MATEmática                    Matrize s, dete rm inantes e sis tema s lineare s                                           ...
30. a      I.xxx12  3121 3 − 3333 33                   3 33 33333                     3 333 33 −                 × × × 12 ...
214 1 34 2 2 + = = xx ( )                    s Iyxz2 0 + =      s ()xzy2 2 2 = −            s xzy 2 2 4 ⋅ = ( I           ...
10. d                      11. 3 1 2 2 0 3 1 3 7 2 3 2 3 1 0 7 2 3 xyzxyz− + + −   =   − + +1 2                           ...
3x– 1 = 2 s x= 3y+ 2 = –3      s y= –53z+ 1 = 7 s z= 2Logo: xy = 3 · (–5) · 2 = –30        z12. xpqxyr z x y−++ −− + + +++...
20. cABA⋅= − ====⋅⋅            ==   BA       = − ====⋅=                                  =010000010100sBBO ≠BABA⋅=AA 0 B 0...
PQRSxy ⋅=      z        ⋅⋅ Q S y Q S y Rx⋅ s1111111111111t                   R x ⋅⋅ R x =QS z                             ...
= 112sen2sen )   ( ( co ( x x⋅⋅⋅            ( cos ) ) s ) x x                  ⋅ 1 2 =                                    ...
AA⋅= − − ==== =            ⋅−             =    −A         >           == + t10101210011210++ +    − + − + + +++++ −− === =...
36.                                         −n        Annnn= − − − − − − − − =111111111111  ()( (                    ...
32.  Trata  se            -    de    um    deter minan e                                        t    de                   ...
13. (–1) · (–1) + 2 · 131123142−− + (1)· (–1) + 4 ·               4                             4     123112114−−−− == –(–...
xxx ( ⋅−⋅−−−= + 312277077000 ()29. dAplicandoLapl e             ac na segundacoluna,temos:                                ...
S = {–3 < a < 5}9
Capítulo 3Complementos de                                               Assim:M= −− === = 91111                           ...
2. a                                             detB= – 6 (triangul r                                                    ...
12. b                                                     s x + 8 – 3 – 6x– 2x+ 2 = 0 s                                   ...
mapa a a ⋅= − ⋅+ ⋅=cos()s ( s n )                                                                  m p ⋅+28. d            ...
s     loglogde 33132113132             t(          ⋅⋅         2 1 1 ⋅⋅ 2 1 1                                     1 3 =    ...
b)A2 + 2AB – B = 0 s B = A2 + 2AB s B = A(A+ 2B)ss det(B) det[A(A+ 2B)]s        =s det(B) det(A) det(A+ 2B)        =      ...
∴ S = {(sen(a);cos(a))}       14. a)SPD (sistemaescalonadodo primeirotipo)a bbb + = = = a 5 2 12 6 s               e      ...
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2º mat emática

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2º mat emática

  1. 1. MATEmática Matrize s, dete rm inantes e sis tema s lineare s 1Capítulo 1MatrizesConexões 56786666478886421234 ⋅⋅ 6 8 6 6 7 8 6 + + + + + + + + + + 7 6 6 4 8 8 = +512213261218244142432812112870607440+ = îFranciel foi aprovada e emmat m t a,geografia biologia,masnão emhistória. e á ic eExercícios complementares13. A 28a linhacomeçarácom 28, na primeiracoluna. Assim, falta 16 elementosnessalinha: 28 + m 16 = 4414. At= –A xy x z 10330203130−−0 000 z y 2 0 00 ===== ===== = = = 000 −−−− − = = ==I.x= –x s 2x= 0 s x= 0I 2 = –y s y= –2 I.II I –1 = –z s z= 1 .∴ x+ y+ z= 0 – 2 + 1 = –115. O primeiromembrodaequaçãoé a somadoscemprimeirostermosdeumaPA cujoprimeirotermo é iguala X e cujarazãoé iguala X.SXXXX 1001002101505050= + ⋅= ⋅=(). 100Então:5.050 · X = 5050005050..l X X = 1001.0 · a.s 5016. a)A vendeu16 automóveis tipo1 emmarço(a ). do 13B vendeu 20 automóv eisdo tipo1 emmarço(b ).13a + b13 = 16 + 20 = 36 13Portanto,A e B vender m,junta 36 automóv a s, eisdo tipo1 emmarço.b)ConcessionáriaA ConcessionáriaBMê 1: 12 + 15 = 27 M s 20 + 16 = 36 s ê 1:Mê 2: 8 + 12 = 20 Mê 2: 16 + 10 = 26 s sMê 3: 16 + 24 = 40 M s 20 + 10 = 30 s ê 3:Mê 4: 20 + 36 = 56 M s 24 + 26 = 50 s ê 4:Portanto,a concessionáriaA ultra a sou concessionáriaB no volum devendas p s a e (considerando- sesoment os automóveis e do tipo1 e do tipo2)nosmese demarçoe abril. s c)AB+ = + + + + + + + + 1 220816162020241516121024103626)n o d)C=+=++ + + 1281620241512243612D= C = 20162024241610102626 = + + 29. d (A+ B) = (A+ B) · (A+ B) s (A+ B) = A2 + AB + BA + B2 2 2 Se o produtodeA porB é comuta vo,(A· B = B · A), podemos ti escrev r: e (A+ B) = A2 + AB + AB + B2 s 2 s (A+ B) = A2 + 2AB + B2 2
  2. 2. 30. a I.xxx12 3121 3 − 3333 33 3 33 33333 3 333 33 − × × × 12 ,11,,10100 x3 0 23 1× ,,1 I Am× n · Bn × p – C m× p = 0m× I. p Observandoa ordemdasmatri e a únicapossibilidade z s, é:ABCDD233121212210× × × × ⋅−= × s×××− = 121210C d⋅−0 0 1 2 0 −−o d m =00 1 x rm 1 0 1 0 x1010011200xx oe r e 11 0 s 1200+ −++++++++++++++======= x s 1200+ −− +++++++======xxx −− xx = s s 1 + x– 2 = 0 s x= 1 2222 b⋅⋅ 31. Aabbabbabababb 00=2b b 0 2 b + =2b b aa = a = ab aab A2 = A s s abababbabb 0+2a a = b ds ababb 00+ = = = bb= Is 222 b bb be bb r 222 ab ) ( Substituindoem( )temos: I, a = a s a – a = 0 s a(a– 1)= 0 s 2 2 s a = 0 (Nãoconvém.)ou a – 1 = 0 s a = 1 Logo: a = 1 e b = 0 32. e A · At= I 12121001xy y z =0u⋅ 2 2 0 1 y 11 s s ⋅ z x b oa 1 1 0 x =22 0 0 s 14221001 + + + + 10x yb= 0 u yx y z z 222 01 zz yx ox z x y
  3. 3. 214 1 34 2 2 + = = xx ( ) s Iyxz2 0 + = s ()xzy2 2 2 = − s xzy 2 2 4 ⋅ = ( I 2 I)y + z = 1( I) 2 2 IISubstituindo( ) Iem( I : I)34 4 3 2 2 2 2 ⋅ = = zyzys ( V I )Substituindo (IV) e m (III): a = (–1) + 1 = 1; a = (–1) + 2 = –1; a = 0 31 3 32 3 33y y 2 3 1 + = s 43 1 2 2 Logo: A = − − − − 011101110 y= s y 34 = 2 4. b Sendom , nije p ele en s ij ij m to deM, N e P, resp ti a e t temos: ec v m n e,∴ x + y = 34 34 64 32 23 32 23 32 23 11 11 11 22 22 22 M N P mn p m n p + = + = + = i s N mnp 2 2 s 32 23 7 32 23 4 13 ⋅ + ⋅ = ⋅ + ⋅ + = 72 4 yyx() s 9 4 42 9 4 62 xyyx+ = 32 x 32 + = =Tarefa proposta + = 4 ()()I I 2 I Fazendo( I ( )vem: I ) I, –1. c 5y– 5x= 20 s y– x= 4M a a a a a a = ) em I I I): 11 5. e 12 21 22 31 32 p – q11 = 2 – (–3) = 5 11 p – q12 = –1 – 1 = –2 12 p – q21 = 3 – 1 = 2 21a = 2(1 – 1) = 0 a 11 12 p – q22 = 1 – 3 = –2 22 = 2· 1+ 2= 4 Logo, a distânciaentre asmatri e p e q é 5. z sa = 2 · 2 + 1 = 5 a22 21 = 2(2– 2)= 0 6. ca = 2 · 3 + 1 = 7 a32 31 O país1 exportou1,2 + 3,1 = 4,3 bilhões. = 2· 3+ 2= 8 O país2 exportou2,1 + 2,5 = 4,6 bilhões.Portanto: O país3 exportou0,9 + 3,2 = 4,1 bilhões.M = ao = 8 4 5 0 7 8 n: 2 0 t O país1 importou2,1 + 0,9 = 3,0 bilhões. O país2 importou1,2 + 3,2 = 4,4 bilhões. O país3 importou3,1 + 2,5 = 5,6 bilhões.2. b 7. bA aaaa= = 0 11 12 21 − − = + + + 1 22 1 4 3 4 2 2 xxxxx 22 x+ 1 = –1 s x= –2a = (–1) + 1 · 1 · 1 = 11 1 8. b 1 a = (–1) + 1 · 2 21 2 xa a a a = − − − + − = 22 22422482 · 1 = –2a = (–1) + 2 · 1 · 2 = 12 1 a – 2 = 2 s a = 4 s a = –2 ou a = 2 2 2 –2 a = (–1) + 2 · 22 2 2· 2= 4 –2a = 4 s a = –2Logo: A = − − 12 4a = –8 s a = –2 24 –2 + a = 2 s a = 4 s a = 2 ou a = –2 2 2 Logo: a = –23. A a a a a a a a a a = 9. d a a a 12 13 21 22 23 a a 11 a + a + a + … + a = (1+ 1)+ (2+ 2)+ (3+ 3)+ … + 2n = 11 22 33 nn 31 32 33 e a i ji jiji j= = 2 + 4 + 6 + … + 2n − ≠ = + ( ) 10 , A sequ n ê cia umaPA. é ⋅ ()1 2 s se, seEntão:a = 0; a = (–1) + 2 = 11 12 1 S aan n n = + –1; a = (–1) + 3 = 13 1 1a = (–1) + 1 = –1; a s S n n n = + ⋅ ()2 2 2 s S n n n n n = + ⋅ = + 2 1 2 2 () 21 2 22 = 0; a = (–1) + 3 = 23 2 –1
  4. 4. 10. d 11. 3 1 2 2 0 3 1 3 7 2 3 2 3 1 0 7 2 3 xyzxyz− + + − = − + +1 2 −a a 32 t = 23 = 2– 3= –1
  5. 5. 3x– 1 = 2 s x= 3y+ 2 = –3 s y= –53z+ 1 = 7 s z= 2Logo: xy = 3 · (–5) · 2 = –30 z12. xpqxyr z x y−++ −− + + ++++++++ += − + − − − 13262113ppyzqr −−−−−−− zz zz y z x + z z zz2621 z zzx– 1 = –x + 1 s x= 1p = –x – 3 s p = –1 – 3 s p = – 4q= yx+3 = –p s 1 + 3 = –p s p = – 42y+ 6 = –2y – 6 s y= –3 ( ) IDe ( ), I temos:q = –3r = z– 2 s + + + + + + 1057152722337 − −−−− yxx x y x −+++++ + + + + + + y y x y −− −== = = =–y = –q s – (–3) = –q s q= −−− − −−−−−−121031522722321 − − −−−−−− − –3 573371xy y+= − + = === x–z + 2 = –r Resolv endoo siste a, m temos:z+ 1 = –z – 1 s z= –1 ( I I) x= –2 e y= 1De ( I temos:r = –1 – 2 = I) , w = |x · y| = | –2 · 1| = | –2 | = 2 –3 15. eLogo: p = – 4; q = –3 e r = A loja L1 vendeu 30 unidades do produto P1 e 15 unidades do –3 produtoP2, logoa somadasquantidades dosprodutosdostiposP1 e P2 vendidospelaloja 1 é 45. L13. 23217034212x x z zy y yz 16. b ⋅− . xy x − + + − + z y+ +31 31 27 42 02 =+7734322101 zx y x z −−++ + + +++++ + + y z yz + ++ PQ − = − ==== == =− . d = 4 23232544123664108888= o −1 == 2 ====− 88 883x– y= 7 ( ) I2z+ 1 = 3z– 4 s z= 5x+ 2y= 0 ( I I) = − − − − − − ==== ========= = = −−− = 46142103825125()De ( ) e I ( I , vem: I) 3720xy y−=+ ==== x Logo: (P– 2Q) = −−− t 21255 ss 6214207142xyx x y x− = + = (I = )= 17. b 3A = B + CSubstituindo em ( ) I, 361243xy w w y w ww == = =++ ++++++s z x x z w =− = = w ww + temos:3 · 2 – y= 7 s y= –1 s 333346123xyzwx y wwww + + + − + ++++++ x z ww w=+ wEntão: A=to y 37117401106 ã = : 3x= x+ 4 s 2x= 4 s x= 2Logo, o traçodamatri A z 3y= 2 + y+ 6 s 2y= 8 s y= 4 é: 3 + 4 + 6 = 13 3w = 2w + 3 s w = 314. BAA=− ⋅t32 s 3z= z+ 3 – 1 s 2z= 2 s z= 1 ∴ x+ y+ z+ w = 2 + 4 + 1 + 3 = 10s yx y y x y−+ x x y x 18. b + + 4 6 110571527223 + =3m A3 × 4_ Bp × q= C3 × 5; logo: p = 4 e q = 5 37 = −−−−============ = = −−−− −−−1106412323216214 − 19. 0⋅−9 3012432 121 23ssyxxy y x y−+ x y x 102321512310502100−0 0 . 00 0 g : o p = = + − − + + 1154136031203+ + + 14 322 2 23 1 121 10571527223 371122 4= − + + +++++ −=== = += === −−−−============ = = −−−− −−110641232329332 6920323323017 292552 33172s
  6. 6. 20. cABA⋅= − ====⋅⋅ == BA = − ====⋅= =010000010100sBBO ≠BABA⋅=AA 0 B 0 0 B 0000101000000s== O 0⋅−A A 0 =AA 0 B0 0⋅ =⋅−A2010001000000= −=== == = 1 1 = =1 0 0 0 0 1BBO 22000100010001= ⋅⋅ 0 1 0 0 0 =======≠ sABA+ = − ==== =+++ + += −=== = = ++ = =+010000010101sBBO ≠21. eA4 × 7 · B7 × 9 = C 4 × 9 (quatrolinhase novecolunas)c63 é o ele e to m n dalinha6 e coluna3, logoc63 não existe.22. bABAmnABABAn1212112× × × ⋅ ⋅⋅× × ()[()ln ] i ×Amn1221 se2121211121 1⋅⋅ 1 1 1 1 2 2 ⋅−1 1 = − ==== s − 11 2 2 bbs 221211121 b⋅−1 1 = bb−bb 2 1 − ==== s bs 4222111211121bbbb−−−− 2 − − 2 s 1= 1() 1 1s 4b11 – 2b21 = 2 s 2b11 – b21 = 1 ss b21 = –1 + 2b1123. a(A· B) · C = ()ABnD342 × × ⋅× · C m× 32 2n= 4Então:D3 × 2 · C m× 2Logo: m= 2∴ m= 2 e n = 424. c
  7. 7. PQRSxy ⋅= z ⋅⋅ Q S y Q S y Rx⋅ s1111111111111t R x ⋅⋅ R x =QS z y ssxxyy z t+++1t = + + + t z 1 1121211I.x+ 1 = 2 s x= 1I z+ 1 = 2 s z= 1 I.I I + 1 = t+ 1 s 1 · y+ 1 = t+ 1 s y= t I xy .IV.y+ 1 = z+ ts y+ 1 = 1 + ts y= t25. bI.(V)A3 × 2 · B2 × 1 = C 3 × 1I (F)A5 × 4 · B5 × 2 (Nãoexist o produto.) I . eI I A2 × 3 · B3 × 2 = C 2 × 2 I (V) . 26. a)A2 = A · A = 1 1⋅−−A = 123012111123012111−−1 111 111 1 · A = e == = 103223343002012022101211+ − + + + − + − + + + − − + + − + − − 3321+ ++++++ + + + + += = −−− 274230020 1 1⋅−− b)A · At= 123012111101211321−−1 111 111 1 A · A e = = = 1490261230260140121230121++ + + − + − + + + + + − − + − + − + + + ++++++ + +=11 ++ = 1482851213−−−−33 33 3 33 3 33 c)2A + 3At= = 246024222303633963−−3 333 3 33 ++++++ + + + 333 + + + +== −− = = = 543657785 ==== = = = − 27. A2 × 2 · X = B2 × 2 Paraquea igualdad sej possíve a matri X deve e a l, z serdo tipo2 × 2. SejaXmnpq=ig l ud a Como A · X = B, temos: = ⋅⋅ 21031020 X B 1 3 0 01mnpqs 22331020mpnqpq++tm:=j p 0 1 =130 2 e s ao o s2122312131623mp m p+=+ = = = = psívsss m m o e202000nqnnq+= = = = m ss mp Logo: X=n + = = m n = = m160230 q p ( ) )) x x x e e t ⋅ssen( cos ) xs o 28. a)cos()() )cos( cos( ( x x x s ns nsenaiz r ) ( x oc ( = )= cos()( ( cos ) ) os 2xxxx x+ + ⋅sensensenseensensen()cos() )cos( cos( ( x x x⋅+ ⋅+)22 o = )) ((c ( ) x ⋅ 2 2 ( ) )) x x x )s c(4
  8. 8. = 112sen2sen ) ( ( co ( x x⋅⋅⋅ ( cos ) ) s ) x x ⋅ 1 2 = 1 s= 1221sensen ) ) x11 ( ( x= 2 2b)A(x) A(x) A(x) · = s 32. cs 1221sens ens s n ) ) s ) ) en e ( ( co ( ( AB⋅= − − − − ==== = ⋅−−. ()cox x xs e x x ns e =ss( x s( n ) o) () = y 1 1 3 211011111s y 1 1 z• cos(x) 1 s x= 0 ou x= 2π ( ) = I• sen(2x) sen(x) = ss 2sen(x) cos(x) sen(x) · = ss 2sen(x) cos(x) sen(x) 0 s · – = AB⋅= − ⋅−+ ⋅−+ − ⋅⋅−+ − ⋅−+2111110111()() ) ) ) ) −s sen(x) [2cos(x) 1] = 0 · –i) sen(x) 0 s x= 0 ou x= π ou x= 2π ( I = I) s ( ( (((ii)2cos(x – 1 = 0 s )s cos(x) 12 s = − ⋅⋅ ⋅ 1 1 11)ss x= π3 ou x= 53π (III)S = ( ) [( I ( I ) {0; 2π} I% I ) I I= 5 ] 0 129. bAB⋅= − ====⋅−B s AB⋅= − − + − ++++++211011s == . ) )= + + + ++++3014211060302410– 1 = 06361 = =BA⋅= − ====⋅−A + 3 1 + − − + == + 0 = s AB⋅=B 1 00 ) 11 +++211030146104300000 −− 0 ====== = 7430 0 == 33. bABBA− ==== = −−−=== = 63617430= == == − 1791 XABXC− = + + 23 s s 3(X– A)= 2(B + X)+ 6C s 3X – 3A = 2B + 2X + 6C y⋅⋅ s s X = 3A + 2B + 6C s30. 1123411xy−yy 0 . yy y 3 = − ======s 0 sX= −======+++++++++ −++++++63932420246126 s − sX= − + + − + + − + + ++++++62243469212306ss 3413811+− +++++++= −====== x s () y sX== 3 + A281233s 3441421238139+ − = = = + = − = − = − xxxyyy 34. c A=3 +2 A 8 2081512B=66 = C 32Xxy=1 C “ 33ss3 ssss 01 A · B = Xs 8⋅ ⋅31. b s 208151232 21 05 0 1 1 85xy 8 5 =011 2 s 22⋅⋅2130112−2 2 22 . 3 1 2 = +. 31 z xs 0 4 1 0 xy y z 4 ⋅ 20382153122 + ⋅⋅+ ⋅⋅ ⋅ 3 2 =⋅ 32xy 8 1 81 s5s 232xyz z x − +++++++= +++++++s y y z++ s 76692x =0 5xy 1y 8 1 5 12s 232xyz y x y z z++= − + = + +++++ 35. dx= –y–2y + y+ 3z= 2y3z= 3ys z= yxy z x++= y z − ++ − = − + − = − yyy y 1111 y y
  9. 9. AA⋅= − − ==== = ⋅− = −A > == + t10101210011210++ + − + − + + +++++ −− === = 10020020142225 += == (A· At–3I)· X = B 2225300312−−2 2 −2 222 2 2 2222 22 2222 2 ⋅⋅ 3 ) · =3xy· X I I s s −− −−− − − − − ⋅⋅ I y = I x · xy· 122212xy s −−− + + + + + + + = = = = = = = xy y22212s x s −− = − += = = = xy y21222+ x s –3x = 3 s x= –1 e y= 0 Logo: x+ y= –1 + 0 = –15
  10. 10. 36. −n Annnn= − − − − − − − − =111111111111  ()( ( )) Bpppp nnnn 123123123123 =n 11122223333 1332Û 11 313 ()(( )) Û 11  3 1  3a)Paran par: 2110010xx > 2 – x – x> 0 s x + x– 2 < 0 xs 2 2S = –1 + 1 – 1 + 1 – 1 … + 1 = 0Paran ímpar:S = –1 + 1 – 1 + 1 … – 1 = –1 1– 2b)An × n · Bn × p = C n × p S = ]–2; 1[c42 = a · b12 + a · b22 + … + a · bn2 s 41 42 4n 16. Aplicando o teoremade Laplacena quinta coluna,temos:s c42 = 1 · 21 + 1 · 22 + 1 · 23 + … 1 · 2n s 2132111023403210252321110234032 ⋅−⋅−= ⋅−+s cnn42232222= + + + + +SomadosprimeirostermosdaPGs 55s caqqn42111= −−() ss 409421212.()=−− n s ()11025s – 4.094 = 2 – 2n + 1 s Aplicando o teore a de Laplac na segunda m e coluna,temos:s 2n + 1 = 4.096 s 221123432125 ⋅⋅−⋅−= + ()s 2n + 1 = 212 s n = 11 linhas 12Capítulo 2DeterminantesConexõesA área podeserobtidapor: A = – 4 · (15– 4 + 24 + 9 – 4 – 40)= – 4 · 0 = 0A = AABC + ACDA s 29. as A = DD1222+ O segundo deter minan e a combinaçãolinear t éD1 = 011201131−− e D2 131111011− do primeiro (L = –L 1 + L2). Portan 2 to, os determinan e sãoiguais. t s∴ A = 9262+ = 7,5 u.a. 30. Trata- edeumdeterminante Vandermonde. s de Suaresoluçãoé dadapor: (3– 2)· (– 4 – 2)· (– 4 – 3)= 1 · (– 6)· (–7) = 42 31. Trata- edeumdeterminante Vandermonde. s de Então: (k – k)· (k – k)· (k – k ) · (1 – k)· (1 – k ) · (1 2 –1 –1 2 2Exercícios complementares – k )= 0 –1 I.k – k = 0 213. d k(k– 1)= 0 s k = 0 ou k = 1I.ad– bc = 0 s ad= bcI abdcadbc001022=+ I. I k – k = 0 s 101 kkkk−= − I –1 . 2 s = 0 s k2 = 1 s k = ±1 I I –1 – k = 0 s 101 kkkk−= − Ik. 2 23 s= 0 sk = 1 sk= 1 3Substituindo I em I , concluímos que o I IV.1 – k = 0 s k = 1 deter minan e t vale:2bc + bc = 3bc V. 1 – k = 0 s k = 1 s k = ±1 2 2 VI.1 – k = 0 s 1 –101kkk=− s= 0 s k = 1 –1 14. a De I,I I I V e VI,temos: I I IV, , ,00100cos( s n ) e ( cos ) x y= ) e (s n) (x y S = {–1; 0; 1} = cos(x) cos(y – sen(x) sen(y = · ) · ) 6= + = = cos()cosxyπ312 15. a
  11. 11. 32. Trata se - de um deter minan e t de s c = –3 ou c = 5 Vander monde. 10. aTemos,então: xxx1110010101010−−− = s 1 – x = 0 s x= ±1 2(x– 2)· (x– 3 – 2)· (x– 3 – x)· (1– 2)· (1– x)· (1– x+ 3)< 0 ss (x– 2)· (x– 5)· (–3)· (–1) · (1– x)· (4– x)< 0 s 11. cs (x– 2)· (x– 5)· (1– x)· (4– x)· 3 < 0 s 2x· log x · 3 – 8x· log x· 3 = 0 s 2 2 2s (x– 2)· (x– 5)· (1– x)· (4– x)< 0 s 2x· 2 · log x– 8x· log x= 0 s 2 2Tratase de uma inequação- roduto, - p cuja s 2x+ 1 · log x– 23x· log x= 0 s 2 2 resoluçãoé dadapor: s log x· ()22 2 13xx+− = 0 sI.x– 2 = 0 s x= 2 s log x= 0 s x= 1 ou ()22 2 13xx+− = 0 sI x– 5 = 0 s x= 5 I. s x+ 1 – 3x= 0 s x= 12I I – x= 0 s x= 1 I1. 11232+ =IV.4 – x= 0 s x= 4 12. b xxx x x − + =11311113110s x x x+ ––+ + + – – – + – + – – + + + – – – – + + – – + 1 1 224455IIIIIIIVI _ II _ III _ IVS = {x3 ® | 1 < x< 2 ou 4 < x< 5} s (x – 1)· x+ 3x+ x– x – 3(x– 1)– (x+ 1)= 0 s 2 3 s x – x+ 3x+ x– x – 3x+ 3 – x– 1 = 0 s 3 3Tarefa proposta s x= 21. e 7xaax011011=x+ ax– ax = 1 s 2s ax – (a+ 1)x+ 1 = 0 2(a+ 1) – 4a = 0 s 2s a2 + 2a + 1 – 4a = 0 s a – 2a + 1 = 0 s 2s (a– 1) = 0 s a = 1 22. a)2 · 5 – 4 · 1 = 10 – 4 = 6b)5 · 0 – 3(–1) = 0 – (–3) = 33. a)sen 20° – (–cos 20°)= 2 2= sen 20° + cos 20° = 1 2 2b)sen75° · cos75° + sen75° · cos75° == 2sen75° · cos75° = sen(2· 75°) == sen150° = 124. eMaaaa=Ma aa a= 11122122s M=− − 11 22 1142detM=−1142 = 1 · 2 – 4 · (–1) = 2 + 4 = 65. aa – b + (–a2 + b2)= a – b2 – a + b2 = 0 2 2 2 2 6. d Aaaaa=2 ° = 2 11122122s A=12 2 22 1 2345 detA=2345 = 2 · 5 – 4 · 3 s detA= 10 – 12 s detA= –2 7. 121142101121410−− = = 4– 4+ 0+ 4+ 0– 2= 4– 2= 2 8. 124221511122524xxxx=s s 2 + 4x+ 20x– 8 – 10 – 2x = 24 s 2 s –2x2 + 24x– 16 – 24 = 0 s s 2x – 24x+ 40 = 0 s 2 s x – 12x+ 20 = 0 s 2 s x= 2 ou x= 10 S = {2; 10} 9. d111191311119ccc = 0 – 2 0 = = 27 + c + c – 9 – c2 – 3 = 0 s s c2 – 2c – 15 = 0 s
  12. 12. 13. (–1) · (–1) + 2 · 131123142−− + (1)· (–1) + 4 · 4 4 123112114−−−− == –(– 4 + 9 – 4 + 2 + 6 – 12) + (4 + 3 + 4 – 3 – 8 – 2) 22. e = bc – (b – 4ac)= bc s –(b2 – 4ac)= 0 s 2= –(–3) + (–2) = 3 – 2 = 1 s b2 – 4ac= 0 (Δ = 0)14. xxxx x x 4444440= s x + 4x + 16x– 4x x x x 3 2 2 Portanto, o gráfico da função tangencia o eixo Ox. – 4x – 4x = 0 s 2 2 23. a ab x2222 xxx 22− = ++++++ −−++ + += −− eeee + ++s x – 8x + 16x= 0 s x(x – 8x+ 16)= 0 s 3 2 2s x= 0 ou x – 8x+ 16 = 0 s Soma= 8 2 = + + −− + = −−eeeeee 2022022424 x xxxA somadasraízes 0 + 8 = 8. é = + + − + − = −−eeee 224xxxx 222215. 11213130xx= x + 6 – 13 – 3x= x – 3x– 2 2 = = 441 7 24. b ⋅⋅ene 2232 s3023xx=11213130302xx s x – 3x– 7 = 3xs x= 2 dxx= ns x – 6x– 7 = 0 s x= 7 ou x= –1 2 ssdxx= ⋅⋅ene 232216. b nAB⋅= − ====⋅−−. == 7 o = − ======21343122401711 u s d = 2x· 2 · 23xs d = 24x+ 1 log d = log 24x+ 2 2 1 = (4x+ 1)· log 2 = 2 = 4x+ 1 ⋅ 25. Aplicando o teorema de Laplace na última coluna,temos:det )AB= − = 40171144 ( (–1) · 2100021190047502181110− 10 Observandoa quarta coluna,paraa aplicaçãode Laplace, podemos concluir que o determinan e t val zero. e17. ax– x = 0 s x – ax= 0 s x(x– a)= 0 s x = 0 2 2 26. c ou x= a M – k· I − = − 30451001k=Paraduasraíze reaise iguais,temos:a = 0 s18. Aplicando Laplace na terceira linha, = − − 304500kk= temos: = − + − ++++ + ( 3045kk + () )detA= 3 · (–1) · 10011011a− = 3 · (–1) · 7 (1+ 1)s det – k · I = 0 s –(3 + k)· (5– k)= 0 s (M ) s (3+ k)· (5– k)= 0 s k = –3 ou k = 5s detA= –3 · 2 = – 6 27. FazendoC 1 = C 1 – C 2; C 2 = C 2 – C 3; C 3 = C 3 – C 4,19. p(x) 6 + 2x+ 2 = 2x+ 8 = temos:a)P(5)= 2 · 5 + 8 = 18 kg abbabbab a b −−− = − ⋅000000000 ()b)30 = 2x+ 8 s 2x= 22 s x= 11 anos20. a b a a 30310324330043330 xxx xxx = s8 · 3x – 4 · 3x = 0 s 4 · 3x= 0 s 3x= 0 ∅ 28. Por setratar determinante umamatriz do de de Vander monde,temos:S= (x– 2)· (3– 2)· (3– x)· (1– 2)· (1– x)· (1– 3)= 0 s s (x– 2)· (3– x)· (1– x)= 0 s21. 21341102411012−−− =nnnn s –2n + n2 – s x= 2 ou x= 3 ou x= 1 n + 3n – 4n = 12 s S = {1; 2; 3}s n2 – 4n – 12 = 0 s n = 6 ou n = –2 8S = {–2; 6}
  13. 13. xxx ( ⋅−⋅−−−= + 312277077000 ()29. dAplicandoLapl e ac na segundacoluna,temos: x− ) 11 s3 · (–1) · 13014142121205141421 + 3 4 ⋅−⋅()= ssxxxx− ) ( ⋅=3270() ( ⋅−⋅− )= (–3) · 11 + 2 · (–66)= –16530. AplicandoLaplace primeira na linha,temos:x· (–1) · xx120300216= s 2 s x=3 ou x= 2 ou x= 7 ou x= 0 Logo, o triângulo é retângulo, pois: 222 ) 732 ( = ) (+s x· 2x = 16 s 2sx = 8 s ∴ A = 2323 = ⋅ 3s x= 2Ou seja:α = 2 s α2 = 22 = 431. Por setratardo determinante umamatriz de deVandermonde,temos: 34. d(5– 7)· (x– 7)· (x– 5)= 0 s Por Vander monde:s (–2) · (x– 7)· (x– 5)= 0 s (log 20 – log 2) · (log 200 – log 2) · (logs (x– 7)· (x– 5)= 0 s 200 – log 20) · · (log 2.000 – log 2)( logs x= 7 ou x= 5 2.000 – log20)(log2.000 – log200)= =.0 ⋅⋅ . 0 0 ⋅⋅ . 0 0 ⋅logloglog20220022002S = {5; 7}32. bAaaaa=; 7= } 11122122 0– 0 0 0a 421= ⋅+++++ += =sen(11)s ππ 0 11 + en ⋅⋅ g o ⋅⋅ l g g o ⋅logloglog20002200020200 l gaxx 12= x ⋅−[] − = − sen(12)sen() en( = s ) 02...000g0 = 20 02 = (log10)· (log100)· (log10)· (log1.000) · (log100)· (log10)= = 1 · 2 · 1 · 3 · 2 · 1 = 12 ⋅−[] 35. Por Jacobi,vem:axx = 21 =sen(21)s ( en ) 1111111+ b11111+ a1111+ c111000b0001a000c + + + × × × – 1 – 1 – 1 = Aplicando o teore ade Lapl ena primeira m aca 40= ⋅+ + = ( =sen(22)s ππ coluna,temos: det( ⋅−⋅= + 11000000 abcabc 22 02n = en )= 11Logo: Axx=−− s( =en 10sen( s n ) ) e ( 36. p(x) (3– x)· (a– x)· (1– x)+ 4 · (3– x)= 0 =101414− = =sen()s ( s n ) e 2xxx ss en ) e ( s n( x ))=±12 s s p(x) (3– x)· [(a– x)· (1– x)+ 4] = 0 s =∴ S = −−−− 1167656665676116ππππππππ;;;;;;;;;;;;; s (3– x)= 0 s s x= 3 (únicaraizreal) ou (a– x)· (1– x)+ 4 = 0 s s a – ax– x+ x + 4 = 0 s 2 s x – (a+ 1)x+ (a+ 4)= 0 233. b Devemos ter Δ < 0, para que não exist m aFazendoC 1 = C 2 – C 1, C 2 = C 2 – C 3 e C 3 = C 3 – C 4, outrasraíze reais.Assim: s temos: Δ = (a+ 1) – 4 · (a+ 4)< 0 s 2 xxx −−−−−− =3322770227700770000 x s a + 2a + 1 – 4a – 16 < 0 s 2 Aplicando Laplace na primeira coluna, s a – 2a – 15 < 0 2 temos: 5– 3
  14. 14. S = {–3 < a < 5}9
  15. 15. Capítulo 3Complementos de Assim:M= −− === = 91111 == |M| = – 9 + 11 = 2determinantes M− = 112Conexões 30. e detA= –2 + 2 + 3 = 3a)A = 3152e exs detA= 6 – 5 = 1 sd det(B ) = det(2A) 1detB= 23 · detAs 1detB= –1 sA–1 = 2153−−3 3 33 33 24 sb)B = 5101−1 1 s detB= 5 11 11 s detB=124B–1 = 15150555 5 1 Â = 151501 = “ 31. a Aplicandoo teore am deBinet(P.8): det(A· B–1 )= det(4A) det –1 )( ) P.8 · (B Ic)C = −− 2346s detC= –12 + 12 = 0 Como A é uma matri de ordem n, pode- e z s colocaro 4 emevid nê cia (P.4).Não exist C –1 (a matri C não é invertív l e z e ), n vez s e emdet(4A),logo: poisdetC= 0.Exercícios complementares det(4 d t fatores nn= ) e d AA ⋅⋅⋅⋅⋅= ⋅44444 eet 413. eB = k· A Aa = ⋅4det t = detB= det( · A)s (B) ks k3 · detA= 96 s k3 · 1,5 = 96 s n, sss kkk 9615644== = 3314. detA= det(A) t eAaaaa=A a = a= aa aa a== a 111221224278 det(d tBBb= = e − 111) Substituindoessesvalore em( )temos: s I, det(4414 ABabab ⋅= ⋅⋅= ⋅− )|A| = |A t = 32 – 14 = 18 |15. e 1 nndet (3A· 2B)= det (3A) det(2B) 32 · detA· 22 · = · detB== 9 · 2 · 4 · 3 = 21616. d 32. c Pabcd=1 d M · P = I s 4o in e 2det(2A· At)= 4k s det(2A) det t)= 4k s · (A 1301711001de · P ⋅ ⋅s 23 · detA· detA= 4k s 8 · d · d = 4k s 2d2 = kdet (3B) 162 s 3d · detB= 162 s 3d · 2 = 162 s = s oM = 3 1 1 07abcd 0 7 =31 1 1 ss 3d = 81 s 3d = 34 s d = 4∴ 2 · 42 = k s k = 32 abacbd33771001++ · P I e () = e= I s , I.aaaccc3137037037== + = + = = − ===== =sss = I bbbdd300711== + = = = I ss I. esLogo: k + d = 3629. bM · A – 2B = 0 s M · A = 2B s Logo: P= −=== = = = 30371 = === ⋅⋅ A somados ele en sda diagonalprincipa é m to l abcdA 2 =B b d =bd =3412161422 c A cA 3 + 1 = 4. Tarefa propostaI.316421431627abab a ab b+= + = t2 = + = 3 (+ s 1. c SejadetA= d. Dividindo umafileiraporx(x≠ 0)∴ a = 9 eb = –11 e multiplicandoumafileirapory(y≠ 0), temos: det=⋅dxy dyx s dxy s ⋅I 324223221cdcdcdcd+= + = +=t = + = += I. + s∴ c = 1 ed = –1 10
  16. 16. 2. a detB= – 6 (triangul r a inferior)Aaaaaa a a a a 11121314212223243132333441 a a a a a= det · B) = detA· detB= 36 (A 4224344aa24222 44124 3423comoa = 2i – j,vem: ij 8. cAA=− detA= 2sen (x) 2cos (x) 2 + 2 s −−−− − − − − − − − − − − −=10123210543276541edet s detA= 2[sen (x) 2cos (x)] 2 · 1 = 2 2 + 2 = 0012321054327654−− t 1414 ⋅⋅ 2 d t 5e⋅=AA)Por Jacobi,vem:1– 1– 230532170246540– 104– 283– detde ( 525 5 e =2d ( t 41 120246 6= + + × × 1 – 2 5–Aplicando o teore a de Laplac na m e = ⋅⋅⋅= 1165detdetAAx primeiralinha:detA= (–1) · A13 s x s detA = (–1) · (–1) · 4 42284412660−− −= e P.5 = ⋅= ⋅= = 11611622 ()detde AAdetA= detA = 0 t 55 t3. aAabc = = b= d 23426 det t)= 2 · detB t a e c t 5 s (A 9. dComo det(A)= detA,temos: t I.(F)Um contra x m l e e p o:detA= 2 detB 23460=4. d I . I (V) mbncp=⋅4111 detBmanbpc=3111 aaaaaa a nnnnn1112122231122 a a 00000 ⋅⋅  nn  a⋅detAa · I I 2121211+ ( ⋅−( = − = I (V) . ) )Como detA= 2, temos:24111= ⋅ambncp detB= 2121+ ( ⋅−( ⋅detAs s 11112ambncp= s ) ) manbpc11112= −∴ detB= 31232 e B = = − ⋅−d t s detB= 1 · detA 10. c det 3)= det I s (M (82) s (detM3 = 82 · det( )s ) I 25. a s (detM3 = 64 · 1 s )PG(a;b; c; d),então: s detM=643 sb = aq; c = aq e d = aq 2 3 s detM= 4detMabcda aqaqaq== 23Por P.5: detM= 0 11. d Multiplicandoa primeiralinhapor a, a segundalinha 6. d porb e a terceiralinhaporc, temos: y x z=−⋅= − ss(P.4) 111= ⋅123691212312323412x z y abcaaabcbbabcccabcaabbcc 232323232323ss12323442341234xy x z=− = (P.3) z y 11 7. d detA= – 6 (triangul r a superior)
  17. 17. 12. b s x + 8 – 3 – 6x– 2x+ 2 = 0 s 2[detM] = 25 s 2 s x – 8x+ 7 = 0 s x= 1 ou x= 7 2s detM= ±5 ( ) I 23. cdetM= 3x+ 12 + 12 – 27 – 4x– 4 = –x – 7 ( I I) 101120154364251025485−− = − − + = − + = −De ( I ( ), I ) I vem: e–x – 7 = 5 s –x = 12 s x= –12 ou–x – 7 = –5 s –x = 2 s x= –2 Logo, o determinan e t dainversaserá:−548–12 + (–2) = –14 24. detA≠ 013. A= − ======1111detA= 2 3x– 6x ≠ 0 s x≠ 0 e x≠ 12 2det 2)= (detA2 = 22 = 4 (A ) 25. Sendoabcdtm inversa,temos: ein ra14. b 10011001ce ⋅ 0 1 0 01abcd ⋅det 2 · B2)= det 2)· det 2)s (A (A (Bs det 2 · B2)= (detA2 · (de (A ) tB) s 2 tm 0 1 =01 0 dr 0s det 2 · B2)= (–1) · (–1) = 1 (A 2 215. ba = 1; b = 2; x= 3 e y= 4A== 1 b ; 1234e detA= –2 abcd1a = r 1001 0b m 0c indet(AB) detA· detB= detA· detA s = t 26. A = 2153−−3 3 s detA= –1 (existe –1 ) 33 33 As det(AB) (de ) s det(AB) 4 = tA2 = Aplicando o dispositivo prático (página 41),16. d temos:1 · 2 · 3 · 4 · 5 + (–1) · (1· 2 · 3 · 4 · 5 · 6)= 3152−−2 2 22 22= 120 + (–1) · (720)== 120 – 720 = – 600 27. c17. d a)(F)detB= 4Q 3 = –2Q 2 s det(Q )= det 3 (–2Q2)s Pelodispositivoprático(página41):s (detQ) = (–2) · (de 3 4 tQ) , dividindo ambos os 2 BA− = − ==== = = = ==1134014 = = == ≠ membros por(detQ) , temos:detQ = 16 2 b)(F)detA= 2; detB= 4 ∴ detA≠ detB18. dsen()cos ) e ( cos ) e ( s2xxx x (s n) (s n) e x 1011=nn()cos() ⋅=1110111 cos()xxx 2 c)(V)AB⋅=) ( )⋅⋅ V ) (V =) ( )110213041708 ) V= sen(x) [cos(x) 1 – cos(x) cos (x)] · + – 2 == sen(x) [1 – cos (x)] · 2 == sen(x) sen (x) sen (x) · = BA⋅=A( )⋅⋅ 2 319. bA2 = –2A ts V A (V =A( )130411021708 ) Vs det 2)= det (A (–2A )s (detA2 = (–2) · det(A)s t ) 3 ts (detA2 = –8 · detAs detA= –8 ) ∴ AB = BA20. dA–1 · B · A = D ss det –1 · B · A)= detDs (As det –1 )· detB· detA= detDs (A ⋅⋅ d)(F)det(A· B) = detA· detB= 2 · 4 ≠ 0 e) (F)s 15detde d t t e ABA = s BBBA213041304115016= ⋅=34 0 01⋅⋅ 3 3 4 3 01 0 1 =34s detB= 5 21. detA≠ 3 =≠ = 6 – 12 – 4x≠ 0 s x≠ −32 22. e 12 detA= 0 s
  18. 18. mapa a a ⋅= − ⋅+ ⋅=cos()s ( s n ) m p ⋅+28. d ⋅⋅ b d b ca⋅⋅ b d b en ) e ( cos ) = oe ( 10csabcdabcd c a =bd c c a 12011201s b naqanaqa+ ⋅= − ⋅+ ⋅=cos()sen ) e ( co ( 01c ⋅s aabccdacbdcd2222++ =+ + o ( s n ) s ) =o sTemos:a = a + 2c s c = 029. aI.–log x · logx– 1 – 3logx≠ 0 2 Resolv endo ambos os siste a m s,–2[log x] – 3logx– 1 ≠ 0 2 encontra mo s:Δ = 9– 8= 1 m= cos(a);n = –sen (a);p = sen(a) q = cos(a) eI logxx I. x≠−≠ ≠ − 12101010 ss 12 Assim,substi uindoemI,vem: t s ⋅e logxxx≠−≠ ≠ − 110110 ss 130. AXBC –1 = B Xaaaa=−− s, cos() en ) e ( cos ) s b b b s n im s (s n) ( co ( b b ) eMultiplicandoà esquerdaporA–1 e à direitapor C, temos:A–1 · A · X · B · C –1 · C = A–1 · B · C s ()sen ) ( cos )−o ) ( − ss (ss I X · B · I A–1 · B · C · = sSendo I matri identidadede mesm ordemque a z a Xababb=⋅+ ⋅⋅cos()cos()sen( sen ) en( coss( s n( cos( sen( c A, B e C.X · B = A–1 · B · C ) (s ) ) e ) ) )Multip licandoà direitaporB –1 , temos:X · B · B–1 = A–1 · B · C · B –1 s s ( a a ⋅⋅− ( cos ) e ( s n ⋅⋅+ ⋅⋅ ⋅ as X · I A–1 · B · C · B –1 s X = A–1 · B · C · B –1 = os() en a b b− bbaabab cos ) ) ( cos ) (s n) e ( ) a31. d ⋅ ⋅ asdet(2A) det 2)s = (As 22 · detA= (de ) tA2Dividindo ambos os membros por detA, temos: s Xabababab=−−−−− − a acos() en ) e ( cos c s (s n) (o )s detA= 4 Observação: A informaçãosen (a)· cos (a)≠32. 0 pode ser excluída do enunciado do SeAaaaaB=−− . A 2 = =cos()sen ) e ( co ( e ( s n ) s ) ccos ) (s proble a, mas, nesse caso, a resolução m en() en ) s ( cos ) b b−c( temos: ( b b − o) s , implicariaumadiscussãomuitolonga.A · X= B 33. dMultip licando à esquerda por A–1 (que existe, A · A–1 = I s A · B = I s 2 2 isso⋅ uã⋅ poisdetA≠ 0), temos:A–1 · A · X = A–1 · B s I X = A–1 · B s X = A–1 · B (I) · 130143101001pq dcs m 3 1 3 0 3 0 4 =31 1 4sssCalcul mo A–1 : e s sAmnpq =mq1 e A · A–1 = I − nl pe cos() en ) e ( cos ) a a n q− I ⋅⋅ 113041001+0 qiss m pqq 1 dcs u p u =o it I.qq414= = s s (s n) (a a m p − · I · = =I I 1301340+ = + = pqp ss413112pp=− = − s I. ∴ s qp− = − − ======= + = 12412112415 = 1001 ss mapanaqa ⋅+ m ⋅⋅+ ⋅−⋅cos()sen( cos( s n ) enn( c ) ) e (s ) 34. a)(A+ B) · (A– B)= A2 – AB + BA – B2 b)O produtoAB deveserigualao produto BA. ( a a aqa+⋅−⋅+ os()s ( cos ) p n en ) ⋅⋅ s()s = e c)detde ( d t ) e2AAAA− = − ⋅=(1 t) e 1d t d)detdetBA=1= ( s 1001 )enPara que a igualdad se verifique, é preciso e 35. b que: log [det 3 (2A )]= log (detA )s –1 27 –1 s log [25 · det(2A )]= log 3 [det –1 · A–1 )]s 3 –1 3 (2
  19. 19. s loglogde 33132113132 t( ⋅⋅ 2 1 1 ⋅⋅ 2 1 1 1 3 = 1 3 − det)AAs s logdetlogdet 321321AA 22 ⋅⋅ 2 3 1s 3313 33 11 = 1 2 s 32132 detdetAA=d s 13 etÛ s 212155detdetAA= s 3 s 220 · detA= det A s 220 = det A s 3 2 s detA= 1.024 = 210 36. a)AB = BA s AB · B–1 = BA · B–1 s A · I B = · A · B–1 s s A = B · A · B–1 s B–1 · A = B–1 · B · A · B–1 s s B–1 · A = I A · B–1 s B–1 · A = A · B–1 · ∴ A · B–1 = B–1 · A (c.q.d.)13
  20. 20. b)A2 + 2AB – B = 0 s B = A2 + 2AB s B = A(A+ 2B)ss det(B) det[A(A+ 2B)]s =s det(B) det(A) det(A+ 2B) = ·Do enunciado,sabemosqueB é inversíve Logo, det l. (B) 0. ≠Assim:det ) det(A+ 2B)≠ 0 s det ) 0 e det + 2B)≠ 0 (A · (A ≠ (AEntão,sedet ) 0, A é inversíve (A ≠ l.(c.q.d.)Capítulo 4 sistemas linearesConexões1. – 5 05 – 2 1 2 y x r t s2. S1(r;s):yxyx= + = − 52S2(r;t):yxyx= + = − 52S3(t;s):yxyx= − = − 223. S1(r;s):yxyx= + = − 52ss x+ 5 = –2x s 3x= –5 s x= − 53∴ S= − 53 10 3 ; (Asretassãoconcorren e ) t s.S2(r;t):yxyx= + = − 52 ss x+ 5 = x– 2 s 0x= –7 (F)∴ S = ∅ (Asretassãopara e a distinta ) l l s s.S3(t;s):yxyx= − = − 22ss x– 2 = –2x s 3x= 2 s x= 23∴ S = 23 43 ; − (Asretassãoconcorren e ) t s. Exercícios complementares 13. D a a a a a = − = − − cos()sen()sen()cos()cos()s2 en() aD s = −1 2 ⋅D a a a a x= − − = = − sen(2)sen()cos(2)cos()sen(2)cos()sen()cos()aaaa + ⋅ 2s s Dx= –sen (a)D a a a a a y= − = − cos()sen(2)sen()cos(2)cos(2)cos()sen2sen()⋅ − ⋅ aaa Dy= – cos(a)xDD a a x= = − − = sen()sen()1yDD a a y= = − − = cos()cos()1
  21. 21. ∴ S = {(sen(a);cos(a))} 14. a)SPD (sistemaescalonadodo primeirotipo)a bbb + = = = a 5 2 12 6 s e Substitui eb naprimeira s - igualdade: a + 6 = 5 s a = –1 ∴ S = {(–1; 6)} b)SPD (sist m escalonadodo primeirotipo) e a 2z= 4 s z= 2 Substitui eznasegundaigualdad s - e: 3y+ 4 · 2 = 14 s 3y= 6 s y= 2 Substitui eye zna primeira s - igualdade: x+ 2 · 2 – 2 = 0 s x= –2 ∴ S = {(–2; 2; 2)} c)SP (sist m escalonadodo segundotipo) I e a p é a variáv l e livre. n + 2p= 5 s n = 5 – 2p Substitui eo valorden naprimeira s - equação. –m + 2 · (5– 2p)+ 3p= 4 s s –m + 10 – 4p+ 3p= 4 s s –m – p = – 6 s s m= 6 – p ∴ S = {(6– p; 5 – 2p; p),comp 3 ® } 15. d Sendoy= 0, subs tuímos ti essevalornasegundaequaçãoe obtemos = 2. x Substituindonaprimeiraequação,temos: (λ + 1)· 2 + 0 = 0 s λ + 1 = 0 s λ = –1 16. d Antes Hoje DepoisEu y 2x aTu x y 2x

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