Bài tập toán cao cấp Tập 1 part 9 pot

5.4. Ph´ep biˆe ´ ndˆo ’ i tuyˆe ´ n t´ınh 223 Nhu . vˆa . y L(x)=−x v`a do d´o x l`a vecto . riˆeng ´u . ng v´o . i gi´a tri . riˆeng λ = −1. 2) 1 + Ta c´o ma trˆa . ncu ’ a ph´ep biˆe ´ ndˆo ’ il`a  54 89  . Phu . o . ng tr`ınh d ˘a . c tru . ng c´o da . ng      5 −λ 4 89−λ      =0⇔ λ 2 − 14λ +13=0 ⇔  λ 1 =1, λ 2 =13. 2 + Ca ’ hai gi´a tri . λ =1v`aλ =13dˆe ` u l`a c´ac gi´a tri . riˆeng. 3 + Dˆe ’ t`ım to . adˆo . cu ’ a c´ac vecto . riˆeng ta c´o hai hˆe . phu . o . ng tr`ınh tuyˆe ´ n t´ınh (I)  (5 − λ 1 )ξ 1 +4ξ 2 =0, 8ξ 1 +(9− λ 1 )ξ 2 =0. (II)  (5 −λ 2 )ξ 1 +4ξ 2 =0, 8ξ 1 +(9− λ 2 )ξ 2 =0. i) V`ı λ 1 =1nˆen hˆe . (I) c´o da . ng 4ξ 1 +4ξ 2 =0, 8ξ 1 +8ξ 2 =0. T`u . d´o suy ra ξ 2 = −ξ 1 ,dod´o nghiˆe . mcu ’ ahˆe . n`ay c´o da . ng ξ 1 = α 1 , ξ 2 = −α 1 , trong d´o α 1 l`a da . ilu . o . . ng t`uy ´y. V`ı vecto . riˆeng kh´ac khˆong nˆen c´ac vecto . ´u . ng v´o . i gi´a tri . riˆeng λ 1 = 1 l`a c´ac vecto . u(α 1 , −α 1 ), trong d´o α 1 =0l`at`uy ´y. ii) Tu . o . ng tu . . khi λ 2 =13hˆe . (I I) tro . ’ th`anh −8ξ 1 +4ξ 2 =0, 8ξ 1 −4ξ 2 =0, 224 Chu . o . ng 5. Khˆong gian Euclide R n t´u . cl`aξ 2 =2ξ 1 .D˘a . t ξ 1 = β ⇒ ξ 2 =2β.Vˆa . yhˆe . (I I) c´o nghiˆe . ml`a ξ 1 = β, ξ 2 =2β.V`ı vecto . riˆeng kh´ac khˆong nˆen c´ac vecto . riˆeng ´u . ng v´o . i gi´a tri . λ 2 = 13 l`a c´ac vecto . v(β,2β).  V´ı d u . 6. T`ım gi´a tri . riˆeng v`a vecto . riˆeng cu ’ a ph´ep biˆe ´ ndˆo ’ i tuyˆe ´ n t´ınh L v´o . i ma trˆa . n A =  12 54  . Gia ’ i. D ath´u . cd˘a . c tru . ng cu ’ a ph´ep biˆe ´ ndˆo ’ i L P (λ)=      1 −λ 2 54− λ      = λ 2 − 5λ −6. N´o c´o hai nghiˆe . m thu . . c λ 1 =6,λ 2 = −1. C´ac vecto . d ˘a . c tru . ng d u . o . . c t`ım t `u . hai hˆe . phu . o . ng tr`ınh  (1 −λ i )ξ 1 +2ξ 2 =0, 5ξ 1 +(4− λ i )ξ 2 =0, i =1, 2. V`ıd i . nh th ´u . ccu ’ ahˆe . = 0 nˆen mˆo ˜ ihˆe . chı ’ thu vˆe ` mˆo . tphu . o . ng tr`ınh. 1 + V´o . i λ 1 = 6 ta c´o 5ξ 1 − 2ξ 2 =0⇒ ξ 1 ξ 2 = 2 5 v`a do d´o ta c´o thˆe ’ lˆa ´ y vecto . riˆeng tu . o . ng ´u . ng l`a u =(2, 5) (ho˘a . cmo . i vecto . αu, α ∈ R, α =0) 2 + V´o . i λ 2 = −1 ta c´o ξ 1 +ξ 2 =0⇒ ξ 1 ξ 2 = −1 v`a vecto . riˆeng tu . o . ng ´u . ng l`a v =(1, −1) (hay mo . i vecto . da . ng βv, β = 0).  V´ı d u . 7. T`ım c´ac gi´a tri . riˆeng v`a vecto . riˆeng cu ’ aph´ep biˆe ´ ndˆo ’ i tuyˆe ´ n t´ınh L trˆen R 3 v´o . i ma trˆa . n theo co . so . ’ ch´ınh t˘a ´ cl`a A =    114 20−4 −11 5    5.4. Ph´ep biˆe ´ ndˆo ’ i tuyˆe ´ n t´ınh 225 Gia ’ i. Ta c´o dath´u . cd˘a . c tru . ng cu ’ a ma trˆa . n A l`a det(A −λE)=        1 −λ 14 2 −λ −4 −115−λ        = −λ 3 +6λ 2 −11λ +6 v`a det(A −λE)=0⇐⇒    λ 1 =1, λ 2 =2, λ 3 =3. Gia ’ su . ’ x =(ξ 1 ,ξ 2 ,ξ 3 ) = 0 l`a vecto . riˆeng ´u . ng v´o . i gi´a tri . riˆeng λ. Khi d ´o x l`a nghiˆe . mcu ’ ahˆe . thuˆa ` n nhˆa ´ t (1 −λ)ξ 1 + ξ 2 +4ξ 3 =0, 2ξ 1 − λξ 2 − 4ξ 3 =0, −ξ 1 + ξ 2 +(5− λ)ξ 3 =0.      (*) 1 + Khi λ = 1 ta c´o (∗) ⇒ ξ 2 +4ξ 3 =0, 2ξ 1 − ξ 2 − 4ξ 3 =0, −ξ 1 + ξ 2 +4ξ 3 =0.      ⇒ nghiˆe . mtˆo ’ ng qu´at l`a (0, −4α, α), α =0t`uy ´y. Vˆa . yv´o . i gi´a tri . riˆeng λ 1 = 1 ta c´o c´ac vecto . riˆeng ´u . ng v´o . in´ol`a (0, −4α, α), α ∈ R, α =0. 2 + Khi λ = 2 ta c´o (∗) ⇒ −ξ 1 + ξ 2 +4ξ 3 =0, 2ξ 1 − 2ξ 2 − 4ξ 3 =0, −ξ 1 + ξ 2 +3ξ 3 =0      ⇒ hˆe . c´o nghiˆe . mtˆo ’ ng qu´at l`a (β,β,0), β =0v`adod ´o vecto . riˆeng ´u . ng v´o . i λ =2l`a(β,β,0), β =0. 226 Chu . o . ng 5. Khˆong gian Euclide R n 3 + Khi λ = 3, thu . . chiˆe . ntu . o . ng tu . . nhu . o . ’ 1 + v`a 2 + ta thu du . o . . c vecto . riˆeng tu . o . ng ´u . ng (2γ,0,γ), γ =0t`uy ´y.  V´ı d u . 8. T`ım gi´a tri . riˆeng v`a vecto . riˆeng cu ’ a ph´ep bdtt v´o . i ma trˆa . n A =    7 −12 6 10 −19 10 12 −24 13    . Gia ’ i. Phu . o . ng tr`ınh d˘a . c tru . ng P (λ)=        7 −λ −12 6 10 −19 −λ 10 12 −24 13 − λ        =0 c´o nghiˆe . m λ 1 = λ 2 =1,λ 1 = −1. C´ac vecto . d ˘a . c tru . ng d u . o . . c x´ac d i . nh t`u . hai hˆe . phu . o . ng tr`ınh (7 −λ i )ξ − 12η +6ζ =0, 10ξ −(19 + λ i )η +10ζ =0, 12ξ −24η + (13 − λ i )ζ =0; i =1, 2. 1 + Khi λ = 1 ta c´o 6ξ − 12η +6ζ =0, 10ξ − 20η +10ζ =0, 12ξ − 24η +12ζ =0. Ha . ng cu ’ a ma trˆa . n (go . i l`a ma trˆa . nd ˘a . c tru . ng) (A − λ 1 E)cu ’ ahˆe . n`ay l`a b˘a ` ng r = 1. Do d´ohˆe . tu . o . ng du . o . ng v´o . imˆo . tphu . o . ng tr`ınh ξ − 2η + ζ =0. T`u . d´o suy r˘a ` ng hˆe . c´o hai nghiˆe . mdˆo . clˆa . p tuyˆe ´ n t´ınh, ch˘a ’ ng ha . n u =(4, 5, 6), v =(3, 5, 7). 5.4. Ph´ep biˆe ´ ndˆo ’ i tuyˆe ´ n t´ınh 227 2 + Khi λ 2 = −1 ta c´o 8ξ − 12η +6ζ =0, 10ξ −18η +10ζ =0, 12ξ −24η +14ζ =0. Ha . ng cu ’ a ma trˆa . n(A−λ 3 E)cu ’ ahˆe . b˘a ` ng r = 2. Do d´ohˆe . tu . o . ng d u . o . ng v´o . ihˆe . hai phu . o . ng tr`ınh. Nghiˆe . m riˆeng cu ’ a n´o c´o da . ng w =(3, 5, 6). Nhu . vˆa . y u, v, w l`a c´ac vecto . riˆeng cu ’ a ph´ep bdtt d˜a cho. V´ı d u . 9. Cho ma trˆa . n A =  01 10  . T`ım ph´ep biˆe ´ nd ˆo ’ i tuyˆe ´ n t´ınh L tu . o . ng ´u . ng v´o . i ma trˆa . nd´o. Gia ’ i. Gia ’ su . ’ x = ae 1 + be 2 l`a vecto . t`uy ´y cu ’ am˘a . t ph˘a ’ ng. Dˆe ’ t`ım ph´ep biˆe ´ nd ˆo ’ i tuyˆe ´ n t´ınh ta cˆa ` nchı ’ r˜o a ’ nh y = Ax.Tac´o y =  01 10  a b  =  b a  = be 1 + ae 2 . Nhu . vˆa . y ph´ep biˆe ´ nd ˆo ’ i L c´o t´ınh chˆa ´ t l`a: thay dˆo ’ i vai tr`o cu ’ a c´ac to . a dˆo . cu ’ amˆo ˜ i vecto . x ∈ R 2 .T`u . d´o suy r˘a ` ng L l`a ph´ep pha ’ nxa . gu . o . ng dˆo ´ iv´o . idu . `o . ng phˆan gi´ac th´u . nhˆa ´ t.  B ` AI T ˆ A . P Trong c´ac b`ai to´an (1 - 11) h˜ay ch´u . ng to ’ ph´ep biˆe ´ ndˆo ’ id˜a c h o l `a ph´ep bdtt v`a t`ım ma trˆa . ncu ’ ach´ung theo co . so . ’ ch´ınh t˘a ´ c. 1. Ph´ep biˆe ´ ndˆo ’ i L l`a ph´ep quay mo . i vecto . cu ’ am˘a . t ph˘a ’ ng xOy xung quanh gˆo ´ cto . adˆo . mˆo . t g´oc ϕ ngu . o . . cchiˆe ` u kim dˆo ` ng hˆo ` . (DS. A L =  cos ϕ −sin ϕ sin ϕ cos ϕ  ) 228 Chu . o . ng 5. Khˆong gian Euclide R n 2. Ph´ep biˆe ´ ndˆo ’ i L l`a ph´ep quay khˆong gian thu . . cbachiˆe ` umˆo . t g´oc ϕ xung quanh tru . c Oz. (D S.    cos ϕ −sin ϕ 0 sin ϕ cos ϕ 0 001    ) 3. Ph´ep biˆe ´ nd ˆo ’ i L l`a ph´ep chiˆe ´ u vuˆong g´oc vecto . a ∈ R 3 lˆen m˘a . t ph˘a ’ ng xOy. (D S.    100 010 000    ) 4. Ph´ep biˆe ´ nd ˆo ’ i L l`a t´ıch vecto . y =[a, x], trong d ´o a = a 1 x 1 + a 2 x 2 + a 3 x 3 l`a vecto . cˆo ´ d i . nh cu ’ a R 3 . (D S.    0 −a 3 a 2 a 3 0 −a 1 −a 2 a 1 0    ) Chı ’ dˆa ˜ n. Su . ’ du . ng ph´ep biˆe ’ udiˆe ˜ n t´ıch vecto . du . ´o . ida . ng di . nh th´u . c. 5. Ph´ep biˆe ´ nd ˆo ’ i L l`a ph´ep biˆe ´ ndˆo ’ idˆo ` ng nhˆa ´ t trong khˆong gian n-chiˆe ` u R n trong mo . ico . so . ’ . (DS. E =       10 0 01 0 . . . . . . . . . . . . 00 1       ) 6. L l`a ph´ep biˆe ´ nd ˆo ’ idˆo ` ng da . ng L(x)=αx trong khˆong gian n-chiˆe ` u. (DS.       α 0 0 0 α 0 . . . . . . . . . . . . 00 α       ) 5.4. Ph´ep biˆe ´ ndˆo ’ i tuyˆe ´ n t´ınh 229 7. Ph´ep biˆe ´ ndˆo ’ i L c´o da . ng L(x)=x 2 e 1 + x 3 e 2 + x 4 e 3 + x 1 e 4 trong d´o x = x 1 e 1 + x 2 e 2 + x 3 e 3 + x 4 e 4 . (DS.      0100 0010 0001 1000      ) 8. Ph´ep biˆe ´ nd ˆo ’ i L l`a ph´ep chiˆe ´ u vuˆong g´oc khˆong gian 3-chiˆe ` ulˆen tru . c∆lˆa . pv´o . i c´ac tru . cto . adˆo . nh˜u . ng g´oc b˘a ` ng nhau, t´u . cl`a(  Ox,∆) = (  Oy, ∆)=(  Oz, ∆) = α. (DS.        1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3        ) Chı ’ dˆa ˜ n. Su . ’ du . ng t´ınh chˆa ´ tcu ’ a cosin chı ’ phu . o . ng cu ’ a g´oc bˆa ´ tk`y cos 2 α + cos 2 α + cos 2 α =1. 9. Ph´ep biˆe ´ ndˆo ’ i L l`a ph´ep chiˆe ´ u R 3 theo phu . o . ng song song v´o . im˘a . t ph˘a ’ ng vecto . e 2 ,e 3 lˆen tru . cto . adˆo . cu ’ a vecto . e 1 (DS.    100 000 000    ) 10. Ph´ep biˆe ´ nd ˆo ’ i L l`a ph´ep quay R 3 mˆo . t g´oc ϕ = 2π 3 xung quanh d u . `o . ng th˘a ’ ng cho trong R 3 bo . ’ iphu . o . ng tr`ınh x 1 = x 2 = x 3 . (D S. a)    001 100 010    nˆe ´ u quay t `u . e 1 dˆe ´ n e 2 , b)    010 001 100    nˆe ´ u quay t `u . e 2 dˆe ´ n e 1 ) 230 Chu . o . ng 5. Khˆong gian Euclide R n Trong c´ac b`ai to´an (12-22) cho hai co . so . ’ (e):e 1 ,e 2 , ,e n v`a (E):E 1 , E 2 , ,E n cu ’ a khˆong gian R n v`a ma trˆa . n A L cu ’ a ph´ep biˆe ´ n dˆo ’ i tuyˆe ´ n t´ınh L trong co . so . ’ (e). T`ım ma trˆa . ncu ’ a L trong co . so . ’ (E). Phu . o . ng ph´ap chung l`a: (i) t`ım ma trˆa . n chuyˆe ’ n T t`u . co . so . ’ (e)dˆe ´ nco . so . ’ (E); (ii) T`ım ma trˆa . n T −1 ; (iii) T`ım B L = T −1 AT . 11. A L =  17 6 68  , E 1 = e 1 − 2e 2 , E 2 =2e 1 + e 2 ). (DS.  50 020  ) 12. A L =  −31 2 −1  , E 1 = e 2 , E 2 = e 1 + e 2 .(DS.  −23 1 −2  ) 13. A L =  24 −33  , E 1 = e 2 − 2e 1 , E 2 =2e 1 − 4e 2 .(DS.  −314 −38  ) 14. A L =  10 2 −4  , E 1 =3e 1 +2e 2 , E 2 =2e 1 +2e 2 .(DS.  56 −6 −8  ) 15. A L =    0 −21 310 2 −11    , E 1 =3e 1 + e 2 +2e 3 , E 2 =2e 1 + e 2 +2e 3 , E 3 = −e 1 +2e 2 +5e 3 .(DS.    −85 −59 18 121 84 −25 −13 −93    ) 16. A L =    15 −11 5 20 −15 8 8 −76    , E 1 =2e 1 +3e 2 + e 3 , E 2 =3e 1 +4e 2 + e 3 , E 3 = e 1 +2e 2 +2e 3 .(DS.    100 020 003    ) 17. A L =    2 −10 01−1 00 1    . E 1 =2e 1 + e 2 − e 3 , E 2 =2e 1 −e 2 +2e 3 , 5.4. Ph´ep biˆe ´ ndˆo ’ i tuyˆe ´ n t´ınh 231 E 3 =3e 1 + e 3 .(DS.    −211 7 −414 8 5 −15 −8    ) 18. A L =  21 03  , e 1 =3E 1 −E 2 , e 2 = E 1 + E 2 .(DS.  33 02  ) 19. A L =  −14 50  , e 1 = E 1 + E 2 , e 2 =2E 1 .(DS.  27 2 −3  ) 20. A L =    12−3 031 −12 5    , e 1 = E 1 , e 2 =3E 1 + E 2 , e 3 =2E 1 + E 2 +2E 3 . (D S.    −118−3 −18 0 −210 2    ) 21. A L =    2 −10 01−1 00 1    , e 1 =2E 1 + E 2 −E 3 , e 2 =2E 1 −E 2 +2E 3 , e 3 =3E 1 + E 2 .(DS.    3 −10 −8 −18 5 2 −13 −7    ) 22. Trong c´ac ph´ep biˆe ´ nd ˆo ’ i sau dˆay t `u . R 3 → R 3 ph´ep biˆe ´ ndˆo ’ i n`ao l`a tuyˆe ´ n t´ınh (gia ’ thiˆe ´ t x =(x 1 ,x 2 ,x 3 ) ∈ R 3 ) 1) L(x 1 ,x 2 ,x 3 )=(x 1 +2x 2 +3x 3 , 4x 1 +5x 2 +6x 3 , 7x 1 +8x 2 +9x 3 ); 2) L(x 1 ,x 2 ,x 3 )=(x 1 +3x 2 +4, 5x 3 ;6x 1 +7x 2 +9x 3 ;10,5x 1 +12x 2 + 13x 3 ) 3) L(x 1 ,x 2 ,x 3 )=(x 2 + x 3 ,x 1 + x 3 ,x 1 + x 2 ). 4) L(x 1 ,x 2 ,x 3 )=(x 1 ,x 2 +1,x 3 + 2). 5) L(x 1 ,x 2 ,x 3 )=(x 2 + x 3 , 2x 1 + x 3 , 3x 1 − x 2 + x 3 ). 6) L(x 1 ,x 2 ,x 3 )=(2x 1 + x 2 ,x 1 + x 3 ,x 2 3 ). 7) L(x 1 ,x 2 ,x 3 )=(x 1 − x 2 − x 3 ,x 3 ,x 2 ). 232 Chu . o . ng 5. Khˆong gian Euclide R n (DS. 1), 2), 3), 5), 7) l`a ph´ep bdtt; 4), 6) - khˆong) 23. T`ım phu . o . ng tr`ınh d˘a . c tru . ng v`a sˆo ´ d˘a . c tru . ng cu ’ a ph´ep bdtt L nˆe ´ u 1) L(e 1 )=2e 1 ; L(e 2 )=5e 1 +3e 2 ; L(e 3 )=3e 1 +4e 2 − 6e 3 , trong d´o e 1 ,e 2 ,e 3 l`a co . so . ’ cu ’ a khˆong gian. (D S. (λ + 6)(λ −2)(λ −3) = 0) 2) L(e 1 )=−e 1 , L(e 2 )=2e 1 +5e 2 , L(e 3 )=2e 1 − e 2 +3e 3 +5e 4 , L(e 4 )=e 1 +7e 2 +4e 3 +6e 4 , trong d´o e 1 ,e 2 ,e 3 ,e 4 l`a co . so . ’ cu ’ a khˆong gian. (D S. (λ + 1)(λ − 5)(λ 2 − 9λ −2) = 0) 3) L(e 1 )=2e 1 +2e 3 , L(e 2 )=2e 1 +2e 2 , L(e 3 )=−2e 2 +2e 3 ; e 1 ,e 2 ,e 3 l`a co . so . ’ cu ’ a khˆong gian. (DS. λ 3 −6λ 2 +12λ =0) 24. Gia ’ su . ’ trong co . so . ’ e = {e 1 ,e 2 } ph´ep bdtt L c´o ma trˆa . nl`a A L =  35 −14  c`on trong co . so . ’ E = {E 1 , E 2 }, E 1 = e 1 − e 2 , E 2 = e 1 +2e 2 ph´ep bdtt L ∗ c´o ma trˆa . n A L ∗ =  0 −2 11  . T`ım ma trˆa . ncu ’ a c´ac ph´ep biˆe ´ nd ˆo ’ i: 1) L + L ∗ trong co . so . ’ e 1 , e 2 ; 2) L + L ∗ trong co . so . ’ E 1 , E 2 . (DS. 1) 1 3  10 13 514  ;2) 1 3  113 −323  ) 25. Gia ’ su . ’ trong co . so . ’ e = {e 1 ,e 2 ,e 3 } ph´ep bdtt L c´o ma trˆa . n A L =    20−2 11 0 30−1    [...]... = o Ta t` c´c hˆ sˆ cua (6 .11 ) theo cˆng th´.c (6 .9) Ta c´ ım a e o ’ o u ´ −2 D1 ,1 α 21 = ( 1) 3 = 1, =− 1 2 α 31 = ( 1) 4 α32 = ( 1) 5 D2 ,1 = ∆2 −2 1 3 1 D2,2 =− ∆2 2 1 =− , 2 2 1 −2 1 2 = 0 6 .1 Dang to`n phu.o.ng a 243 Nhu vˆy a x1 x2 x3  1  = y1 + y2 − y3 , 2  = y2 ,    = y3 a V´ du 4 Du.a dang to`n phu.o.ng ı ϕ(x1 , x2, x3) = 2x2 + 3x1x2 + 4x1 x3 + x2 + x2 1 2 3 ` vˆ dang ch´ t˘c e... u1 − u2 − 4u3 ,   2 1 x2 = u1 + u2 − 2u3 ,  2   x3 = u3 ’ ’ e e ı ıch Dˆ kiˆm tra ta t´nh t´ C T AC   1 1 1 0 0  2   1 1 C T AC = − 0  1 0   2 2 2 −4 −2 1 1 2 Ta c´ o     1 1 0 0 1 − 1 −4    2  1    1  1  = 0 − 0  2  −2  4 2 0 0 −8 0 0 0 1 D´ l` ma trˆn cua dang ch´ t˘c thu du.o.c o a a ’ ınh ´ a 6 .1 Dang to`n phu.o.ng a 6 .1. 2 2 41 a Phu.o.ng ph´p Jacobi... (6.4), (6.5)  1 0  1 1 0 0 1 ϕ(·) = u2 − u2 − 8u2 1 3 4 2 ’ ´ ` ’ trˆn cua ph´p biˆn dˆi ho.p ta cˆn nhˆn c´c ma trˆn cua a ’ e e o a a a a v` (6.6) Ta c´ a o    1   1 1 − −4 1 0 0 0 1 − −3   2  2     1  0 1 2 =  0 0 1 0  −2 = C 1    2 0 0 1 1 0 0 1 0 0 1 ´ ’ ´ ` Do ph´p biˆn dˆi khˆng suy biˆn du.a dang ϕ vˆ dang ch´ t˘c l` e e o e ınh ´ a a o e  1  x1 = u1 − u2 − 4u3... 1 = a 11 = 0, ∆2 = a 11 a12 a 21 a22 a 11 a12 a1n a 21 a22 a2n = 0, , ∆n = = 0 an1 an2 ann (6.7) ’ Cu thˆ ta c´ o e - ´ Dinh l´ Nˆu dang to`n phu.o.ng y e a n ϕ(x1, , xn ) = aij xi xj i,j =1 ` ’ thoa m˜n diˆu kiˆn v`.a nˆu: ∆i = 0 ∀ i = 1, n th` tˆn a e e u ı ` o e ’ ´ ´ ´ e o e ı o e u a dˆi tuyˆn t´nh khˆng suy biˆn t` c´c biˆn x1, , xn y1, , yn sao cho ϕ(·) = 1 2... x3 = y3 v` thu du.o.c a ϕ(·) = y1(y1 + y2) + 2y1y3 + 4(y1 + y2)y3 2 = y1 + y1 y2 + 6y1 y3 + 4y2 y3 ´ Xuˆt ph´t t` dang to`n phu.o.ng m´.i thu du.o.c, tu.o.ng tu nhu trong a a u a o v´ du 1 ta c´ ı o 1 ϕ(·) = y1 + y2 + 3y3 2 1 = y1 + y2 + 3y3 2 2 − 2 1 y2 + 3y3 2 2 + 4y2 y3 1 − y 2 + y2y3 − 9y 3 4 ´ ’ ´ e e e o o e Thu.c hiˆn ph´p biˆn dˆi khˆng suy biˆn 1 z1 = y1 + y2 + 3y3 , 2 z2 = y2, z3 = y3... d˜ cho a ’ a  2  A = −2 1 u ı v´.i c´c dinh th´.c con ch´nh o a 1 = 2, c´ dang o  −2 1  3 1 1 1 ∆2 = 2, ∆3 = 1 o a a e ınh ´ a Khi d´ dang to`n phu.o.ng d˜ cho du.a du.o.c vˆ dang ch´ t˘c ` 1 2 2 2 ϕ(·) = 2y1 + y2 + y3 (6 .10 ) 2 ` a a e Ta t`m ph´p bdtt du.a dang to`n phu.o.ng d˜ cho vˆ dang (6 .10 ) ı e N´ c´ dang o o  x1 = y1 + α 21 y2 + α 31 y3,  (6 .11 ) y2 + α32 y3, x2 =   y3... v´.i c´c hˆ sˆ du.o.c x´c dinh theo (6 .9) Ap dung (6 .9) ta thu du.o.c o a e o ´ a α 21 = ( 1) 3 D1 ,1 1 α 31 = ( 1) 4 D2 ,1 ∆2 α32 = ( 1) 4 D2,2 ∆2 3 3 = −2 = − , 2 4 3 2 2 1 0 = 8, = 1 − 4 2 2 3 0 2 = 12 =− 1 − 4 ´ ’ Vˆy ph´p biˆn dˆi l` a e e o a  3  x1 = y1 − y2 + 8y3 ,  4 y2 − 12 y3 , x2 =   y3 x3 = 6 .1. 3 ’ ´ o a e Phu.o.ng ph´p biˆn dˆi tru.c giao ´ ’ a a o u e V` ma trˆn A cua dang to`n... a e a e a a o.ng u.ng v´.i ph´p biˆn dˆi tru.c giao cua c´c biˆn x1 v` x2: ´ ’ ´ ’ a ´ o e e o e a giao N´ tu o 1 x1 = √ x1 − 26 5 x2 = √ x1 + 26 5 √ x2 , 26 1 √ X2 26 6 .1 Dang to`n phu.o.ng a 247 T` d´ ta c´ u o o 2 5 1 5 1 ϕ(·) = 27 √ x1 − √ x2 − 10 √ x1 − √ x2 26 26 26 26 2 1 5 2 2 + 3 √ x1 + √ x2 = 2x1 + 28x2 26 26 5 1 √ x1 + √ x2 26 26 ’ ` ´ o u o e a a ı Nhˆn x´t Hˆ th´.c cuˆi c`ng c´ thˆ... −4 −2 1 ´ a y (DS α, β v´.i α = 0, β = 0 bˆt k`) o −2 1 1 1         1 2 −2 −2 0 6         34 A = 1 0 3  (DS  1  α, 1 β, −7 γ, 1 3 0 1 1 5 ´t k`) α = 0, β = 0, γ = 0 bˆ y a       1 0 2 −2 0       ´ 35 A = 0 3 0 (DS  0  α, 1 β; α = 0, β = 0 bˆt k`) a y 0 0 0 1 0 33 A = ´ ´ ’ e ınh 36 Cho ph´p biˆn dˆi tuyˆn t´ L : R2 → R2 nhu sau e e o L : (x1, x2 ) −→ (5x1 +... = {E1 , E2}, E1 = 2e1 + e2 , E2 = e1 − e2 ph´p bdtt o ∗ o a L c´ ma trˆn AL∗ = 3 −2 1 0 ´ ’ T` ma trˆn cua c´c ph´p biˆn dˆi ım a ’ a e e o 1 1 ’ 1) L ◦ L∗ trong co so e (DS ) 0 2 1 7 ’ 2) L∗ ◦ L trong co so e (DS ) 0 2 1 1 −2 ’ ) 3) L ◦ L∗ trong co so E (DS 3 −7 4 Chu.o.ng 5 Khˆng gian Euclide o 234 Rn 1 7 10 ’ ) 4) L∗ ◦ L trong co so E (DS 3 1 −4 2 1 ’ trong co so E = 5 −3 ∗ ∗ ’ a {E1 , . 12 6 10 19 10 12 −24 13    . Gia ’ i. Phu . o . ng tr`ınh d˘a . c tru . ng P (λ)=        7 −λ 12 6 10 19 −λ 10 12 −24 13 − λ        =0 c´o nghiˆe . m λ 1 = λ 2 =1, λ 1 = 1. . A L =    12 −3 0 31 12 5    , e 1 = E 1 , e 2 =3E 1 + E 2 , e 3 =2E 1 + E 2 +2E 3 . (D S.    11 8−3 18 0 − 210 2    ) 21. A L =    2 10 01 1 00 1    , e 1 =2E 1 + E 2 −E 3 , e 2 =2E 1 −E 2 +2E 3 , e 3 =3E 1 +. y 3 .      (6 .11 ) Ta t`ım c´ac hˆe . sˆo ´ cu ’ a (6 .11 ) theo cˆong th ´u . c (6 .9) . Ta c´o α 21 =( 1) 3 D 1, 1 ∆ 1 = − −2 2 =1, α 31 =( 1) 4 D 2 ,1 ∆ 2 =      − 21 3 1      2 = − 1 2 , α 32 =( 1) 5 D 2,2 ∆ 2 =