Bài tập toán cao cấp Tập 1 part 10 pdf

6.1. Da . ng to`an phu . o . ng 251 t´u . cl`aho . c´ac vecto . riˆeng phu . thuˆo . c hai tham sˆo ´ α v`a β. Ta lˆa ´ y ra hai vecto . tru . . c giao n`ao d´ocu ’ aho . u =2(α + β)e 1 + αe 2 + βe 3 . Ch˘a ’ ng ha . nd˘a . t α =0,β =1th`ıthudu . o . . c vecto . riˆeng u 1 (2, 0, 1) v`a sau khi chuˆa ’ n h´oa ta d u . o . . c E 1 =  2 √ 5 , 0, 1 √ 5  . D ˆe ’ c´o vecto . th ´u . hai u 2 ta cˆa ` ncho . n α v`a β sao cho u 1 ,u 2  =0t´u . c l`a 2 · 2(α + β)+β =0⇔ 4α +5β =0. Ta c´o thˆe ’ cho . n α =5,β = −4v`at`u . (6.16) ta c´o u 2 (2, 5, −4) v`a sau khi chuˆa ’ n h´oa ta c´o E 2 =  2 3 √ 5 , √ 5 3 , −4 3 √ 5  . b) Gia ’ su . ’ λ = −2. Ta c´o 8ξ 1 +2ξ 2 +2ξ 3 =0, 2ξ 1 +5ξ 2 − 4ξ 3 =0, 2ξ 1 − 4ξ 2 +5ξ 3 =0      Ha . ng cu ’ a ma trˆa . ncu ’ ahˆe . b˘a ` ng 2 nˆen hˆe . co . ba ’ nchı ’ gˆo ` mmˆo . t nghiˆe . m. Ch˘a ’ ng ha . n gia ’ i hai phu . o . ng tr`ınh cuˆo ´ i ta c´o ξ 2 = ξ 3 v`a ξ 1 = − ξ 2 2 v`a do d ´o ξ 1 = − ξ 1 2 = − ξ 3 2 . D ˘a . t ξ 1 = α ta c´o ho . vecto . riˆeng phu . thuˆo . cmˆo . t tham sˆo ´ u 3 (α, −2α, −2α),α∈ R 252 Chu . o . ng 6. Da . ng to`an phu . o . ng v`a ´u . ng du . ng v`a sau khi chuˆa ’ n h´oa ta du . o . . c E 3 =  1 3 , −2 3 , −2 3  . R˜o r`ang l`a E 1 , E 2 , E 3 l`a mˆo . tco . so . ’ tru . . c chuˆa ’ ncu ’ a khˆong gian R 3 v`a ma trˆa . n chuyˆe ’ nvˆe ` co . so . ’ m´o . i n`ay l`a ma trˆa . n tru . . c giao da . ng T =         2 √ 5 2 3 √ 5 1 3 0 √ 5 3 − 2 3 1 √ 5 − 4 3 √ 5 − 2 3         T`u . d ´othudu . o . . cda . ng to`an phu . o . ng trong co . so . ’ m´o . i E 1 , E 2 , E 3 l`a B = T T AT =    70 0 07 0 00−2    t´u . cl`a ϕ(·)=7x  1 2 +7x  2 2 − 2x  3 2 , trong d ´o x 1 = 2 √ 5 x  1 + 2 3 √ 5 x  2 + 1 3 x  3 , x 2 = √ 5 3 x  2 − 2 3 x  3 , x 3 = 1 √ 5 x  1 − 4 3 √ 5 x  2 − 2 3 x  3 .  V´ı d u . 8. T`ım ph´ep biˆe ´ ndˆo ’ i tru . . c giao du . ada . ng to`an phu . o . ng ϕ(x 1 ,x 2 ,x 3 )=x 2 1 − 8x 1 x 2 − 16x 1 x 3 +7x 2 2 −8x 2 x 3 + x 2 3 vˆe ` da . ng ch´ınh t˘a ´ c. 6.1. Da . ng to`an phu . o . ng 253 Gia ’ i. 1 + Ma trˆa . ncu ’ ada . ng to`an phu . o . ng l`a A =    1 −4 −8 −47−4 −8 −41    T`u . d ´o ta c´o |A − λE| =        1 − λ −4 −8 −47− λ −4 −8 −41− λ        =0⇔ λ 1 = −9,λ 2 = λ 3 =9 2 + T`ım c´ac vecto . riˆeng D ˆe ’ t`ım to . adˆo . cu ’ a c´ac vecto . riˆeng ta cˆa ` n gia ’ i c´ac hˆe . phu . o . ng tr`ınh thuˆa ` n nhˆa ´ t (1 − λ i )ξ 1 −4ξ 2 − 8ξ 3 =0, −4ξ 1 +(7− λ i )ξ 2 − 4ξ 3 =0, −8ξ 1 − 4ξ 2 +(1− λ i )ξ 3 =0      (6.17) lˆa ` nlu . o . . tv´o . i λ 1 = −9, λ 2 = λ 3 =9. a) Gia ’ su . ’ λ 1 = −9. Khi d´ot`u . (6.17) ta c´o 10ξ 1 − 4ξ 2 − 8ξ 3 =0, −4ξ 1 +16ξ 2 − 4ξ 3 =0, −8ξ 1 −4ξ 2 +10ξ 3 =0      hay l`a 5ξ 1 − 2ξ 2 − 4ξ 3 =0, ξ 1 − 4ξ 2 + ξ 3 =0, 4ξ 1 +2ξ 2 − 5ξ 3 =0.      V`ıha . ng cu ’ a ma trˆa . ncu ’ ahˆe . b˘a ` ng 2 nˆen hˆe . c´o nghiˆe . m kh´ac 0. Ta gia ’ i hˆe . hai phu . o . ng tr`ınh d ˆa ` u 5ξ 1 − 2ξ 2 − 4ξ 3 =0, ξ 1 −4ξ 2 + ξ 3 =0 254 Chu . o . ng 6. Da . ng to`an phu . o . ng v`a ´u . ng du . ng v`a thu du . o . . c nghiˆe . mtˆo ’ ng qu´at l`a u(2α, α, 2α),α∈ R. D ´o l`a ho . vecto . riˆeng (phu . thuˆo . cmˆo . t tham sˆo ´ )´u . ng v´o . i gi´a tri . riˆeng λ 1 = −9. Sau khi chuˆa ’ n h´oa ta thu du . o . . c E 1 =  2 3 , 1 3 , 2 3  . b) Gia ’ su . ’ λ 2 = λ 3 = 9. Khi d´ot`u . (6.17) thu du . o . . chˆe . phu . o . ng tr`ınh thuˆa ` n nhˆa ´ t −8ξ 1 − 4ξ 2 − 8ξ 3 =0, −4ξ 1 − 2ξ 2 − 4ξ 3 =0, −8ξ 1 − 4ξ 2 − 8ξ 3 =0      hay l`a 2ξ 1 + ξ 2 +2ξ 3 =0, 2ξ 1 + ξ 2 +2ξ 3 =0, 2ξ 2 + ξ 2 +2ξ 3 =0.      Ha . ng cu ’ a ma trˆa . ncu ’ ahˆe . b˘a ` ng 1 nˆen hˆe . nghiˆe . mco . ba ’ ncu ’ a n´o gˆo ` m hai nghiˆe . m. Nghiˆe . mtˆo ’ ng qu´at cu ’ ahˆe . c´o da . ng v(α, −2α − 2β,β),α,β∈ R,α 2 + β 2 =0. (6.18) T`u . nghiˆe . mtˆo ’ ng qu´at n`ay ta r´ut ra hai vecto . riˆeng tru . . c giao v 1 v`a v 2 tu . o . ng ´u . ng v´o . i gi´a tri . riˆeng λ 2 = λ 3 =9. Dˆe ’ c´o v 1 ta cho α =1,β =0 v`a thu du . o . . c v 1 =(1, −2, 0). (6.19) Dˆe ’ t`ım v 2 ta cˆa ` nx´acdi . nh α v`a β trong (6.18) sao cho tho ’ a m˜an diˆe ` u kiˆe . n tru . . c giao gi˜u . a v 1 v`a v 2 ,t´u . cl`av 1 ,v 2  =0. T`u . (6.18) v`a (6.19) ta c´o α +4α +4β =0⇔ α = − 4 5 β. 6.1. Da . ng to`an phu . o . ng 255 Do vˆa . y, ta c´o thˆe ’ lˆa ´ y β = 5 v`a khi d´ot`u . (6.18) suy ra v 2 =(−4, −2, 5). Sau khi chuˆa ’ n h´oa v 1 v`a v 2 ta thu du . o . . c E ∈ =  1 √ 5 , − 2 √ 5 , 0  E 3 =  −4 3 √ 5 , − 2 3 √ 5 , √ 5 3  . (Lu . u´yr˘a ` ng E 1 ⊥E 2 , E 1 ⊥E 3 v`ı E 1 v`a E 2 , E 3 l`a c´ac vecto . riˆeng tu . o . ng ´u . ng v´o . i hai gi´a tri . riˆeng kh´ac nhau nˆen ch´ung tru . . c giao v´o . i nhau). 3 + X´ac di . nh ph´ep biˆe ´ ndˆo ’ i tru . . c giao. Trong co . so . ’ tru . . cchuˆa ’ nv`u . a thu du . o . . c E 1 , E 2 , E 3 da . ng to`an phu . o . ng d˜achodu . o . . cdu . avˆe ` da . ng ch´ınh t˘a ´ c ϕ(·)=−9y 2 1 +9y 2 2 +9y 2 3 nh`o . ma trˆa . n tru . . c giao T =         2 3 1 √ 5 − 4 3 √ 5 1 3 − 2 √ 5 − 2 3 √ 5 2 3 0 √ 5 3         v´o . i ph´ep biˆe ´ ndˆo ’ itu . o . ng ´u . ng l`a x 1 = 2 3 y 1 + 1 √ 5 y 2 − 4 3 √ 5 y 3 , x 2 = 1 3 y 1 − 2 √ 5 y 2 − 2 3 √ 5 y 3 , x 3 = 2 3 y 1 + √ 5 3 y 3 .  256 Chu . o . ng 6. Da . ng to`an phu . o . ng v`a ´u . ng du . ng B ` AI T ˆ A . P Trong c´ac b`ai to´an sau dˆay h˜ay viˆe ´ t ma trˆa . ncu ’ ada . ng to`an phu . o . ng c´o biˆe ’ uth´u . cto . adˆo . sau trong khˆong gian R 3 (1-4) v`a trong R 4 (5-6). 1. x 2 1 + x 2 2 −3x 1 x 2 .(DS.    1 −3/20 −3/210 000    ) 2. 2x 2 1 +3x 2 2 − x 2 3 +4x 1 x 2 − 6x 1 x 3 +10x 2 x 3 . (DS.    22−3 235 −35−1    ) 3. x 2 1 + x 2 2 + x 2 3 +4x 1 x 2 +4x 2 x 3 +4x 1 x 3 . (DS.    122 212 221    ) 4. 4x 2 1 + x 2 2 +9x 2 3 − 4x 1 x 2 − 6x 2 x 3 +12x 1 x 3 . (D S.    4 −26 −21−3 6 −39    ) 5. 4x 2 1 + x 2 3 − 2x 2 4 − x 1 x 2 +8x 1 x 4 − 5x 2 x 4 . (DS.           4 − 1 2 04 − 1 2 00− 5 2 0010 4 − 5 2 0 −2           ) 6. 2x 1 x 2 − 6x 1 x 3 − 6x 2 x 4 +2x 3 x 4 . 6.1. Da . ng to`an phu . o . ng 257 (DS.      01−30 100−3 −30 0 1 0 −31 0      ) Trong c´ac b`ai to´an 7-8, t`ım ma trˆa . ncu ’ amˆo ˜ ida . ng to`an phu . o . ng 7.  x 1 x 2   13 45  x 1 x 2  (D S.    1 7 2 7 2 5    ) 8.  x 1 x 2 x 3     −10 2 241 30−1       x 1 x 2 x 3    .(D S.        −11 5 2 14 1 2 5 2 1 2 −1        ) 9. Cho c´ac da . ng to`an phu . o . ng sau dˆay du . o . . cviˆe ´ tdu . ´o . ida . ng ma trˆa . n. H˜ay viˆe ´ t c´ac da . ng to`an phu . o . ng d ´o d u . ´o . ida . ng thˆong thu . `o . ng 1)  x 1 x 2 x 3     30−1 0 −21 −11 2       x 1 x 2 x 3    (D S. 3x 2 1 − 2x 1 x 3 − 2x 2 2 +2x 2 x 3 +2x 2 3 ) 2)  x 1 x 2 x 3     41 0 13−1 0 −1 −1       x 1 x 2 x 3    . (D S. 4x 2 1 +2x 1 x 2 +3x 2 2 − 2x 2 x 3 − x 2 3 ) 10. Viˆe ´ t c´ac da . ng to`an phu . o . ng sau dˆay du . ´o . ida . ng ma trˆa . n. 1) 3x 2 1 + x 2 2 − x 1 x 2 .(DS.  x 1 x 2     3 − 1 2 − 1 2 1     x 1 x 2  ) 2) x 2 1 + x 2 3 −2x 1 x 2 +5x 1 x 3 . 258 Chu . o . ng 6. Da . ng to`an phu . o . ng v`a ´u . ng du . ng (DS.  x 1 x 2 x 3       1 −1 5 2 −100 5 2 01         x 1 x 2 x 3    ) 3) 2x 2 1 +3x 2 2 − 2x 2 3 + x 1 x 2 +2x 1 x 3 +3x 2 x 3 . (DS.  x 1 x 2 x 3        2 1 2 1 1 2 3 3 2 1 3 2 −2          x 1 x 2 x 3    ) Trong c´ac b`ai to´an sau d ˆay (11-14) t`ım da . ng to`an phu . o . ng thu d u . o . . ct`u . da . ng d ˜achobo . ’ i ph´ep biˆe ´ nd ˆo ’ id˜achı ’ ra 11. ϕ(x 1 ,x 2 )=3x 2 1 − x 2 2 +4x 1 x 2 ; x 1 =2y 1 − y 2 , x 2 = y 1 + y 2 . (DS. ϕ 1 (y 1 ,y 2 )=19y 2 1 − 2y 2 2 − 10y 1 y 2 ) 12. ϕ(x 1 ,x 2 ,x 3 )=2x 2 1 +3x 2 2 − x 2 3 + x 1 x 2 ; x 1 = −y 1 +2y 2 ; x 2 =3y 1 + y 2 + y 3 ; x 3 = −2y 1 − y 2 . (D S. ϕ 1 (y 1 ,y 2 ,y 3 )=22y 2 1 +12y 2 2 +3y 2 3 +11y 1 y 2 +17y 1 y 3 +8y 2 y 3 ) 13. ϕ 1 (x 1 ,x 2 ,x 3 )=2x 2 2 +4x 2 3 −2x 1 x 2 + x 2 x 3 ; x 1 = y 1 + y 2 − y 3 , x 2 = y 1 − y 2 + y 3 , x 3 = y 3 + y 2 . (D S. ϕ 1 (y 1 ,y 2 ,y 3 )=7y 2 2 +9y 2 3 −3y 1 y 2 +5y 1 y 3 )) 14. ϕ(x 1 ,x 2 ,x 3 )=x 2 1 −2x 2 2 + x 2 3 − 4x 1 x 2 − 2x 2 x 3 ; x 1 = y 1 +2y 2 , x 2 = y 2 , x 3 = y 2 − y 3 . (DS. ϕ 1 (y 1 ,y 2 ,y 3 )=y 2 1 − 7y 2 2 + y 2 3 ) D`ung phu . o . ng ph´ap Lagrange du . a c´ac da . ng to`an phu . o . ng sau vˆe ` da . ng ch´ınh t˘a ´ c (15-19) 15. 2x 2 1 +3x 2 2 +4x 2 3 − 2x 1 x 2 +4x 1 x 3 − 3x 2 x 3 . (DS. ϕ =2(x 1 − 1 2 x 2 + x 3 ) 2 + 5 2 (x 2 − 3 5 x 3 ) 2 + 11 10 x 2 3 ⇒ ϕ 1 (y 1 ,y 2 ,y 3 )=2y 2 1 + 5 2 y 2 2 + 11 10 y 2 3 ) 6.1. Da . ng to`an phu . o . ng 259 16. ϕ(x 1 ,x 2 ,x 3 )=9x 2 1 +6x 2 2 +6x 2 3 − 6x 1 x 2 − 6x 1 x 3 +12x 2 x 3 . (DS. ϕ( ·)=(3x 1 − x 2 − x 3 ) 2 +5(x 2 + x 3 ) 2 ⇒ ϕ(y 1 ,y 2 ,y 3 )=y 2 1 +5y 2 2 +0· y 2 3 ) 17. x 2 1 +5x 2 2 − 4x 2 3 +2x 1 x 2 − 4x 1 x 3 . (D S. ϕ =(x 1 + x 2 − 2x 3 ) 2 +4x 2 2 −8x 2 3 ⇒ ϕ(y 1 ,y 2 ,y 3 )=y 2 1 +4y 2 2 −8y 2 3 ) 18. x 1 x 2 +2x 1 x 3 +4x 2 x 3 . (Chı ’ dˆa ˜ nv`aD´ap sˆo ´ :D`ung ph´ep biˆe ´ ndˆo ’ i phu .          x 1 = y 1 x 2 = y 1 + y 2 x 3 = y 3 v`a thu du . o . . cda . ng c´o ch´u . a b`ınh phu . o . ng ϕ(·)=(y 1 + 1 2 y 2 +3y 3 ) 2 − 1 4 (y 2 − 2y 3 ) 2 − 8y 2 3 ⇒ ϕ 1 = z 2 1 − 1 4 z 2 2 − 8z 2 3 ) 19. 4x 1 x 2 − 5x 2 x 3 (Chı ’ dˆa ˜ nv`aD´ap sˆo ´ :D`ung ph´ep biˆe ´ ndˆo ’ i phu .            x 1 = 1 2 (y 1 − y 2 ) x 2 = 1 2 (y 1 + y 2 ) x 3 = y 3 v`a thu du . o . . c ϕ(·)=  y 1 − 5 4 y 3  2 −  y 2 + 5 4 y 3  2 = z 2 1 − z 2 2 +0· z 2 3 ) D`ung phu . o . ng ph´ap Jacobi d ˆe ’ du . a c´ac da . ng to`an phu . o . ng vˆe ` da . ng ch´ınh t˘a ´ c (20-25) 20. 3x 2 1 +4x 1 x 2 − 2x 1 x 3 +2x 2 2 − 2x 2 x 3 +6x 2 3 . 260 Chu . o . ng 6. Da . ng to`an phu . o . ng v`a ´u . ng du . ng (DS. ϕ( ·)=3y 2 1 + 2 3 y 2 2 + 11 2 y 2 3 nh`o . ph´ep biˆe ´ nd ˆo ’ i          x 1 = y 1 − 2 3 y 2 , x 2 = y 2 + 1 2 y 3 , x 3 = y 3 .) 21. 5x 2 1 −2x 1 x 2 +4x 1 x 3 +5x 2 2 +4x 2 x 3 +3x 2 3 . (D S. ϕ(·)=5y 2 1 + 24 5 y 2 2 + y 2 3 nh`o . ph´ep biˆe ´ nd ˆo ’ i          x 1 = y 1 + 1 5 y 2 − 1 2 y 3 , x 2 = y 2 − 1 2 y 3 , x 3 = y 3 .) 22. x 2 1 +5x 2 2 +2x 3 2 +4x 1 x 2 +2x 1 x 3 +4x 2 x 3 . (DS. ϕ( ·)=y 2 1 + y 2 2 + y 2 3 nh`o . ph´ep biˆe ´ ndˆo ’ i:          x 1 = y 1 −2y 2 − y 3 , x 2 = y 2 , x 3 = y 3 .) 23. 5x 2 1 + x 2 2 +3x 2 3 +4x 1 x 2 − 2x 1 x 3 − 2x 2 x 3 . (DS. ϕ( ·)= 1 5 y 2 1 +5y 2 2 + y 2 3 nh`o . ph´ep biˆe ´ nd ˆo ’ i          x 1 = y 1 − 2 5 y 2 − y 3 , x 2 = y 2 +3y 3 , x 3 = y 3 .) 24. x 2 1 + x 2 2 +5x 2 3 −4x 1 x 2 − 2x 1 x 3 +4x 2 x 3 . [...]...  x1 = √ y1 + y2   14 14 2 2 (DS ⇒ ϕ(·) = 12 y1 − 2y2 )  3 5  y1 − √ y2  x2 = 14 14 √ 30 2x2 − 4 5x1x2 + 3x2 1 2 √  2 5  y2  x1 = y1 + 2 2 3√ 3 (DS ⇒ ϕ(·) = 7y1 − 2y2 ) 2  5 y1 + y2 x2 = − 3 3 31 ϕ(x1, x2) = 4x1 x2  1 1  x1 = √ y1 − √ y2 2 2 2 2 (DS 1 1  ⇒ ϕ(y1, y2 ) = 2y1 − 2y2 ) x2 = √ y1 + √ y2  2 2 2 2 32 3x1 + 6x1x2 + 3x2  1 1 x1 = √ y1 − √ y2,  2 2 2 (DS ⇒ ϕ(·) = 6y1 ) 1 1 ... t˘c d´ a o 26 2x2 − 4x1x2 + 5x2 1 2  2 1 x1 = √ y1 + √ y2   5 5 2 2 (DS 1 2  ⇒ ϕ(·) = y1 + 6y2 ) x2 = − √ y1 + √ y2 5 5 27 x2 + x2 + 4x1 x2 1 2  1 1 x1 = √ y1 + √ y2 ,  2 2 2 2 (DS ⇒ ϕ(·) = 3y1 − y2 ) 1 1  x2 = √ y1 − √ y2  2 2 28 5x2 + 12 x1 x2 1  3 2 x1 = √ y1 − √ y2,  13 13 2 2 (DS ⇒ ϕ(·) = 9y1 − 4y2 ) 2 3  x2 = √ y1 + √ y2  13 13 √ 29 7x2 + 3x2 + 6 5x1x2 1 2 2 61 262 Chu.o.ng 6 Dang... y1 + √ y2  2 2 2 2 2 33 6x1 + 5x2 + 7x3 − 4x1 x2 + 4x1 x3  2 1 2  x1 = y1 − y2 + y3,    3 3 3   1 2 2 2 2 2 (DS x2 = − y1 + y2 + y3 , ⇒ ϕ(·) = 9y1 + 6y2 + 3y3 ) 3 3 3     2 2 1  x3 = y1 + y2 − y3  3 3 3 √ 34 2x2 + x2 + 3x2 − 4 2x2x3 1 2 3 1 (DS x1 = y1, x2 = √ y2 + 3 2 2 2 ϕ(·) = 2y1 + 5y2 − y3 ) 2 y3, x3 = − 3 1 2 y2 + √ y3 ; 3 3 35 2x2 + 5x2 + 2x2 − 4x1 x2 − 2x1 x3 + 4x2x3 1 2 3 1. .. c´ 1 − ξ2 = 0, − 1 + ξ2 = 0 ⇒ 1 = ξ2 v` o a o ´.ng ch´ u.ng v´.i λ2 = 0 x´c dinh bo.i vecto riˆng ’ a ınh ´ o e hu o v(β, β), β∈R ’ v` chuˆn h´a ta du.o.c a a o 1 1 E2 = √ , √ 2 2 ’ ’ ´ ’ ’ a a a e u e Nhu vˆy ta d˜ chuyˆn t` co so e1, e2 dˆn co so tru.c chuˆn E1 , E2, o trong d´ 1 E1 = √ e1 − 2 1 E2 = √ e1 + 2 ’ ’ bo.i ma trˆn chuyˆn a e 1 √ e2, 2 1 √ e2 2  1 1 √ √  2 2   T = 1  1 √... ) 1 + 0 · ξ2 + 0 · ξ3 = 0, 0 · 1 + (20 − λi )ξ2 − 20ξ3 = 0, 0 · 1 − 20ξ2 + (20 − λi )ξ3 = 0 v´.i 1 = 9, λ2 = 40, λ3 = 0 o o a) V´.i 1 = 9 ta c´ o 0 · 1 + 0 · ξ2 + 0 · ξ3 = 0, 0 · 1 + 11 ξ2 − 20ξ3 = 0, 0 · 1 − 20ξ2 + 11 ξ3 = 0 a e ´ o T` d´ thu du.o.c vecto riˆng u.ng v´.i 1 = 9 l` u o u(α, 0, 0), α ∈ R, α = 0 ’ v` sau khi chuˆn h´a ta du.o.c a a o E1 = (1, 0, 0) o b) V´.i λ2 = 40 ta c´ o 31 1. .. phu e a a (17 − λi ) 1 + 6ξ2 = 0, 6 1 + (8 − λi )ξ2 = 0 ` a lˆn lu.o.t v´.i 1 = 20 v` λ2 = 5 a o i 1 = 20 ta c´ V´ o o −3 1 + 6ξ2 = 0 6 1 − 12 ξ2 = 0 ⇒ 1 = 2ξ2 e ´ o o o Do d´ vecto riˆng u.ng v´.i 1 = 20 c´ dang u(2α, α), α∈R ’ v` sau khi chuˆn h´a ta du.o.c a a o 1 2 E1 = √ , √ 5 5 265 a a´ Chu.o.ng 6 Dang to`n phu.o.ng v` u.ng dung 266 V´.i λ2 = 5 ta c´ o o 12 1 + 6ξ2 = 0, 6 1 + 3ξ2 = 0... 1 ( 2)2 11 x2 + 2y 2 − 3z 2 + 2x + 8y + 18 z − 54 = 0 x2 y2 Z 2 + − = 1) 36 18 12 12 2x2 + y 2 − 4xy − 4yz = 0 ` a (DS Hypecboloid 1- tˆng; x2 y2 2 +z = ) a o (DS M˘t n´n, 4 2 13 2x2 + 2y 2 + 3z 2 + 4xy + 2xz + 2yz − 4x + 6y − 2z + 3 = 0 √ a o (DS M˘t parabˆloid eliptic, 2x 2 + 5y 2 − 5 2z = 0) 14 2x2 + 2y 2 + 3z 2 − 2xz − 2yz − 16 = 0 x2 y2 z2 a + 2 = 1) (DS M˘t elipxoid, 2 + √ 4 2 (2 2)2 = 1) ... ınh o a ’ o a a11x2 + 2a12xy + a22y 2 + 2a13x + 2a23y + a33 = 0 (6.20) ’ ` ´ ’ a e Tˆng cua ba sˆ hang dˆu tiˆn o o ϕ(x, y) = a11x2 + 2a12xy + a22y 2 (6. 21) ´ ’ a e a a a l` dang to`n phu.o.ng cua c´c biˆn x v` y v` du.o.c goi l` dang to`n a a a o.ng u.ng v´.i phu.o.ng tr` (6.20) Ma trˆn cua dang to`n phu.o.ng phu ´ o ınh a ’ a n`y c´ dang a o A= a 11 a12 a12 a22 ´ ’ ı o e ı a 1+ Nˆu detA >...6 .1 Dang to`n phu.o.ng a 2 2 2 ´ ’ (DS ϕ(·) = y1 − 3y2 + 4y3 nh` ph´p biˆn dˆi o e e o  x = y + 2y + y ,  1 1 2 3   x = y2,  2    x3 = y3 ) 25 3x2 + 2x1 x2 − 2x1 x3 + x2 + 4x2 x3 1 3 1 2 2 2 ´ ’ (DS ϕ(·) = 3y1 − y2 + 17 y3 nh` ph´p biˆn dˆi o e e o 3  x = y − 1 y − 2y ,  1 2 3  1  3 y2 + 7y3 ,  x2 =   x = y3 ) 3 ´ ’ a ım e e o ... hˆ phu a (1 − λi ) 1 − ξ2 = 0, − 1 + (1 − λi )ξ2 = 0 ` a lˆn lu.o.t v´.i 1 = 2 v` λ2 = 0 a o o V´.i 1 = 2 ta c´ o − 1 − ξ2 = 0, − 1 − ξ2 = 0 ⇒ 1 = −ξ2 ’ v` do d´ hu.´.ng ch´ tu.o.ng u.ng v´.i 1 = 2 du.o.c x´c dinh bo.i vecto a o o ınh ´ o a riˆng e u = (α, −α), α∈R ’ ` ınh o a e ınh ´ a 6.2 Du.a phu.o.ng tr` tˆng qu´t vˆ dang ch´ t˘c 269 ’ v` sau khi chuˆn h´a ta c´ a a o o 1 1 E1 = √ , − . y 2 . (DS. ϕ 1 (y 1 ,y 2 ) =19 y 2 1 − 2y 2 2 − 10 y 1 y 2 ) 12 . ϕ(x 1 ,x 2 ,x 3 )=2x 2 1 +3x 2 2 − x 2 3 + x 1 x 2 ; x 1 = −y 1 +2y 2 ; x 2 =3y 1 + y 2 + y 3 ; x 3 = −2y 1 − y 2 . (D S. ϕ 1 (y 1 ,y 2 ,y 3 )=22y 2 1 +12 y 2 2 +3y 2 3 +11 y 1 y 2 +17 y 1 y 3 +8y 2 y 3 ) 13 =2(x 1 − 1 2 x 2 + x 3 ) 2 + 5 2 (x 2 − 3 5 x 3 ) 2 + 11 10 x 2 3 ⇒ ϕ 1 (y 1 ,y 2 ,y 3 )=2y 2 1 + 5 2 y 2 2 + 11 10 y 2 3 ) 6 .1. Da . ng to`an phu . o . ng 259 16 . ϕ(x 1 ,x 2 ,x 3 )=9x 2 1 +6x 2 2 +6x 2 3 −. thu . `o . ng 1)  x 1 x 2 x 3     30 1 0 − 21 11 2       x 1 x 2 x 3    (D S. 3x 2 1 − 2x 1 x 3 − 2x 2 2 +2x 2 x 3 +2x 2 3 ) 2)  x 1 x 2 x 3     41 0 13 1 0 1 1       x 1 x 2 x 3    . (D S.