −1 1 0 −1 0 1 1 1 5 0 0 −4 4 (3 4 1 0 2 (2 0 5 −4 2 0 3 1 −1 −1 4 −5 1 0 0 1 5 (3 (1 (1) 1 10 11 (1 (2) 2 −2 0 1 1 3 9 2 6 8 (1 (3) 3 5 −1 0 4 2 8 2 R2 → 2 R2 + 4 R1 (2 R2 → R2 − 4 R1 (2 R1 → 2 R1 + R2 (1 (4) . (5) 1 2 −3 −2 4 1 (1 3 −2 2 0 1 −2 0 0 0 1 2 3 1 0 24 21 1 0 0 1 −2 0 −8 −7 0 0 0 1 2 3 1 2 −1 2 1 ( 2 0 0 3 −6 1 0 0 0 0 0 1 0 0 137 2 0 1 0 − 3 4 0 0 1 3 1 2 −1 3 (3 0 1 2 2 0 0 3 4 1 3 −1 0 11 −5 0 0 0 0 0 0 1 0 0 0 www.GooL.co.il © 0 −1 0 0 1 1 0 1 0 0 1 1 0 0 0 0 ¯ – ¯ 1 0 0 0 2 1 0 0 2 (4 3 0 0 1 3 5 (5 1 −5 −4 0 1 1 0 0 0 8 1 (6 0 0 0 0 0 0 0 0 −1 11 00 00 00 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 2 1 3 5 (7 1 1 −5 −4 00 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 3 −2 2 3 (8 1 −1 2 −1 0 −2 0 2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 1 1 1 0 1 1 + i 1 1 + i (9 2i , 0 0 1 + i 0 0 0 0 F= F= (6) ( x, y ) = (5 − 2t , t ) (2 ( x, y ) = (1, 2) (1 φ (4 φ (3 ( x, y, z ) = ( −1 − 7t , 2 + 2t , t ) (6 ( x1 , x2 , x3 ) = (1, −3, −2 ) (5 ( x, y ) = (−1,1) (7 φ (8 ( x, y, z ) = ( 2,1, −1) (12 1 + 2t , t ) (9 3 φ (11 k = −2 . (7) k ≠ 1, k ≠ −2 . (1 ( x, y, z, t ) = ( −a + 2b,1 + 2a − 2b, a, b ) (10 k =1 . k = −2 . k ≠ 1, k ≠ −2 . (2 k = −1 . k = 1, k = −0.4 . k ≠ 1, k ≠ −0.4 . (4 ( x, y ) = ( k =1 . k= 4 7 . k = ±1, k = −2 . . k = −1, k = −3, k = 2 . www.GooL.co.il © ¯ – ¯ . k ≠ −1, k ≠ 4 7 . (3 k ≠ ±1, k ≠ −2 . (5 k ≠ −1, k ≠ −3, k ≠ 2 . (6 9 k =1 . k ≠ 1 . (3 k ≠3 . k = 3 . (2 k =1 . k ≠ ±1 . k = −1 . (1 (8) (9) . a = 2, b = −3 . a = 2, b ≠ −3 . . a = −6, b = 2.5 . . a≠2 a ≠ −6 a = 2, b = 2 . b = 0, c = 1.5, d = 3 . a ≠ 2 . (1 b ≠ 2.5 . (2 a = 2, b ≠ 2 . (3 ab + 2c ≠ d . (10) (11) ( z1 , z2 , z3 ) = (2,3, −1) , ( z1 , z2 , z3 ) = (( −1 + i )t + 1 + i, 3, t ) (2 ( x1 , x2 , x3 ) = (0, 3, 0) (1 F= F= 1 −1 1 1 2 2 k + 4 k − 4 . 0 −10 0 0 −k 2 + k + 2 4 − k 2 −1 1 1 1 2 2 3 −7 k + 1 k − 1 . (12) 4 −6 k + 2 4 k = 2 .3 k = −1 .2 . k ≠ 2 , k ≠ −1 .1 . ( x, y, z ) = (1 + 0.2t , 0.8t , t ) . k = −2 . . , k=2 . k = −2 . . x4 = 60 − x5 , x2 = 100 − x3 + x5 , x1 = 100 + x3 − x5 . k = ±2 . x5 . x5 = 60 , x4 = 0 , x2 = 160 − x3 , x1 = 40 + x3 . x3 . (13) x3 . . 40 . I1 = − 5 7 6 . , I 2 = , I3 = 22 22 11 www.GooL.co.il © . I1 = 255 97 158 . (14) , I2 = , I3 = 317 317 317 ¯ – ¯ 10 2 – . A4×6 , B 4×6 , C 6×2 , D 4×2 , E 6×4 : (1) . . B + AB (5 E ( B − A) (10 AE − B (4 AC − D (3 AB (2 A + B (1 E B (8 ( E + A ) D (7 E ( B + A) (6 T E ( AC ) (9 T :¯ , x, y, z (2) x + 2 y 3x − 2 y 2 − 2 z 5 + z 2 x − 5 y 2 x + 8 y = −4 − 3z −12 z : 4 0 4 A = 1 2 , B = 0 −1 1 1 1 0 I2 = , I3 = 0 0 1 0 (3) 1 4 2 4 1 1 1 1 4 2 , C = , D = 1 0 −1 , E = −1 0 1 −2 4 1 5 4 2 10 4 1 −1 0 0 1 0 0 1 :( 2tr ( D 2 − 2 E ) (5 2 D + 4 EI 3 (4 DABC (10 ¯ b E − D + I 3 (2 5C (3 tr ( C T C ) (9 I 2 BC (8 ) E + D (1 1T 1 A + C (7 4C T + A (6 2 4 x,A ¯ (4) . Ax = b 2 x − 3 y + z + t = 1 (2 " 2 x + y − z = 3 (1 4x + y + 2z = 4 x + 2 y − 4z = 5 y + z +t =1 6x + 4 y + z = 2 x − 4 z − 2 y = 10 www.GooL.co.il © ¯ – ¯ 11 : 4 −2 4 A = 1 −1 1 1 −6 3 : AT x = 2 x + 3b (5 ¯ Ax = x (4 x x = y z ¯ ¯ AT = A A + AT .2 . ( A − B ) 2 .3 . :¯ B AB 2 .2 . AB 3 .1 B AB + B .2 . . A4 B 4 = B 4 A 4 ¯ ¯ . . A . . AB = BA ¯ . A BABABA .1 . AB = − BA ¯ :¯ AAT .1 . B A2 − B 2 .2 . A . A2 + B .3 . . (6) . A − AT .3 . :¯ Ax = b (1 A :¯ . 1 b = 2 3 Ax = − kx + b (3 Ax = 4 x + b (2 . AT = − A (5) . A . AB .1 . A, B, AB : ¯ . ¯ . ¯ (7) 4 1.5 (3 2 1 5 2 (2 7 3 1 2 (1 3 4 2 −1 1 (6 3 −2 2 5 −3 4 2 1 1 (5 0 2 −1 5 2 3 1 0 2 (4 4 −1 8 2 1 3 1 1 4 2 2 −1 4 0 1 (9 2 3 2 −1 2 −2 0 www.GooL.co.il © 1 1 0 1 4 (8 2 −1 0 1 1 1 3 − 1 −2 4 2 ¯ – ¯ 1 1 1 1 0 0 0 (7 2 0 0 2 3 0 2 3 4 12 1 1 −1 2 . 5 −7 k + 3 : ¯ 3 −1 k + 3 1 1 . 1 1 k 1 1 1 k 1 1 1 k 1 1 1 k 1 1 1 k . (8) k k 1 1 : ¯ 1 1 ¯ ¯ . :¯ ¯ x + 4 y + 2 z + 4t = 1 (2 (9) 2 x − y + z = 3 (1 x + 2y − z = 0 3x − 2 y + 2 z = 5 y + z +t =1 5 x − 3 y + 4 z = 11 x + 3 y − z − 2t = 0 :X ¯ n ¯ . (10) P −1 X T P = A (3 A−1 XC = A−1 DC (2 AXC = D (1 T −1 T T −1 −1 −1 ABC X BA C = AB (6 ( A − AX ) = X C (5 C ( A + X ) D −2 = I (4 −1 . B 2 X ( 2B ) = B + I ¯ −1 . BYB = B + B ¯ T 5 AT B ( I + 2 A ) −2 = ( 7 A) 1 2 .B = 4 9 . 1 0 2 . B = 4 −1 8 2 1 3 . 2 3 . A −1 = 4 7 . X −1 Y −2 B . A2 − 5 A − 2 I = 0 .I A: A−1 A ¯ . ( A − 3I )( A + 2 I ) = 0 .I www.GooL.co.il © ¯ – A: ¯ A: A−1 A ¯ ¯ . (11) A: ¯ . 13 −1 3 0 . A = 3 −1 0 , p( x) = x3 − 4 x 2 − 20 x + 48 : −2 −2 6 . .1 . p( A) A A−1 ¯ ¯ A ( )1 .2 . I . A4 = 0 A: .¯ . . D −1 AD = C ¯ ¯D ¯ . ¯¯ . P −1 AP = B . −1 : Q BQ = C ¯¯ .¯ A¯¯ I−A ¯ , ¯ A C (13) ¯ :¯ ¯) C (12) B * ** B A :¯ .( .¯ ( www.GooL.co.il © ¯ – ¯ )3 14 2 – (5 (4 4 × 2 (3 (2 4 × 6 (1 (1) 6 × 6 (10 6 × 4 (9 (8 6 × 2 (7 6 × 6 (6 ( x, y, z ) = ( 2,1, −1) 18 12 8 (4 −2 0 2 24 8 16 (2 5 20 10 (3 4 −3 −1 1 2 −2 20 5 25 0 −1 −10 (2) 5 5 3 (1 (3) 0 0 0 8 3 9 8 16 (6 17 6 7 21 230 (5 −32 82 −22 (10 48 87 75 −48 108 −36 63 (9 0 (7 13 (8 2.25 1.5 8 17 1 1.25 1.75 −8 −2 −10 (4) 2 1 −1 A = 1 2 −4 4 4 1 1 2 −3 4 12 A= 0 11 1 −2 −4 (4 + k ) x − 2 y + 4 z = 1 (3 x + (k − 1) y + z = 2 x − 6 y + (3 + k ) z = 3 3 b = 5 (1 2 x y x= z t 1 0 1 0 x x = y z 1 4 b = (2 1 10 −2 y + 4 z = 1 (2 x − 5y + z = 2 x − 6y − z = 3 4x − 2 y + 4z = 1 x− y+z =2 x − 6 y + 3z == 3 (1 (5) 2 x + y + z = 3 (5 3x − 2 y + 4 z = 0 (4 −2 x − 3 y − 6 z = 6 x − 2y + z = 0 4x + y + z = 9 x − 6 y + 2z = 0 1,2,3 . www.GooL.co.il © ¯ – ¯ 2. 1,2,3 . (6) 15 (7) 1 −1.5 (3 4 −2 3 −2 (2 −7 5 1 (1 2 1.5 −0.5 2 −1 0 (6 2 −3 1 −1 −1 1 8 −1 −3 (5 −5 1 2 −10 1 4 −11 2 2 (4 4 −1 0 6 −1 −1 7 −2 3 −1 (9 −10 3 −5 2 −10 3 −4 1.5 −1 2 −1 4 7 −10 −20 4 (8 6 −1 −2 3 3 −5 −8 2 3 −1 −1 2 1 1 − 2 0 0 k = 1, k = −4 (2 0 0 0 1 2 −1 3 0 1 3 −1 4 0 (7 0 0 1 4 . k ≠ 1, k ≠ −2 (1 (8) ( x, y , z , t ) = ( −13, 4, −5, 2) (2 . ( x, y , z ) = (1, 2,3) (1 (9) . CD 2 − A .4 . ( P −1 ) AT P T .3 T . D .2 BAT C ( B −1 )T BC T .6 . A−1 DC −1 .1 . (10) . ( A + C −1 ) A .5 −1 22 86 38 Y = 64 246 114 . 60 238 100 .B −1 5 −1 X = 4 . −2 1 . A−1 = 1 A − 1 I . 6 6 1 264 450 B= . 245 448 768 A−1 = 0.5 A − 2.5 I . (11) 0 0 0 121 5 = − B + B + I .2 , f ( B) = 0 0 0 .1 . 48 12 12 0 0 0 ( I − A) −1 = I + A + A2 + A3 . (12) www.GooL.co.il © ¯ – ¯ 16 3 :( / – ) (1) 4 −1.5 (3 2 −1 a b (1 c d 2 1 1 (6 3 −2 5 0 2 0 4 1 4 1 5 2 (2 −7 3 2 1 1 (5 0 2 −1 1 0 0 1 0 2 (4 4 1 8 2 0 3 00 72 00 4 −1 5 (9 4 0 1 1 −2 5 2 4 0 −7 3 −5 0 2 5 (8 0 −6 0 3 −7 4 0 5 44 1 1 1 1 0 2 2 2 1 1 0 0 0 0 (3 4 5 3 3 −1 0 −2 (6 5 −5 −1 −8 0 2 3 9 004 1 0 0 2 7 0 (7 0 0 4 0 7 5 0 (11 1 9 8 3 4 (10 3 0 −5 0 2 0 3 0 0 2 −4 2 1 5 9 1 0 3 0 4 2 −1 1 7 0 2 4 0 0 0 −8 −3 2 1 0 0 . 1 −1 −3 1 0 2 −1 2 8 3 −1 −2 0 0 3 3 (2) 1 3 3 12 0 2 5 4 −1 −2 −1 1 3 0 0 0 −4 (2 −5 −3 −1 2 −1 0 −2 (5 4 −5 −1 −8 0 2 3 9 0 −3 1 0 0 5 2 7 1 3 −2 −5 3 5 2 2 0 2 (1 7 4 2 1 2 −1 1 3 −1 0 −2 (4 1 5 −5 −1 −8 −2 −6 2 3 9 3 7 −3 8 −7 3 5 5 2 7 : (3) 2 5 4 6 12 10 6 −2 −4 −6 7 7 www.GooL.co.il © 1 (3 −1 2 3 3 4 3 3 5 4 6 0 0 3 4 7 0 (2 0 6 3 ¯ – ¯ 2 5 −3 −1 (1 3 0 1 −3 −6 0 −4 9 6 15 −7 −2 17 : , 12 15 18 (3 13 16 19 14 17 20 1 2 3 (2 4 5 6 5 7 9 (4) 1 0 2 (1 7 0 12 3 0 2 sin 2 x cos 2 x 1 (6 a a + x a + y (5 y + z 2 2 sin y cos y 1 b b+ x b+ y x c c+ x c+ y 1 sin 2 z cos 2 1 1 z+x y 1 y + x (4 z 1 5 0 1 −12 (7 3 −1 4 1 −4 18 4 −14 4 3 5 −2 0 −4 1 −3 11 06 −6 −4 2 −21 2 34 56 1 2 −5 7 −4 2.5 −1 −1.5 −11 2 −6 9 −1 3 4 ab : 0 g + 3d 0 h + 3e 0 i + 3 f 0 1 3a a + 3d (3 3b b + 3e 3c c + 3 f 0 0 2a − 3d 2b − 3e 2c − 3 f g + 4a (2 h + 4b i + 4c 2d 2e 2f .d g e h a b c c f =4 : i g+d h+e i+ f 2d (1 2e 2f 1 a a2 1 b b 2 = (b − a )(c − a )(c − b) ¯ ¯ 1c 1 1 1 1 x y z t x2 y2 z2 t2 c © ¯ – . (6) 2 x3 y3 = ( y − x)( z − x)(t − x)( z − y )(t − y )(t − z ) ¯ ¯ z3 t3 www.GooL.co.il (5) ¯ . 18 , .n ¯ (7) : 1 aij = 0 1 1 1 M 1 i + j = n + 1 (3 i = j +1 j aij = n 0 1 (6 3 36L 6 M MO M 3 6 L 3(n − 1) i = 1, j = n i> j i = j (4 a aij = b c : a = 3, b = 1, c = 2 ¯ . n = 20 i= j (* 7 i = j +1 j = i +1 .. . ¯ (1 i = j ≠1 i< j a aij = b 1 1 L 1 (5 2 2 L 2 2 3 L 3 M M O M 2 3 L n 1 1 1 M 1 11L 33L i = j =1 1 0 aij = j − j (2 :(7 * .2 . .1 : a b cd e a b c d e f g h i j f g h i j k l m n o+ k l m n o p q r s t p q r s t 2a + 1 −2b 1 x y (8) : −a − 1 3b c − 1 d − x e − y .| A |= 4, | B | = 2 , 3 A: B (9) | −2 A2 AT adjB | (4 | − A−2 BT A3 | (3 | 4 A2 B 3 | (2 | ABA−1 BT | (1 .| A |=| B | : ¯ . | A | = 2 , 2 AB + 3 I = 0 , 4 ¯ ( PQ )−1 APQ = B : B A: . (10) . .| B | . A + 3B = 0 , B 2 − 2 A−1 = 0 , 3 ¯ B A: . | A |, | B | : www.GooL.co.il © ¯ – ¯ . 19 . adj ( Anxn ) =| A |n −1 .2 .| A |= 0 ¯ | A−1 | = 1 .1 : ¯ | A| . A¯ . 2 AB = BT A2 , | A | = 128 , n .n 1 . det B − n A2 n : 3 ¯ 1 3 . : . ¯ x + 2 z + 5t = 8 (3 −2 x − 6 y = −8 5 x + 3 y − 7 z + 4t = 5 2 x + 5 y + 44 z = 51 . A: . det ( Anxn ) = 2, det ( Bnxn ) = : . x + z = 3 (2 4 x + y + 8 z = 21 2 x + 3z = 8 (11) x + 2 y = 5 (1 3 x + 4 y = 11 : ¯ (12) kx + y + z + t + r = 1 x + ky + z + t + r = 1 x + y + kz + t + r = 1 x + y + z + kt + r = 1 x + y + z + t + kr = 1 ? ?x = k . ¯ 1 2 ?x = k . 1 5 ¯ x= y= z =t =r . A −1 ¯ ¯ ¯ ¯ . ¯ adj ( A) 1 1 A= 0 1 . k 2 1 1 (2 A = 0 2 −1 5 2 3 4 2 4 (3 2 −1 0 1 1 1 3 −1 −2 www.GooL.co.il © ¯ – ¯ 1 2 (1 A= 3 4 (13) 20 : (14) −9 26 −1 14 10 13 −7 87 4 0 . A = 71 35 3 0 0 2 0 0 0 17 2 0 0 0 0 ( A )1,5 −1 A −1 ¯ , ( adjA)1,5 (2 (1 : ¯ | A |= 1 A ¯ . (15) . A −1 . ¯ .¯ A .¯ AT adj ( A) .¯ ? AB (5 CD (4 AD (3 .¯ C, D . ¯ A . A, B : . A + B (2 C + D (1 : ¯ :¯ 4 0 −7k 2 3 −5 ¯ . k¯ (16) 075 0 0 3k 0 0 2 4k k 9 + k 042 −1 0 −8 −3 2 : . ( −1, 0) , (0,5) , (1, −4) , (2,1) .2 . (17) (0, 0), (5, 2) , (6,5) , (11, 6) .1 . (0, 0, 0), (1, 0, −2), (1, 2, 4), (7,1, 0) : . . (3,3, −2), ( −1,3,1), (1,1, −1) : . . (1, 2), (3, 4), (5,8) : . . ¯: www.GooL.co.il © ¯ – ¯ 21 3 – .9 (10 . 300 (9 .234 (8 .24 (7 . 14 (6 . 3 (5 . 1 (4 . 1 (3 . 29 (2 . ad − bc (1 (1) .6 (3 .114 (2 .120 (1 (3) . (−1) .104 (6 .44 (5 .24 (4 .3 (3 .0 (2 .0 (1 (2) n (3n +1) 2 (3 . (−1) n −1 n ! (2 . n! (1 (7) .6 (11 .9 (3 .16 (2 . 8 (1 (5) . 2 ⋅ 3n − 2 (6 .1 (5 . (a − b)n −1[a + (n − 1)b] (4 . Dn = aDn −1 − bcDn − 2 , D2 = a 2 − bc , D3 = a 3 − 2abc . . 211 (4 . 8 (3 . 213 (2 .4 (1 (9) . x = 1, y = 2 (1 (11) . k = −2 . .0 (8) . D20 = 221 − 1 .2 . Dn = 2n +1 − 1 .1. .7 . . | A |= 18, | B |= −2 / 3 . .81/32 . (10) . 4n . . k ≠ 1, k ≠ −4 . (7 (12) . x = y = z = t = 1 (3 . x = 1, y = 1, z = 2 (2 . 8 −1 −3 (2 adj ( A) = A−1 = −5 1 2 −10 1 4 . 4 −2 (1 (13) adj ( A) = −3 1 1 −2 A −1 = 1.5 −0.5 7 −10 −20 4 −7 10 20 −4 (3 −2 3 6 −1 2 −3 −6 1 A−1 = , adj ( A) = 3 −5 −8 2 −3 5 8 −2 3 −1 −1 2 1 −2 −3 1 . k = 0 (16) . ¯ (5 . (4 . (3 . (2 . (1 (15) . 3 x − y + 4 z + 2 = 0 . 22 . .2 . www.GooL.co.il © ¯ – ¯ .0.5 (2 .240 (1 (14) .14 .2. .13 .1. (17) 22 4 – : .R .R ¯ ( f :R → R ) : R3 Pn [ R ] ¯ n .R M n [ R] ¯ n .R Rn ¯ n F [ R] ¯ (1) W . W = {( a, b, c) | a + b + c = 0} . . W = {( a, b, c) | a = c} . . W = {( a, b, c) | a = 3b} . . W = {( a, b, c) | a < b < c} . . W = {(a, b, c) | a = c 2 } . . ¯ , W = {(a, b, c) | b = a + d , c = a + 2d } . b ,a c . c ¯ W = {(a, b, c) | b = a ⋅ q, c = a ⋅ q 2 } . b ,a : M n [ R] ¯ (2) W . W = { A | A = AT } , ¯. .B ¯ ¯ ¯ W. ¯ W. . W = { A | AB = BA} , . W = { A | | A |= 0} , ¯. ¯ ¯. W. ¯ ¯ W. . , ¯ ¯. . W = { A | A2 = A} , ¯ ¯ ¯ W. ¯ ¯ W. ¯ B . W = { A | AB = 0} www.GooL.co.il © ¯ – ¯ 23 . W = { A | tr ( A) = 0} , ¯. ¯ . ¯ . Pn [ R ] W. ¯ ¯ ¯ ¯ W. ¯ (3) W . W = { p ( x) | p (4) = 0} , ¯. ¯4 ¯ .4≤ n≤7 ¯n ¯ W. ¯ W. ¯ .x W. ¯ ¯ .4 ≥ ¯ ¯ . W = { p ( x ) | deg( p ) ≤ 4} , W. ¯ . ¯ ¯ W. W = { p ( x) | p (0) = 1} . . F [ R] ¯ (4) W . W = { f ( x) | f (− x) = f ( x )} x¯, ¯ W. ¯ ¯ W. ¯ ¯ W. . x¯, ¯ . . W = { f ( x) | f ( x ) ≤ M } ¯. ¯ ¯ W. ¯ ¯ W. ¯. . .( [0,1] 1 ) W = f ( x) | ∫ f ( x) dx = 4 . 0 f .( x ¯ f ) W = { f ( x) | f '( x) = 0} . .( x ¯ f ) W = { f ( x) | f '( x) = 1} . . W = { f ( x) | f ( x) = f ( x + 1)} . W = {( z1 , z2 , z3 ) | z2 = z1 , z3 = z1 + z1} : C3 .R .C www.GooL.co.il © ¯ – ¯ (5) . ¯ . 24 , , : (6) . u1 = (4,1,1,5) , u2 = (0,11, −5,3) , u3 = (2, −5,3,1) , u4 = (1,3, −1,2) ? u4 u1 u1 ? Sp{u4 } ? u2 .3 u3 u1 ? Sp{u1 , u2 } ¯ .2 {u1, u4} ? ¯ .1 . .1 . u3 .2 {u1, u2 , u3} ? .3 . ? u2 ¯ u4 u1 ? Sp{u1 , u2 } ¯ ¯ .1 . u4 .2 {u1, u2 , u4} ? . .3 ¯ . v = ( 4,12, k , −2k ) ? u2 . u1 v k ¯ .1 . Sp{u1 , u2} v k ¯ .2 {u1 , u2 , v} k ¯ .3 . v = ( a , b, c , d ) ? u2 . v a , b, c , d .1 v a , b, c , d .1 {u1 , u2 , v} a , b, c , d .1 u1 ? Sp{u1 , u2} ? . u3 ¯ (2, −3,3,1) u2 , u1 ? www.GooL.co.il © ¯ ¯ – . ¯ 25 ¯ . u4 ¯ (7,10, −2,11) u3 , u2 , u1 . ? : (7) 4 1 0 11 2 −5 1 3 .A= , B = −5 3 , C = 3 1 , D = −1 2 1 5 . M 2 [ R] . .1 ¯ ¯ .2 ? Sp { B, C} ¯ .3 A : (8) p1 ( x) = 4 + x + x 2 + 5 x3 , p2 ( x) = 11x − 5 x 2 + 3x 3 , . p3 ( x) = 2 − 5 x + 3x 2 + x3 , P4 ( x) = 1 + 3x − x 2 + 2 x3 .1 . P3[ R ] ¯ ¯ .2 . ? Sp { p1 , p4 } : .3 p2 a , b, c ¯ (9) . {(c,2,4),(2.4, a,2),(c, b,6),(b, 2, a)} {u, v, w} .V [F ] ¯ ¯ ¯ ¯ (10) , : {u − v, u − w, u + v − 2w} . {u + 2v + 3w,4u + 5v + 6w,7u + 8v + 9w} . {u + v, v + w, w} {(1, i, i − 1),(i + 1, i − 1, −2)} C3 (11) .R www.GooL.co.il © ¯ – ¯ . . .C . 26 : R3 (12) { (1,0,1), (0,0,1) } (1 { (1,1,2), (1,2,3), (3,3,4), (2,2,1) } (2 { (1,2,3), (4,5,6), (7,8,9) } (3 : M 2 x 2 [ R] (13) 1 2 5 6 9 1 , , (1 3 4 7 8 2 3 1 2 5 6 9 1 5 6 5 16 , , , , (2 3 4 7 8 2 3 7 2 7 8 1 1 0 1 0 0 0 0 , , , (3 1 1 1 1 1 1 0 1 : P2 ( R) (14) { 1 + x, x 2 + 2 x + 3 } (1 { 1 + x, x 2 + 2 x + 3,2 x + 4 x3 , x − x 3 } (2 { 1 + 2 x + 3x 3 , 4 + 5 x + 6 x 2 ,7 + 8 x + 10 x 2 } (3 . T = {(1, 2,3), (4,5,6), (7,8,9),(2,3,4)} : R 3 (15) . R3 .T T , T' T' www.GooL.co.il © ¯ – ¯ . . . 27 ¯ : ¯ x − y + z + w = 0 x − y + z + w = 0 (3 x + 2 z − w = 0 (2 2 x − 2 y + 2 z + 2w = 0 x + y + 3z − 3w = 0 3 (16) x + y − z + 2w = 0 (1 3x − y + 7 z + 4w = 0 −5 x + 3 y − 15 z − 6w = 0 .(1 ¯ " W .(2 ¯ " U . (3 ¯ " V .V .U ∩ V U,W (2 . U ∪ V ( (1 ( .U ∩ V ( . U = {( a, b, c, d ) ∈ R 4 | a = c, b = d } (17) . U = {( a, b, c, d ) ∈ R 4 | c = a + b, d = b + c} (18) . U = {v ∈ R 4 | v ⋅ (1, −1,1, −1) = 0} (19) . U = { A ∈ M 2 x 2 [ R ] | A = AT } (20) 1 −1 0 0 . U = A ∈ M 2 x 2[ R] | A ⋅ = 1 1 0 0 (21) .U .U .U .U .U { . U = p( x) ∈ P3[ R ] | p(1) = 0 .U www.GooL.co.il © ¯ – ¯ } (22) 28 : R4 ¯ (23) U = span {(1,1, −1,2), (3, −1,7,4), (−5,3, −15, −6)} V = span {(1, −1,1,1), (1,0,2, −1), (1,1,3, −3), (5,1,5,8)} .U , . .V , . .U ∪ V . .U ∩ V . : M 2 x 2 [ R] ¯ (24) 1 1 4 1 2 −1 . U = span , , −1 2 −1 1 1 −3 .U : P3[ R ] ¯ (25) . U = span {1 + x − x 2 + 2 x 3 , 4 + x − x 2 + x 3 , 2 − x + x 2 − 3 x 3} .U , (26) :(rank) 1 3 5 (2 1 2 1 6 2 5 3 1 −1 −2 2 1 −2 3 5 −4 −1 www.GooL.co.il © ¯ – ¯ 11 4 0 11 −5 2 −5 3 1 3 −1 5 (1 3 1 2 29 , : R3 (27) B1 = {(1,1, 0), (0,1, 0), (0,1,1)} , B2 = { (1, 0,1), (0,1,1), (0,0,1) } . [ v ]B . B1 . . [ v ]B . B2 . 1 2 B . M B2 . B2 . M B2 B1 B1 . . B1 1 B2 . : ( B B [ M ]B2 = [ M ]B1 1 2 ) −1 (3 B [ M ]B2 ⋅ [v]B = [v]B (2 1 2 1 . B [ M ]B1 ⋅ [v]B = [v]B (1 2 1 2 : P2 [ R ] (27) B1 = {1 + x, x, x + x 2 } , B2 = { 1 + x 2 , x + x 2 , x 2 } . [ v ]B . B1 . . [ v ]B . B2 . 1 2 B . M B2 . B2 1 B1 . : M 2[ R] (28) 1 1 0 1 0 0 0 0 B = , , , 0 0 1 0 1 1 0 1 1 0 0 1 0 0 0 0 E = , , , 0 0 0 0 1 0 0 1 . [ v ]B .B . . [ v ]E .E . E . M B .E www.GooL.co.il © . B ¯ – ¯ 30 5 – ¯, , ¯ : ¯ (1) . . . . . ¯ . ¯ . ¯ . . .¯ ¯, ¯ ¯, . ¯ ¯ . . , P −1 AP = D D . A2009 A−1 . ¯P ¯ ¯ . . ¯ . . ¯ . . I 1 0 1 1 1 0 A = 0 1 0 (3 A = 0 1 0 (2 1 0 1 0 0 2 2 −1 A= (6 1 4 A 0 2 −1 A = 0 2 −1 (1 0 1 0 3 −2 −1 3 0 A= (5 A = 3 −1 0 (4 4 −1 −2 −2 6 F = C,F = R F = C,F = R .R 5,6 C * . :¯ ¯ . B A .3 . . B . tr ( A) = tr ( B ) .2 . An = PB n P −1 www.GooL.co.il © ¯ – ¯ (2) . A . | A | = | B | .1 P −1 AP = B ¯ (3) 31 6 ( . . – ) ( ) . , (1) ¯ (2) T ( x, y ) = ( x + y , x − y ) ; T : R 2 → R 2 (1 T ( x, y , z ) = ( x + y − 2 z , x + 2 y + z , 2 x + 2 y − 3 z ) ; T : R3 → R3 (2 T ( x, y, z ) = (2 x + z ,| y |) ; T : R 3 → R 2 (3 T ( x, y ) = ( xy, y, z ) ; T : R 2 → R3 (4 T ( x, y, z ) = ( x + 1, x + y, y + z ) T : R 3 → R 3 (5 ( B ∈ M n [ R ]) T ( A) = BA + AB ; T : M n [ R ] → M n [ R] (6 T ( A) = A + AT ; T : M n [ R] → M n [ R ] (7 T ( A) = | A | ⋅I ; T : M n [ R] → M n [ R] (8 T ( A) = A ⋅ AT ; T : M n [ R] → M n [ R] (9 T ( A) = A−1 ; T : M n [ R] → M n [ R] (10 T (a + bx + cx 2 + dx 3 ) = a + bx + cx 2 ; T : P3[ R] → P2 [ R] (11 T ( p ( x) ) = p ( x + 1) ; T : Pn [ R ] → Pn [ R] (12 T ( p ( x) ) = p '( x) + p ''( x) ; T : Pn [ R] → Pn [ R] (13 T ( p ( x) ) = p 2 ( x) ; T : Pn [ R] → P2 n [ R] (14 ( F = C , F = R) www.GooL.co.il © T (z) = z ¯ – ¯ ; T : C[ F ] → C[ F ] (15 32 : (¯ )m (3) T ( x, y ) = ( m 2 x 2 m , y 2 m + x ) ; T : R 2 → R 2 . , . , ¯ (4) ,¯ . . T (1,1,0) = (1, 2,3), T (0,1,1) = (4,5,6), T (0,0,1) = (7, 8,9) ¯ T : R3 → R3 . . T (1,0,1) = (1,1,0), T (0,1,1) = (1, 2,1) , T (0,0,1) = (0,1,1) ¯ T : R3 → R3 . . T (1, 2, −1,0) = (0,1, −1), T ( −1,0,1,1) = (1,0,0), T (0, 4,0, 2) = (2, 2, −2) ¯ T : R 4 → R3 . . T (1) = 4, T ( 4 x + x 2 ) = x , T (1 − x ) = x 2 + 1 ¯ T : P2 [ R ] → P2 [ R] . .T :V → U : . Im T . (5) . . KerT ) rankT . ( nullT : ¯ T ( x, y, z, t ) = ( x + y, y − 4 z + t , 4 x + y + 4 z − t ) , T : R 4 → R3 (1 T ( x, y, z ) = ( x − 4 y − z , x + y, y − z , x + 4 z ) , T : R 3 → R 4 (2 x 1 1 2 3 y T ( x, y, z, t ) = 1 3 5 −2 , T : R 4 → R 3 (3 2 6 10 −4 z t 1 2 1 2 T ( A) = A ⋅ − ⋅ A , T : M 2 [ R] → M 2 [ R] (4 0 3 0 3 T ( p ( x) ) = p ( x + 1) − p ( x + 4) , T : P2 [ R] → P2 [ R] (5 D ( p ( x ) ) = p '( x) , D : P3[ R] → P3[ R] (6 www.GooL.co.il © ¯ – ¯ (6) 33 . {(4,1, 4),(−1, 4,1)} T : R3 → R3 T : R 4 → R3 . {(0,1,1,1),(1,2,3, 4)} dimIm T = dim KerT (7) (8) .T :V → U ¯¯ . (9) . ? T : R 4 → R3 , . (" , ¯ ¯ ," ) .¯ V . " ¯ (9) ¯ ," ¯ ¯ (10) .¯ T ( x, y, z ) = ( x − y + z , y + z, z − x) , T : R3 → R3 (1 T ( x, y , z ) = ( x − y + z , y + z , x + 2 z ) , T : R 3 → R 3 (2 T (a + bx + cx 2 ) = (a + b + c, a − b, b − 2c) , T : P2 [ R] → R 3 (3 a b T = a − b + (c + d ) x + (a − c) x 2 + dx3 , T : M 2 [ R] → P3[ R] (4 c d " S : R3 → R 2 T : R3 → R3 : : (11) T ( x , y , z ) = ( x, 4 x − y , x + 4 y − z ) , S ( x , y , z ) = ( x − z , y ) : ( ) ST (5 TS (4 4 S − 10T (3 4 S (2 S + T (1 S 2 (10 S −1 (9 T −2 (8 T −1 (7 T 2 (6 www.GooL.co.il © ¯ – ¯ 34 7 – ¯ .¯ ¯ ¯: .(4 ) : R3 (1) B1 = {(1,1, 0), (0,1, 0), (0,1,1)} , B2 = { (1, 0,1), (0,1,1), (0, 0,1) } . [ v ]B . B1 . . [ v ]B . B2 . 1 2 . M B2 . B2 B . M B2 1 B1 . . B1 B1 B2 . : ( B B [ M ]B2 = [ M ]B1 1 2 ) −1 (3 B [ M ]B2 ⋅ [v]B = [v]B (2 1 2 1 . B [ M ]B1 ⋅ [v]B = [v]B (1 2 T ( x, y , z ) = ( x + y , y + z , z − x ) , T : R 3 → R 3 . T B 1 2 : . B1 . T B 2 . . B2 1 . : [T ]B ⋅ [v]B = [T (v)]B (2 2 2 . [T ]B ⋅ [v]B = [T (v )]B (1 2 1 B 1 1 B M 1 ⋅ T B ⋅ M 2 = T B (3 B2 1 B1 2 ?¯ . . . . ¯ ? www.GooL.co.il © ¯ ¯ – ¯ . . 35 . R3 . R3 T T B 1 −29 −45 6 = 20 31 −4 13 19 −1 B2 B2 M B1 (2) −1 −9 6 = 1 6 −4 : ¯ 1 5 −2 B M 1 B2 . T B 2 1 2 T ( A) = A 3 4 B1 , T : M 2 [ R] → M 2 [ R ] (3) 1 1 0 1 0 0 0 0 . B = , , , : 0 0 1 0 1 1 0 1 D ( p( x) ) = p '( x) , D : P4 [ R] → P3 [ R] (4) .4 ¯ (5) . Rn . T ( x, y ) = ( x + y , y + z , z − x ) , T : R 2 → R 3 . . T ( x, y, z , t ) = (4 x − y − z + t , x + y + 4 z + t ) , T : R 4 → R 2 . . T ( x, y, z ) = (4 x + y − z, x − y + z ) T : R3 → R 2 B1 = {(1,1, 0), (0,1,1), (0, 0,1)} T B . T B2 ¯ . R2 1 www.GooL.co.il © B2 = {(1, 4), (1,5)} ¯ – ¯ R3 (6) ...
View Full Document

This note was uploaded on 01/26/2012 for the course IEM 101.1911.1 taught by Professor Yuliaglushko during the Spring '11 term at Ben-Gurion University.

Ask a homework question - tutors are online