ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 1 ThS PHNG
DUY QUANG (ch bin) BI GING N THI CAO HC Mn: TON KINH T H NI, 2011
ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 2 Phn 1. Ton
c sng dng trong kinh t NI DUNG TON CAO CP 1 Chuyn 1. Ma trn v nh
thc 1. Ma trn v cc php ton 2. nh thc ca ma trn vung cp n 3. Ma trn
nghch o 4. Hng ca ma trn Chuyn 2. H phng trnh tuyn tnh v ng dng 1.
Khi nim h phng trnh tuyn tnh 2. Phng php gii h phng trnh 3. Mt s m
hnh tuyn tnh trong phn tch kinh t TON CAO CP 2 Chuyn 3. Gii hn, lin
tc, vi tch phn hm mt bin s 1. Gii hn ca dy s 2. Gii hn ca hm s 3.
Hm s lin tc 4o hm, vi phn v ng dng 5. ng dng o hm trong phn tch
kinh t 6. Tch phn hm mt bin sChuyn 4. Php tnh vi phn hm nhiu bin s
v ng dng 1. Gii hn v lin tc 2. o hm ring v vi phn ca hm nhiu bin 3.
ng dng ca o hm ring trong phn tch kinh t 4 Cc tr hm nhiu bin 5. ng
dng cc tr hm nhiu bin trong phn tch kinh t ThS Phng Duy Quang Trng
Khoa C bn Trng i hc Ngoi Thng 3 NG DNG Chuyn 5. p dng Ton cao cp
trong phn tch kinh t Luyn tp B sung bi tp ng dng mi TI LIU THAM KHO
1. Alpha C. Chiang, Fundamental Methods of Mathematical Economics,
McGRAW-HILL Book Copany, 1984. 2. L nh Thy (ch bin), Ton cao cp cho
cc nh kinh t, NXB Thng k, 2004. 3. Bi tp Ton cao cp cho cc nh kinh
t, NXB HKTQD, 2008 4. Nguyn Huy Hong, Ton cao cp T1, T2. NXB Gio dc
Vit Nam, 2010. 5. Nguyn Huy Hong,Hng dn gii bi tp Ton cao cp cho cc
nh kinh t T1, T2. NXB Gio dc Vit Nam, 2010. 6. Ng Vn Th, Nguyn
Quang Dong, M hnh ton kinh t, NXB Thng k, 2005. ThS Phng Duy Quang
Trng Khoa C bn Trng i hc Ngoi Thng 4 A. TON CAO CP 1 Chuyn 1.Ma trn
v nh thc 1. Ma trn v cc php ton 1. Cc khi nim Cho m, n l cc s nguyn
dng nh ngha 1. Ma trn l mt bng s xp theo dng v theo ct. Mt ma trn c
m dng v n ct c gi l ma trn cp mn. Khi cho mt ma trn ta vit bng s
bntrongdungoctrnhocngocvung.Matrncpmncdngtngqut nh sau:||||||
\|mn 2 m 1 mn 2 22 21n 1 12 11a ... a a... ... ... ...a ... a aa
... a ahoc (((((
mn 2 m 1 mn 2 22 21n 1 12 11a ... a a... ... ... ...a ... a aa
... a a Vit tt l A = (aij)n xn hoc A = [aij]n xn V d 1. Cho ma trn
((
=1 7 67 5 2A . A l mt ma trn cp 2 x 3 vi a11 = 2 ; a12 = 5 ; a13
= - 7 ; a21 = 6 ; a22 = 7 ; a23 = 1 nh ngha 2. Hai ma trn c coi l
bng nhau khi v ch khi chng cng cp v cc phn t v tr tng ng ca chng i
mt bng nhau. Ma trn chuyn v ca A l AT : AT = [aji]n xn Ma trn i ca
ma trn A l ma trn A = [- aij]n x n V d 2. Cho ma trn ((((
=0 21 43 1A . Xc nh AT, - A Ta c ((
=0 1 32 4 1AT ; ((((
= 0 21 43 1AThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi
Thng 5 Ma trn khng cp m x n l ma trn m mi phn t u bng 0 :n x m] 0 [
= Khi n = 1 ngi ta gi ma trn A l ma trn ct, cn khi m = 1 ngi ta gi
ma trn A l ma trn dng. Ma trn vung cp n l ma trn c s dng v s ct bng
nhau v bng n. Mt ma trn c s dng v s ct cng bng n c gi l ma trn vung
cp n. Khi cc phn t a11, a22, , ann gi l cc phn t thuc ng cho chnh,
cn cc phn t an1, n 12a, , a1n gi l cc phn t thuc ng cho ph.
Matrntamgiclmatrnvungkhicccphntnmvmtphaca ng cho chnh bng 0. +) Ma
trn A = [aij]n x n c gi l ma trn tam gic trn nu aij = 0 vii > j:
(((((((
= nnn 1 n 1 n 1 nn 2 1 n 2 22n 1 1 n 1 12 11a 0 ... 0 0a a ... 0
0... ... ... ... ...a a ... a 0a a ... a aA+) Ma trn A = [aij]n x n
c gi l ma trn tam gic di nu aij = 0 vii < j: (((((((
= nn 1 n n 2 n 1 n1 n 1 n 2 1 n 1 1 n22 2111a a ... a a0 a ... a
a... ... ... ... ...0 0 ... a a0 0 ... 0 aAV d 4. Cho mt v d v ma
trn vung cp 3, ma trn tam gic trn, tam gic di cp 3. Gii: ((((
=6 1 14 1 25 2 1A; ((((
=6 0 04 1 05 2 1B; ((((
=6 1 10 1 20 0 1CThS Phng Duy Quang Trng Khoa C bn Trng i hc
Ngoi Thng 6 Matrncholmatrnvungcpnmcttcccphntnmngoi ng cho chnh u
bng 0 Ma trn cho c tt c cc phn t thuc ng cho chnh bng 1 c gi l ma
trn n v : (((((((
=1 0 ... 0 00 1 ... 0 0... ... ... ... ...0 0 ... 1 00 0 ... 0
1ETp cc ma trn cp m x n trn trng s thc , k hiu: Matm x n() Tp cc ma
trn vung cp n trn trng s thc , k hiu: Mat n() V d 5. Cho ma trn
((
=1 7 67 5 2Av ((((
=2m 77 56 2Ba) Tm AT v A b) Tm m AT = B Gii: a) Ta c ((((
=1 77 56 2AT v ((
=1 7 67 5 2Ab)1 m 1 mm 77 56 21 77 56 2B A22T = = ((((
=((((
=2. Php ton trn ma trn a) Php cng hai ma trn v php nhn ma trn vi
1 s nh ngha 3. Cho hai ma trn cng cp mn: [ ] [ ]n mijn mijb B ; a A
= =Tng ca hai ma trn A v B l mt ma trn cp mn, k hiu A + B v c xc nh
nh sau:ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 7 [ ]n
mii ijb a B A+ = +TchcamatrnAvimts lmtmatrncpmn,khiu Avc xc nh nh
sau:[ ]n mija . A = Hiu ca A tr B: A B = A + (-B) T nh ngha ta suy
ra cc tnh cht c bn ca php ton tuyn tnhTnh cht 1. Cho A, B, C l cc
ma trn bt k cp mn, ;l cc s bt k ta lun c:1)A + B = B + A 2)(A + B)
+C = A + (B + C) 3)A + 0 = A 4)A + (-A) = 0 5)1.A = A6)(A + B) = A
+ B 7)( + )A = A + A 8)( )A =( B) V d 6. Cho cc ma trn ((
=((
=3 1 22 1 2B ;1 1 04 2 1A . Khi ((
=((
+((
= 11 1 614 7 43 1 22 1 2). 3 (1 1 04 2 1. 2 B 3 A 2V d 7. Cho ma
trn ((
=3 53 1B . Tm ma trn C sao cho 3B 2(B + C) = 2E Gii: Phng trnh
cho ((
=((
((
= = 2 / 1 2 / 52 / 3 2 / 11 00 13 53 1.21E B21CThS Phng Duy
Quang Trng Khoa C bn Trng i hc Ngoi Thng 8 b) Php nhn ma trn vi ma
trn Cho hai ma trn :A = (((((
mn 2 m 1 mn 2 22 21n 1 12 11a ... a a... ... ... ...a ... a aa
... a a; B = ((((((
np 2 n 1 np 2 22 21p 1 12 11b ... b b... ... ... ...b ... b bb
... b b Trong , ma trn A c s ct bng s dng ca ma trn B.nh ngha 4.
Tch ca ma trn A vima trn B lmtma trn cpmp, k hiu l AB v c xc nh nh
sau:AB = (((((
mn 2 m 1 mn 2 22 21n 1 12 11c ... c c... ... ... ...c ... c cc
... c c trong ( ) p ,..., 2 , 1 j ; m ,..., 2 , 1 i ; b a b a ... b
a b a cn1 kkj ik nj in j 2 2 i j 1 1 i ij= = = + + + == Ch 1.Tch AB
tn ti khi v ch khi s ct ca ma trn ng trc bng s dng ca ma trn ng
sau. Ccama trnAB:MatrnABcsdngbngsdngcamatrnng trc v s ct bng s ct
ca ma trn ng sau. Cc phn t ca AB c tnh theo quy tc:Phn t ijcl tch v
hng ca dng th i ca ma trn ng trc v ct th j ca ma trn ng sau. V d 8.
Cho hai ma trn ((
=1 32 1Av ((
=2 3 14 1 0B . Tnh A.B v B.A Gii : ThS Phng Duy Quang Trng Khoa
C bn Trng i hc Ngoi Thng 9 Ta c ((
=((
+ + ++ + +=((
((
=14 6 18 7 22 . 1 4 . 3 3 . 1 1 . 3 1 . 1 0 . 32 . 2 4 . 1 3 . 2
1 . 1 1 . 2 0 . 12 3 14 1 0.1 32 1B . ANhng s ct ca B khc s dng ca
A nn khng tn ti tch BA. V d 9. Cho ma trn ((
=0 2 30 1 2A ; ((((
=1 2 0 30 1 1 21 3 2 1B . Tnh A.B, BA Gii: Ta c ((
=((((
((
=3 7 8 11 7 5 31 2 0 30 1 1 21 3 2 1.0 2 30 1 2B . ACn B.A khng
tn ti Cc tnh cht c bn ca php nhn ma trnTnh cht 2. Gi s php nhn cc
ma trn di y u thc hin c. 1) (AB)C = A(BC) 2) A(B+C) = AB+AC; (B+C)D
=BD +CD 3)(AB) = ( A)B = A( B) 4) AE = A; EB =B c bit , vi ma trn
vung A: AE = EA = A 5)( )TT TAB B A =
Ch2.Phpnhnmatrnkhngctnhchtgiaohon.Nu = B . A thcha chc = Ahoc = B .
V d 10. Cho cc ma trn ((
=((
=0 10 0B ;0 01 0A .Khi ((
=((
=1 00 0A . B ;0 00 1B . AvBA AB V d 11. Cho ((
=((
=1 00 0B ;0 00 1A , ta c ((
=((
((
=0 00 01 00 0.0 00 1B . AThS Phng Duy Quang Trng Khoa C bn Trng
i hc Ngoi Thng 10 c) Lu tha ca ma trn vung: Cho A l ma trn vung cp
n. Ta xc nh A0 = E; An = An -1. A ( n l s nguyn dng) V d 12. Cho
((
=d cb aA . Chng minh rng, ma trn A tho mn phng trnh = + + ) bc
ad ( X ) d a ( X2 Gii: Ta c ((
+((
+ ((
((
= + + 1 00 1). bc ad (d cb a). d a (d cb a.d cb aE ) bc ad ( A )
d a ( A2 = =((
=((
+((
+ ++ +((
+ ++ +0 00 0bc ad 00 bc ad) d a ( d ) d a ( c) d a ( b ) d a (
ad bc c ) d a (b ) d a ( bc a22. (pcm) V d 13. Cho ma trn ((
=1 01 1A . Tnh A2, A3, ..., An (n l s t nhin) Gii: Ta c ((
=((
((
=1 02 11 01 11 01 1A2; ((
=((
((
=1 03 11 01 11 02 1A3; .... ; tng t ta c th d on ((
=1 0n 1An. D dng chng minh c bng quy np cng thc An. nh ngha 5.
Php bin i s cp trn ma trn A = [aij]m x n l cc php bin i c dng i) i
ch 2 dng (ct) cho nhau:) c c ( d dj i j i ii) nhn mt dng (ct) vi mt
s khc 0:) kc ( kdi i iii) nhn mt dng (ct) vi mt s ri cng vo dng
(ct) khc:) c hc ( d hdj i j i+ +Vd15.Chomatrn ((((
=4 2 1 15 2 1 26 4 2 1A .Thchinccphpbiniscp sau: (1) nhn dng 2
vi 2 ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 11 (2)
hon v dng 1 cho dng 2 (3) nhn dng 2 vi 2 cng vo dng 3 nh ngha 6. Ma
trn dng bc thang l ma trn c tnh cht
i)Ccdngkhckhng(tclcmtphntkhc0)nucthluntrncc dng bng khng (tc l hng
c tt c cc phn t bng 0). ii) hai dng khc 0 k nhau th phn t khc 0 u
tin dng di bao gi cng bn phi ct cha phn t khc 0 u tin dng trn.V d
15. Cc ma trn sau l ma trn dng bc thang (((((
=0 0 0 0 05 2 0 0 05 3 1 1 08 6 5 1 1A; (((((
=1 0 0 0 01 1 2 0 01 8 2 1 07 4 3 1 1B; ((((
=0 0 01 2 02 1 1CThS Phng Duy Quang Trng Khoa C bn Trng i hc
Ngoi Thng 12 2. nh thc ca ma trn vung cp n
nhthcgivaitrquantrngtrongistuyntnhkhigiihphngtrnh tuyn tnh, khi
nghin cu dng ton phng, .... 1. Khi nim nh thcCho ma trn A =
(((((
nn 2 n 1 nn 2 22 21n 1 12 11a ... a a... ... ... ...a ... a aa
... a a. Xt phn t aij ca A, b i dng i v ct j ca A ta c ma trn vung
cp n -1, k hiu Mij: gi l ma trn con con ng vi phn t aij. V d 1.
((((
=33 32 3123 22 2113 12 11a a aa a aa a aA . Tm cc ma trn con ng
vi cc phn t ca A nh ngha 1. Cho mt ma trn A vung cp n: A =
(((((
nn 2 n 1 nn 2 22 21n 1 12 11a ... a a... ... ... ...a ... a aa
... a a.nh thc ca A, k hiu det(A) hocAc nh ngha nh sau: * nh thc cp
1: A = [a11] th det(A) = a11 * nh thc cp 2: ((
=22 2112 11a aa aAth 21 12 22 1122 2112 11a a a aa aa a) A det(
= =V d 2. Tnh nh thc2 2 . 6 14 . 114 26 1= =V d 3. Gii phng trnh:04
925 x2= * nh thc cp 3: ThS Phng Duy Quang Trng Khoa C bn Trng i hc
Ngoi Thng 13 32 23 11 33 21 12 31 22 13 32 21 13 31 23 12 33 22
1133 32 3123 22 2113 12 11a . a . a a . a . a a . a . a a . a . a a
. a . a a . a . aa a aa a aa a aA det + + = =Quy tc Sariut: nh thc
cp 3 c 6 s hng, m mi s hng l tch ca 3 phn t m mi dng, mi ct ch c mt
i biu duy nht. *Ccshngmangducng:ccshngmccphntnmtrnngcho chnh hoc cc
phn t nm trn cc nh ca tam gic c 3 nh c mt cnh song song vi ng cho
chnh. * Cc s hng mang du tr: cc s hng m cc phn t nm trn ng cho ph
hoc cc phnt nmtrn ccnhca tamgic c3 nh c mtcnh song song vi ng cho
ph. nh quy tc tnh nh thc cp 3, ngi ta thng dng quy tc Sarrus sau: T
quy tc Sarrus trn, chng ta cn mt quy tc khc tnh nhanh nh thc cp 3:
ghp thm ct th nht v ct th hai vo bn phi nh thc hoc ghp thm dng
thnht vdngth hai xungbndinhthc rinhnccphnt trn cc ng cho nh quy tc
th hin trn hnh: Du +Du - 1 1 1 1 12 2 2 2 23 3 3 3 3a b c a ba b c
a ba b c a b Du -Du + 1 1 12 2 23 3 31 1 12 2 2a b ca b ca b ca b
ca b c Du - Du + ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi
Thng 14 V d 4.Tnh nh thc1 2 21 0 23 2 13= Gii: Ta c= 31.0.1 +
2.(-2).1 + 3.2.(-2) 3.0.2 1.(-2).2) 1.1.(-2) = -10 V d 5. Gii phng
trnh01 2 41 1 11 x x2=Gii: Ta c
== = + =2 x1 x0 2 x 3 x1 2 41 1 11 x x22 nh thc cp n (n3 ):
det(A) =) M det( ) 1 ( aijj in1 jij+= (vi i bt k) hoc det(A) =) M
det( ) 1 ( aijj in1 iij+= (vi j bt k) V d 6. Gii phng trnh :01 2 4
20081 1 1 20091 x x 20100 0 0 20112= 2. Tnh cht ca nh thc A =[aij]n
x n vi) A det(n= Dng i ca nh thc c gi l tng ca 2 dng nu: ( ) ( ) (
)i1 i 2 ij in i1 i 2 ij in i1 i 2 ij in ij ij ija a ....a ....a b b
....b ....b c c ....c ....c ; a b c ( j 1, n) = + = + =Dng i l t hp
hp tuyn tnh ca cc dng khc nu) n , 1 j ( a akjnk1 kk ij= = =. K hiu
= =ni k1 kk k id d ; dk = (ak1 ak2 ... akn) ThS Phng Duy Quang Trng
Khoa C bn Trng i hc Ngoi Thng 15 Tnh cht 1. (Tnh cht chuyn v)
nhthccamatrnvungbngnhthccamatrnchuynvcan: det(AT) = det(A) V d 1.
Cho ((
=d cb aA . CMR det(AT) =det(A)
Ch1.Ttnhchtchuynv,mitnhchtcanhthcngchodngth cngngchoctvngcli.
Do,trongcctnhcht canhthc, chpht biu cho cc dng, cc tnh cht vn gi
nguyn gi tr khi thay ch "dng" bng ch "ct". Tnh cht 2. (Tnh phn
xng). ichhaidngchonhauvginguynvtrccdngcnlithnhthci du. V d 2. Xt d
cb a v b ad c H qu 1. Mt nh thc c hai dng ging nhau th bng khng.
Chng minh Ginhthcchaihngnhnhaul n . ichhaihngtac,theotnh cht 2 ta c
n= - n 0 0 2n n= = Tnhcht3.(Tnhthunnht).Nunhnccphntmtdngnovicng mt
s k th c nh thc mi bng k ln nh thc c nn 2 n 1 nin 2 i 1 in 1 12
11nn 2 n 1 nin 2 i 1 in 1 12 11a.........a...a...a ... a a... ...
... ...a ... a a. ka.........a...a...ka ... ka ka... ... ... ...a
... a a=nh l ny c th pht biu: Nu mt nh thc c mt dng c nhn t chung
th a nhn t chung ra ngoi du nh thc H qu 2. Mt nh thc c hai dng t l
vi nhau th bng khng. Chng minh: Tht vy, nu a h s t l ra ngoi du nh
thc th c mt nh thc c hai dng ging nhau nn n bng khng. ThS Phng Duy
Quang Trng Khoa C bn Trng i hc Ngoi Thng 16 V d 2.19. Chng minh nh
thc sau chia ht cho 17: 9 11 7 64 1 1 2204 35 68 177 6 2 124 = Gii:
Ta cD . 179 11 7 64 1 1 212 2 4 17 6 2 12. 179 11 7 64 1 1 2) 12 .(
17 2 . 17 ) 4 .( 17 1 . 177 6 2 124= = = .V D l nh thc to bi cc s
nguyn nn D cng l s nguyn. Do 174M Tnh cht 3. (Tnh cng tnh). Nu nh
thc c mt dng l tng hai dng th nh thc bng tng ca hai nh thc. 11 12
1n 11 12 1n 11 12 1ni1 i1 i 2 i 2 in in i1 i 2 in i1 i 2 inn1 n2 nn
n1 n2 nn n1 n2 nna a a a a a a a ab c b c b c b b b c c ca a a a a
a a a a= + + + +L L LL L L L L L L L L L L LL L LL L L L L L L L L
L L LL L L Hqu3.Nunhthccmtdnglthptuyntnhcaccdngkhcth nh thc y bng
khng. l h qu ca tnh cht cng tnh v tnh thun nht. H qu 4. Nu cng vo
mt dng mt t hp tuyn tnh ca cc dng khc th nh thc khng i. T cc tnh
cht ca nh thc, ta thng s dng cc php bin i s cp trn ma trn trong qu
trnh tnh nh thc cp n: * i ch 2 dng (ct) cho nhau:) c c ( d dj i j i
, php bin i ny nh thc i du *Nhnmtdng(ct)vimtskhc0:) kc ( kdi
i,phpbininynhthc tng ln k ln. * Nhn mt dng (ct) vi mt s cng vo dng
(ct) khc:) c hc ( d hdj i j i+ + , php bin i ny khng lm thay i gi
tr ca nh thc. ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng
17 V d 4. Tnh nh thc y ' c cx y ' b bx y ' a ax' c ' b ' ac b a3+ +
+= Gii:Nhn dng 1 vi (-x), dng 2 vi (-y) cng vo dng 3 ta c: 00 0 0'
c ' b ' ac b a3 2 1d yd xd3= = + V d 5. Tnh nh thc 2 2 2 22 2 2 22
2 2 22 2 2 24) 3 d ( ) 3 c ( ) 3 b ( ) 3 a () 2 d ( ) 2 c ( ) 2 b (
) 2 a () 1 d ( ) 1 c ( ) 1 b ( ) 1 a (d c b a+ + + ++ + + ++ + + +=
Gii:Nhn dng 1 vi (-1), ri cng ln lt vo dng 2, dng 3, dng 4 c: 9 d 6
9 c 6 9 b 6 9 a 64 d 4 4 c 4 4 b 4 4 a 41 d 2 1 c 2 1 b 2 1 a 2d c
b a2 2 2 2d d4 , 3 , 2 i4i 1+ + + ++ + + ++ + + += + = Sau nhn dng
2 vi (- 2) cng vo dng 3, nhn dng 2 vi (-3) cng vo dng 4 c: 6 6 6 62
2 2 21 d 2 1 c 2 1 b 2 1 a 2d c b a2 2 2 2d d 2d d 343 24 2+ + + +=
+ + = 0 (v c 2 dng t l nhau) V d 6. Tnh nh thc 12a c2c b2b a1 b a
c1 a c b1 c b a4+ + += Gii: ThS Phng Duy Quang Trng Khoa C bn Trng
i hc Ngoi Thng 18 Cng cc ct vo ct 1 ta c: 12a c2c b1 c b a1 b a 1 c
b a1 a c 1 c b a1 c b 1 c b a4+ ++ + ++ + ++ + ++ + += t nhn t
chung ca ct 1 ra ngoi: 012a c2c b11 b a 11 a c 11 c b 1). 1 c b a
(4=+ ++ + + = 3.Cc phng php tnh nh thc Cho nh thc cp n:nm nj 1 nin
ij 1 in 1 j 1 11na ... a ... a... ... ... ... ...a ... a ... a...
... ... ... ...a ... a ... a= a) Phng php khai trin: Phn b i s ca
ijaXa i dng th i v ct th j (dngv ct cha phn t ija ) ca A ta c mt ma
trn con (n - 1), k hiu l ijM . nh thc ca ijMc gi l nh thc con cp n
-1 tng ng vi phn t aij ca A v) M det( ) 1 ( Aijj iij+ =c gi l phn b
i s ca phn t ijaca nh thc d. Cho nh thc cp n l n . Khi nc th tnh
theo hai cch sau: i)Cng thc khai trin theo dng th i : = =+= = n1
jij ijn1 jijj iij nA a ) M det( . ) 1 ( a(1) ii)Cng thc khai trin
theo ct th j: ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng
19 = =+= = n1 iij ijn1 iijj iij nA a ) M det( . ) 1 ( a(2) H qu. i
vi nh thc cp n l n , ta c i) = ==k i khi 0k i khiA ann1 jkj ij(3)
ii) = ==k j khi 0k j khiA ann1 iik ij (4) Chng minh:Nhn xt: Mc ch
ca cng thc (1) hoc (2) l chuyn vic tnh nh thc cp
nvtnhnhthccpn-1,ritcpn-1chuynvcpn-2,,chonnh
thccp3,2.Khipdngcngthc(1)hoc(2),tannchndnghocctc
chanhiuphnt0nhtkhaitrin.Nukhngcdnghocctnhvyta
bininhthcavnhthcmibngnhthcbanunhngcdng hoc ct nh vy. V d 1. Tnh nh
thc a) 0 5 42 1 31 1 23 = b) 4 2 12 1 31 2 13= Gii: a) Khai trin nh
thc theo dng 3 ta c: 7 5 12 02 31 2. ) 1 .( 52 11 1. ) 1 .( 42 3 1
33= = + + = + + b) Khai trin nh thc theo ct 1 ta c: 35 5 30 02 11
2. ) 1 )( 1 (4 21 2. ) 1 .( 34 22 1. ) 1 .( 11 3 1 2 1 13 = = + + =
+ + + V d 2. Tnh nh thc a) 1 2 5 33 1 4 23 1 3 15 0 1 14 = b) 11 4
3 24 1 0 03 0 1 02 0 0 14= Gii:ThS Phng Duy Quang Trng Khoa C bn
Trng i hc Ngoi Thng 20 a)Nhnct1vi(-1)cngvoct2,nhnct1vi(-5)
cngvoct4;rikhai trin nh thc theo ct 1, ta c 14 2 813 1 68 1 414 2
813 1 68 1 4. ) 1 .( 114 2 8 313 1 6 28 1 4 10 0 0 11 1c cc c 542
14 1 = = = ++ + Cng dng 1 vo dng 2, nhn dng 1 vi (-2) cng vo dng 2,
ri khai trin nh thc theo ct 2 ta c: 2030 165 2. ) 1 .( 130 0 165 0
28 1 42 1d dd d 242 13 1= = = +++ b) Nhn ct (-2) vi ct 1 ri cng vi
ct 4 9 4 3 24 1 0 05 0 1 00 0 0 14= Khai trin nh thc theo dng 1 ta
c 9 4 34 1 05 0 19 4 34 1 05 0 1. ) 1 .( 19 4 3 24 1 0 05 0 1 00 0
0 11 14= == + Nhn ct 1 vi 5 cng vo ct 3, khai trin nh thc theo dng
1 ta c 8 16 2424 44 1. ) 1 .( 124 4 34 1 00 0 11 14= = = = + V d 3.
Tnh nh thc ca ma trn tam gic trn v tam gic di ThS Phng Duy Quang
Trng Khoa C bn Trng i hc Ngoi Thng 21 a) nnn 1 n 1 n 1 nn 2 1 n 2
22n 1 1 n 1 12 11na 0 ... 0 0a a ... 0 0... ... ... ... ...a a ...
a 0a a ... a a = b) nn 1 n n 2 n 1 n1 n 1 n 2 1 n 1 1 n22 2111na a
... a a0 a ... a a... ... ... ... ...0 0 ... a a0 0 ... 0 a = Gii:
Ta ch cn xt a) Ln lt khai trin nh thc theo ct 1 : nn 22 11nnn 1 n 1
n 1 nn 2 1 n 2 221 111nnn 1 n 1 n 1 nn 2 1 n 2 22n 1 1 n 1 12 11na
... a . a ...a 0 ... 0a a ... 0... ... ... ...a a ... a. ) 1 .( aa
0 ... 0 0a a ... 0 0... ... ... ... ...a a ... a 0a a ... a a= = =
= + Tng t, ta c nn 22 11nn 1 n n 2 n 1 n1 n 1 n 2 1 n 1 1 n22
2111na ... a aa a ... a a0 a ... a a... ... ... ... ...0 0 ... a a0
0 ... 0 a= = b) Phng php bin i v dng tam gic: Dng cc tnh cht ca nh
thc bin i nh thc a nh thc v nh thc ca ma trn tam gic trn hoc di,
sau p dng cng thc:nn 33 22 11nnn 2 22n 1 12 11a ... a . a . aa ...
0 0... ... ... ...a ... a 0a ... a a=hoc nn 22 11nn 2 n 1 n22 2111a
... a aa ... a a... ... ... ...0 ... a a0 ... 0 a= V d 1. Tnh cc nh
thcThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 22 a) 0 4
3 2 15 0 3 2 15 4 0 2 15 4 3 0 15 4 3 2 15 = b) 4 4 3 2 14 3 3 2 14
3 2 2 14 3 2 1 14 3 2 14b a a a a 1a b a a a 1a a b a a 1a a a b a
1a a a a 1++++= V d 2. Tnh nh thc a) 0 x x x x 1x 0 x x x 1x x 0 x
x 1x x x 0 x 1x x x x 0 11 1 1 1 1 06= b) a x x x x xx a x x x xx x
a x x xx x x a x xx x x x a xx x x x x a6= Gii: a) Nu x = 0, khai
trin nh thc theo dng 1, suy ra06= Nu x 0, nhn ct 1, dng 1 vi x, ri
cng cc dng vo dng 1v t nhn t chung (n -1) ra ngoi ta c: 0 x x x x
xx 0 x x x xx x 0 x x xx x x 0 x xx x x x 0 xx x x x x x.x50 x x x
x xx 0 x x x xx x 0 x x xx x x 0 x xx x x x 0 xx x x x x 0.x12 26=
= Nhn dng 1 vi (-1) ri cng vo cc dng khc ta c: 3 52 26x 5 ) x ( x
.x5x 0 0 0 0 00 x 0 0 0 00 0 x 0 0 00 0 0 x 0 00 0 0 0 x 0x x x x x
x.x5 = == b) Cng cc ct vo ct 1, ri t nhn t chung ra ngoi du nh thc
ta c ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 23 [ ]a
x ... x x 1x a ... x x 1... ... ... ... ... ...x x ... a x 1x x ...
x a 1x x ... x x 1. x 5 aa x ... x x x 5 ax a ... x x x 5 a... ...
... ... ... ...x x ... a x x 5 ax x ... x a x 5 ax x ... x x x 5
a6+ =+++++= Nhn dng 1 vi (-1) v cng vo cc dng 2, dng 3, , dng n ta
c [ ] [ ]6n) x a .( x 5 ax a 0 ... 0 0 00 x a ... 0 0 0... ... ...
... ... ...0 0 ... x a 0 00 0 ... 0 x a 0x x ... x x 1. x 5 a + =+
= ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 24 3. Ma
trn nghch o Trongphnnychngtaxemxtkhinimmatrnnghchocamatrn vung cp
n, iu kin tn ti v cch tm ma trn nghch o 1. nh thc ca tch hai ma trn
vung Cho hai ma trn vung cp n : A = [aij]n x n; B = [bij]n x n nh l
1. nh thc ca tch hai ma trn vung bng tch cc nh thc ca ma trn thnh
phn: det(AB)= det(A)det(B) V d 1. Cho A, B l ma trn vung cp 3 c
det(A) = 2, det(B) = -2. Tnh det(AB), det(A2B); det(2AB). 2. nh
ngha ma trn nghch o nh ngha 1. Cho A l ma trn vung cp n v E l ma
trn n v cp n. Nu c ma trn vung B cp n sao cho A.B = B.A = E th ta
ni ma trn A l kh nghch v B c gi l ma trn nghch o ca ma trn A (hayA
c ma trn nghch o l B), v k hiu A-1 = B. Vd2.a)MatrnA= ((
4 00 1lkhnghchvcmatrnnghchol (((
=4100 1A1. V ta c((
=((
(((
=(((
((
1 00 14 00 1.4100 14100 1.4 00 1. b)Matrn ((
= 0 00 0khngkhnghchvmimatrnvungBcp2uc E . B B . = = . S duy nht
ca ma trn nghch o nh l 2. Ma trn nghch o A-1 ca ma trn vung A nu tn
ti th duy nht 3. S tn ti ca ma trn nghch o nh l 3. Ma trn vung A kh
nghch khi v ch khi det(A) 0. ThS Phng Duy Quang Trng Khoa C bn Trng
i hc Ngoi Thng 25 vA-1 = 1det(A).A = 1det(A). 11 21 n112 22 n21n 2n
nnA A AA A AA A A ( ( ( ( ( LLM M O ML
V d 3. Tm A-1 ca ((((
=((
=1 0 04 1 01 2 1B ;6 23 1AT khi nim v iu kin kh nghch ca ma trn,
ta c mt s tnh cht sau: nh l 4. Gi s A, B l cc ma trn vung cp n. i)
Nu A kh nghch th A-1, AT, kA (k 0), Am (m nguyn dng) cng kh nghch v
(A-1)-1 = A ; (AT)-1 = (A-1)T ; 1 1Ak1) kA ( =; (Am)- 1 = (A-1)m
ii) Nu A, B kh nghch th AB cng kh nghch v (AB)-1 = B-1A-1 iii) Nu A
kh nghch th cc phng trnh A.X = C, X.A = C c nghim duy nht C A X C X
. A1 = =1A . C X C XA= =V d 4. Tm (A2)-1 vi ((
=6 23 1A4. Mt s phng php tm ma trn nghch o a) Phng php nh thc Da
vo nh l 2.12, ta c cc bc tm ma trn nghch o ca ma trn A = [aij]nn nh
sau: Bc 1: Tnh det(A)Nu det(A) = 0 th A khng kh nghch. Nu det(A) 0
th A c ma trn nghch o. Bc 2: Tm ma trn ph hp ca A: A=11 21 n112 22
n21n 2n nnA A AA A AA A A ( ( ( ( ( ( LLM M O ML ThS Phng Duy Quang
Trng Khoa C bn Trng i hc Ngoi Thng 26 trong Aij l phn b i s ca ai
j. Bc 3: Tnh B = 1Adet(A). Khi , ma trn B chnh l ma trn nghch o ca
ma trn A, tc lA-1 = B V d 1. Tm ma trn nghch o ca ma trn a) 1 2A3 4
(= ( b) ((((
=8 0 13 5 23 2 1AGii.a)Bc 1: Ta c det(A) = 1.4 2.3 = -20 .Nn ma
trn A kh nghch vA .) A det(1A1=
Bc 2: Ta lp ma trn ph hpA ca ma trn A. Ta c A11 = (-1)1+ 1.4 =
4; A12 = (- 1)1+ 2. 3 = - 3; A21 = (- 1)2 + 1.2 = - 2; A22 =(- 1)2
+ 2.1 = 1 Nn ((
=1 32 4ABc 3. Tnh ma trn nghch o (((
=((
= =21231 21 32 4.21A .) A det(1A1 Vy (((
=21231 2A1 b)Bc 1. Ta c det(A) = -1 0 nn A kh nghch vA .) A
det(1A1= Bc 2. Ta lp ma trn ph hpA ca ma trn A. Ta c 50 15 2. ) 1 (
A ; 138 13 2. ) 1 ( A ; 408 03 5. ) 1 ( A2 2222 1121 111 = = = = =
=+ + + 20 12 1. ) 1 ( A ; 58 13 1. ) 1 ( A ; 168 03 2. ) 1 ( A3
2232 2221 221= = = = = =+ + + ThS Phng Duy Quang Trng Khoa C bn
Trng i hc Ngoi Thng 27 15 22 1. ) 1 ( A ; 33 23 1. ) 1 ( A ; 93 53
2. ) 1 ( A3 3332 3321 331= = = = = =+ + + Nn ((((
=1 2 53 5 139 16 40ABc 3. Tnh ma trn nghch o ((((
= =1 2 53 5 139 16 40A) A det(1A1 Vy ((((
=1 2 53 5 139 16 40A1 V d 2. Gii phng trnh ma trn sau a) ((
=((
2 9 51 5 3X .4 32 1b) ((((
=((((
1 11 00 1X .8 0 13 5 23 2 1 Gii: a) Ma trn ((
=4 32 1Akh nghch nn phng trnh c nghim duy nht (((
=((
(((
=((
((
=253 20 1 12 9 51 5 3.21231 22 9 51 5 3.4 32 1X1 b) Ma
trn((((
=8 0 13 5 23 2 1Akh nghch nn phng trnh c nghim duy nht ((((
=((((
((((
=((((
((((
=3 68 1625 491 11 00 1.1 2 53 5 139 16 401 11 00 1.8 0 13 5 23 2
1X1 b)Phng php kh Gause-Jordan (Phng php bin i s cp) Thc t ta s p
dng ng thi cc php bin i s cp v dng a A v E v a E v ma trn A-1. T ,
ta c quy tc tm ma trn nghch o bng php bin i s cp (phng php Gauss
Jordan): Bc 1: Vit ma trn n v E cng cp vi ma trn A bn cnh pha phi
ma trn A c ma trn mi k hiu (A|E) ThS Phng Duy Quang Trng Khoa C bn
Trng i hc Ngoi Thng 28 Bc2: Dng cc php bin i s cp trn dng i vima
trn mi ny a dn khi ma trn A v ma trn n v E, cn khi ma trn E thnh ma
trn B, tc l (A|E) (E|B). Khi , B chnh l ma trn nghch o ca A. V d 1.
Tm ma trn nghch o ca ma trn:((((
=8 0 13 5 23 2 1AGii Bc 1: Lp ma trn (A|E) =((((
1 0 00 1 00 0 18 0 13 5 23 2 1 Bc 2: Bin i s cp ((((
((((
((((
+ + + 1 2 50 1 20 0 11 0 03 1 03 2 11 0 10 1 20 0 15 2 03 1 03 2
11 0 00 1 00 0 18 0 13 5 23 2 13 2 2 13 1d d 2 d d 2d d ((((
((((
+ ++ 1 2 53 5 139 16 401 0 00 1 00 0 11 2 53 5 133 6 141 0 00 1
00 2 11 2 32 31 3d d 2 dd d 3d d 3.Vy ((((
=1 2 53 5 139 16 40A1 ThS Phng Duy Quang Trng Khoa C bn Trng i
hc Ngoi Thng 29 4. Hng ca ma trn 1. Khi nim Chomatrn[ ] n} {m, min
k 1 ; a An x mij = .Trcht,tanhclikhinimnh thc con cp k ca ma trn
A.Ly ra k dng v k ct khc nhau . nh thc ca ma trn cp k c cc phn t
thuc giao im ca k dng v k ct c gi l nh thc con cp k ca A , k hiu: (
) n j ... j j 1 ; n i ... i i 1 Dk 2 1 k 2 1j ... j ji ... i ik 2
1k 2 1 < < < < < < trong i1, i2 , , ik l ch s ca
cc dng v j1, j2, , jk l ch s ca cc ct ly ra.V d 1. Cho ma trn
((((
=12 9 2 38 6 3 24 3 1 1A . Cc nh thc con cp 1 ca A chnh l cc phn
t ca A. Ccnhthcconcp2caA,chnghntobiccdng1,2vct1,3l 06 23 1D1312= =
; to bi dng 2, 3 v ct 2, 4 l2012 28 3D2423= = , ... Cc nh thc con
cp 3, chng hn to bi cc dng 1, 2, 3 v ct 1, 3, 4 l 012 9 38 6 24 3
1D134123= = ; to bi dng 1, 2, 3 v ct 1, 2, 4 l012 2 38 3 24 1
1D134123== ; ... nh l 1. Trong ma trn A, nu mi nh thc con cp k ca A
bng 0 th mi nh thc con cp cao hn k cng bng 0.
nhngha1.ChomatrnAcpmxn:A=[aij]mxn .Cpcaonhtcacc nh thc con khc 0 ca
ma trn A c gi l hng ca ma trn A, k hiu r(A) (rank(A)). Nu r(A) = r
th cc nh thc con cp r khc 0 ca A c gi l nh thc con c s ca A. Quy c:
0 }) r({ = Ch 1. T nh ngha, ta suy ra cc tnh cht n gin sau ThS Phng
Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 30 i)n} {m, min ) A (
r 0 ii) r(A) = r(AT) iii) Nu A l ma trn vung cp n th* r(A) = n0 A
hay A khng suy bin * r(A) < n0 A= hay A suy bin V d 2. Tm hng ca
ma trn ((((
=12 9 2 38 6 3 24 3 1 1AGii: Ta c nh thc con cp 2:2012 28
3D2423= = 0 nn r(A) 2. Xt cc nh thc con cp 3: c tt c4 C34=nh thc
con cp 3 ca A 09 2 36 3 23 1 1D123123== ; 012 9 38 6 24 3 1D134123=
= ; 012 2 38 3 24 1 1D124123== ;012 9 28 6 34 3 1D234123==Hay mi nh
thc con cp 3 ca A bng 0. Do r(A) = 2. V d 3. Tm hng ca ma trn sau:
((((((((((
=+++0 ... 0 0 ... 0 0... ... ... ... ... ... ...0 ... 0 0 ... 0
0a ... a a ... 0 0... ... ... ... ... ... ...a ... a a ... a 0a ...
a a ... a aAn r 1 r r rrn 2 1 r 2 r 2 22n 1 1 r 1 r 1 12 11
vi a11a22 arr 0. Gii: ThS Phng Duy Quang Trng Khoa C bn Trng i
hc Ngoi Thng 31 Tacnhthcconcpr :0 a ... a aa ... 0 0... ... ...
...a ... a 0a ... a aDrr 22 11rrr 2 22r 1 12 11r ... 12r ... 12 = =
vminh thc cp cao hn r u cha t nht mt dng ton s khng nn nh thc bng
0. Do vy, r(A) = r.T v d ny ta c kt qu sau: nh l 2 (i) Cc php bin i
s cp khng lm thay i hng ca ma trn ii) Hng ca ma trn dng bc thang
bng s dng khc khng ca ma trn nh l 3 (i) Nu A v B l hai ma trn cng
cp mn bt k, ta lun c: ) B ( r ) A ( r ) B A ( r + +(ii) Vi A v B l
hai ma trn bt k sao cho AB tn ti, ta lun c:) A ( r ) AB ( r v) B (
r ) AB ( r hay} r(B) {r(A), min ) AB ( r (iii) Nu A l ma trn cp m x
n, B l ma trn vung cp n x p thr(A) + r(B) r(AB) + n H qu : Nu A, B
l cc ma trn vung cp n th ta c) AB ( r n ) B ( r ) A ( r + +2. Cc
phng php tm hng ca ma trn a) Phng php nh thc Trc ht, ta chng minh
kt qu: nh l 4. Cho ma trn A = [aij]mxn c mt nh thc con cp r khc 0 l
Dr. Nu mi nh thc con cp r + 1 cha Dr u bng 0 th hng ca A bng r. T
nh l ny, ta c phng php tm hng ca ma trn nh sau: Bc 1: Tm mt nh thc
con cp Dk khc 0 cp k ( { } n , m min k 0 < < ). ThS Phng Duy
Quang Trng Khoa C bn Trng i hc Ngoi Thng 32 Bc 2: Ta tnh cc nh thc
cp k + 1 cha Dk (nu c). Trng hp 1: Nu cc nh thc cp k + 1 u bng 0 th
ta kt lun r(A) = k. Trng hp 2: Nu c mt nh thc cp k + 1 khc 0 th ta
li tnh cc nh thc cpk 2 +cha nh thc cpk 1 +khc 0 ny (nu c).Qu trnh c
tip tc nh vy ta tm c hng ca A . V d 1. Tm hng ca ma trn ((((
=12 9 2 38 6 3 24 3 1 1AGii: Ta c nh thc con cp 2:53 21 1D1212==
0 nn r(A) 2. Xt cc nh thc con cp 3 cha 1212D : c 2 nh thc con cp 3
ca A cha 1212D09 2 36 3 23 1 1D123123== ; ; 012 2 38 3 24 1
1D124123== . Nh vy, mi nh thc con cp 3 cha 1212Du bng 0 nn r(A) =
2. a)Phng php bin i s cp T nh l trn, ta c phng php bin i s cp tm
hng ca A: Bc 1: S dng cc php bin i s cp a ma trn A v dng bc thang
B. Bc 2: m s dng khc khng ca B, s l hng ca A.V d 2. Tm hng ca ma
trn sau: ((((
=2 1 2 14 1 1 22 4 3 1AGii: Dng php bin i s cp a ma trn A v dng
bc thang ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 33
B0 0 0 00 7 7 02 4 3 10 5 5 00 7 7 02 4 3 12 1 2 14 1 1 22 4 3 1A23
22 12 1d71d d 5d d 2d d=((((
((((
((((
=++ + B l ma trn dng bc thang c 2 dng khc 0 nn r(A) = r(B) = 2 V
d 3. Tm m ma trn sau c hng b nht ((((
=m 7 1 11 3 1 13 2 1 1AGii:Ta bin i a ma trn A v dng bc thang Ly
dng 1 cng vo dng 2, dng 1 nhn vi (- 1) cng vo dng 3, ta c: ((((
++ 3 m 5 0 02 5 0 03 2 1 12 13 1d dd d Nhn dng 2 vi (- 1) cng vo
dng 3 ta thu c ma trn dng bc thang: ((((
++ 5 m 0 0 02 5 0 03 2 1 12 13 1d dd d T ma trn dng bc thang, ta
c r(A) nh nht bng 2 khi m 5 = 05 m = V d 4. Tm hng ca ma trn n3 02
1|||
\|((
Gii : Ta c ((
=3 02 1A l ma trn vung cp 2 nn An cng l ma trn vung cp 2. Theo
nh l nhn nh thc ta c det(An) = [det(A)]n = 3n 0. Nn r(An) = 2 ThS
Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 34 Chuyn 2. H
phng trnh tuyn tnh v ng dng 1. Khi nim h phng trnh tuyn tnh 1. Cc
khi nim nh ngha 1. H phng trnh tuyn tnh gm m phng trnh n n l h c
dng 11 1 12 2 1n n 121 1 22 2 2n n 2m1 1 m2 2 mn n ma x+ a x+ ... +
a x= ba x+ a x+ ... + a x=
b..................................................a x+ a x+ ... +
a x = b (I) trong aij(i 1, m; j 1, n) = = , bi(i 1, m) =l cc s thc
cho trc; x1, x2, , xn l n n s cn tm; cc bi(i 1, m) = c gi l cc h s
t do. Nuh(I)csphngtrnhbngsn(m=n)thh(I)cgilh vung.
Nub1=b2==bm=0thh(I)cgilhphngtrnhtuyntnh thun nht.
Nghimcah(I)lmtbns(c1,c2,,cn)saochokhithaythx1 = c1, x2 = c2, , xn =
cn vo (I) th ta c m ng nht thc. C th vit nghim di cc dng sau: (c1,
c2, , cn) hoc (((((
n21c...cc. Gii h (I) l ta i tm tt c cc nghim ca h (I).Ta gi ma
trn A = 11 12 1n21 22 2nm1 m2 mna a... aa a... a...aa... a ( ( ( (
( l ma trn cc h s ca h (I). Ma trn ThS Phng Duy Quang Trng Khoa C
bn Trng i hc Ngoi Thng 35 %A= (((((
m mn 2 m 1 m2 n 2 22 211 n 1 12 11b : a ... a a... ... ... ...
...b : a ... a ab : a ... a a c gi l ma trn b sung ca h (I). nh
ngha 2. Hai h phng trnh c gi l tng ng nu chng c cng tp nghim. Cc
php bin i ca h phng trnh m khng lm thay i tp nghim ca h c gi l php
bin i tng ng ca h phng trnh. Trong qu trnh gii h phng trnh, chng ta
thng dng cc php bin i sau : - i ch hai phng trnh ca h cho nhau -
Nhn hai v ca mt phng trnh vi mt s khc 0 - Nhn hai v ca mt phng trnh
vi mt s tu ri cng vo phng trnh khc v theo v. Ch 1. Cc php bin i tng
ng ca h phng trnh trn chnh l cc php bin i s cp v dng i vi ma trn b
sung ca h . 2. Dng ma trn, dng vc t ca h phng trnh tuyn tnh t X =
12nxx...x ( ( ( ( ( : ma trn ct n,B = 12mbb...b ( ( ( ( ( : ma trn
ct h s t do Khi h phng trnh tuyn tnh (I) c biu din di dng ma trn
A.X =B (II) tn ,.., 1 j ;a...aaCnjj 2j 1j=((((((
= l cc ct ca ma trn A. Khi h (I) c vit di dng x1C1 + x2C2 +
+xnCn = B(III) Nh vy, mt h phng trnh tuyn tnh c th vit tng ng di
dng ma trn hoc dng vc t. ThS Phng Duy Quang Trng Khoa C bn Trng i
hc Ngoi Thng 36 V d 1. Cho h phng trnh tuyn tnh 3 phng trnh 4 n = +
= + + += + 3 x x 3 x x2 x x 2 x x 21 x x x 2 x4 3 2 14 3 2 14 3 2 1
Hcccnlx1,x2,x3,x4nnmatrnnsl (((((
=4321xxxxX;matrnvphi l((((
=321B . Ma trn h s ca h l ((((
=1 3 1 11 2 1 21 1 2 1A; cc vct ct ca A l ((((
=121C1 ; ((((
=112C2; ((((
=321C3 ; ((((
=111C4 ;matrnbsungcah : ((((
=1 : 1 3 1 12 : 1 2 1 21 : 1 1 2 1A~ Khi h c dng ma trn v dng vc
t tng ng l A. X = B ; x1C1 + x2C2 + x3C3 = B 3. H c dng tam gic, h
c dng bc thang nh ngha 3. H n phng trnh, n n c dng== + += + + +n n
nn2 n n 2 2 221 n n 1 2 12 1 11b x a...b x a ... x ab x a ... x a x
a vi a11a22ann 0 c gi l h c dng tam gic Ma trn h s A ca h chnh l ma
trn dng tam gic trn. D thy rng h ny c nghim duy nht. H ny gii bng
cch gii t phng trnh th n tm n xn, ri gii phng trnh th n -1 tm n
xn-1, , qu trnh c tip tc cho n khi tm cnx1.Cchgii
nygilgiingctdilntrntm nghimcah phng trnh. ThS Phng Duy Quang Trng
Khoa C bn Trng i hc Ngoi Thng 37 V d 2. Gii h sau : == = + 1 x2 x 2
x1 x x 2 x33 23 2 1 Gii: T phng trnh th 3 ta c x3 = -1 ; thay vo
phng trnh th 2 ta c x2 = 2 + 2x3 = 0 ; thay x2, x3 vo phng trnh th
nht ta c : x1 = 1 + 2x2 x3 Vy h c nghim duy nht : ((((
=201Xnh ngha 4. H r phng trnh, n n s (r < n) c dng = + += + +
+ += + + + + +r n rn r rr2 n n 2 r r 2 2 221 n n 1 r r 1 2 12 1 11b
x a .... x a....b x a ... x a ... x ab x a ... x a ... x a x a vi
a11a22... arr 0c gi l h dng bc thang Ma trn b sung ca h khi s c bc
thang ((((((((
=0 : 0...b : a.........0...a.........0...00...0... ... ... ...
... ...b : a ... a ... a 0b : a ... a ... a aA~r rm rr2 n 2 r 2 221
n 1 r 1 12 11 Vi h c dng bc thang, t phng trnh th r, ta tnh xr thng
qua cc n xr+1, xr+2, ... , xn. Ri thay vo phng trnh th r -1 tnh xr1
theo cc n xr+1, xr+2, ... , xn qu trnh trn c tip tc cho n x2, x1. V
d 3. Gii h phng trnh = = += + +1 x 2 x2 x x 2 x1 x 2 x x 2 x4 34 3
24 3 2 1 Gii: T phng trnh th 3, ta c x3 = 2x4 1, thay vo phng trnh
2 ta cx2 = - 2(2x4 -1) + x4 + 2 = -3x4 + 4 Cui cng, thay vo phng
trnh th nht thu c x1 = -2x2 + x3 2x4 + 1 = - 2(-3x4 + 4) + 2x4 1
2x4 + 1 = 6x4 8 Vy h c nghim X = (6k 8; - 3k + 4; 2k -1); k tu .
ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 38 2. Phng
php gii h phng trnh 1. iu kin tn ti nghim nh l 1. (nh l
Kronecker-Capeli). H phng trnh tuyn tnh (I) c nghim khi v ch khi
r(A) = r( %A ). V d 1. Tm m h sau c nghim = + += + + = += + +m x x
6 x 8 x 33 x 2 x 4 x 5 x 22 x x 2 x 3 x1 x x x 2 x4 3 2 14 3 2 14 3
2 14 3 2 1 GiiTa c ma trn b sung ca h l (((((
=m : 1 6 8 33 : 2 4 5 22 : 1 2 3 1:! 1 1 2 1A~ Bin i s cp a ma
trn b sung v dng bc thang (((((
(((((
=+ + + 1 m : 0 0 0 04 : 2 1 0 03 : 2 1 1 01 : 1 1 2 1m : 1 6 8
33 : 2 4 5 22 : 1 2 3 11 : 1 1 2 1A~3 2 12 14 3 2d d dd dd d d T y
ta c h c nghim1 m 0 1 m ) A~( r ) A ( r = = = 2. iu kin duy nht
nghim nh l 2.H phng trnh tuyn tnh (I) c nghim duy nht khi v ch
khir(A) = r( %A ) = s n (= n) Chng minh H c nghim duy nhttn ti duy
nht cc s x1o, x2o, ... , xno sao cho x1o C1 + x2oC2 + +xnoCn = B {
} ( ) n ) A~( r ) A ( r n C , ... , C , C rn 2 1= = = (pcm) V d 2.
Tm m h sau c nghim duy nht = + += + += + +23 2 13 2 13 2 1m x x mxm
x mx x1 mx x x ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi
Thng 39 Gii: H c s phng trnh bng s n v bng 3 nn c nghim duy nht0 A
3 ) A ( r = Ta c 2) 1 m )( 2 m (1 1 m1 m 1m 1 1A + = =Nn h c nghim
duy nht 2 m1 m 3. Phng php gii h phng trnh tuyn tnh a)Phng php gii
h Cramer nhngha5.HCramerlhnphngtrnhtuyntnhnn(hvung)cma trn h s A
khng suy bin (det(A) 0). nh l 3. H Cramer lun c nghim duy nht. Cng
thc nghim: xj = Aj (j = 1, n) trong , j l ma trn nhn tAbng cch thay
ct th j bi ct h s t do. V d 3. Gii h sau = += += +4 x x 35 x x7 x 2
x 3 x 43 12 13 2 1 Gii Ta c nh thc ca ma trn h s A l0 71 0 30 1 12
3 4A ==nn h l h Cramer. Do c nghim duy nht: Ax ;Ax ;Ax312111===M284
0 35 1 17 3 4; 351 4 30 5 12 7 4; 01 0 40 1 52 3 73 2 1= = == == Vy
h c nghim (x1= 0/7 =0; x2 = 35/7 = 5; x3 = 28/7 = 4) ThS Phng Duy
Quang Trng Khoa C bn Trng i hc Ngoi Thng 40 V d 4. Tm iu kin h sau
c nghim duy nht = + += + += + +23 2 13 2 13 2 1m mx x xm x mx x1 x
x mx Gii Nhnxt:Cchtrnlcchcs phng trnhbngsn. Nn hcnghim duy nht0 A
Ta c 2) 1 m )( 2 m (m 1 11 m 11 1 mA + = =Nn h c nghim duy nht 1 m2
m b). Phng php gii h tng qut Gi s ta gii h tng qut m phng trnh tuyn
tnh, n n dng: 11 1 12 2 1n n 121 1 22 2 2n n 2m1 1 m2 2 mn n ma x+
a x+ ... + a x= ba x+ a x+ ... + a x=
b..................................................a x+ a x+ ... +
a x = b Cch gii:Tnh r(A), r( %A ) So snh r(A) vi r( %A ): +Nu r(A)
r( %A ) th h (I) v nghim. +Nu r(A) = r( %A ) = s n (= n) th h (I) c
nghim duy nht cho bi cng thc Cramer xj = Aj (j = 1, n) +Nu r(A) =
r( %A ) = r < n th h (4.1) c v s nghim (hay cn gi l h v
nh):ChramtnhthcconcscamatrnA,gis ta chra mtnh
thcconcslDr.Khihphngtrnhchotngngvihgmr phng trnh ca h cho m c h s
ca cc n to nn Dr. r phng trnh ny ThS Phng Duy Quang Trng Khoa C bn
Trng i hc Ngoi Thng 41 c gi l cc phng trnh c bn ca h (I). r n ca h
(I) c h s to thnh r ct ca Dr c gi l cc n c bn, (n r) n cn li c gi l
cc n t do. Ta gii h gm r phng trnh c bn v r n c bn bng cch chuyn (n
r) n t do sang vphivcoinhlccthams, tachCramer.GiihCramer,ta c cng
thc biu din r n c bn qua (n r) n t do. c) Phng php Gauss Ni dung ca
php kh Gauss nh sau: Ta s dng cc php bin i s cp v hng a ma trn b
sung %A v dng ma trn bc thang. Khi h phng trnh cho tng vi h phng
trnh mi c ma trn h s b sung l ma trn bc thang va thu c. Sau ta gii
h mi ny t phng trnh cui cng, thay cc gi tr ca n va tm c vo phng
trnh trn v c tiptcgiichonphngtrnhutintasthucnghimcahphng trnh cho.
V d 5. Gii h = + + = + = + = + +3 x 5 x 4 x 4 x1 x 3 x 2 x 2 x 32 x
x x 3 x1 x 4 x 3 x x 24 3 2 14 3 2 14 3 2 14 3 2 1 Gii Trc ht, tm
hng ca ma trn h s v ma trn b sung Ta c(((((
(((((
(((((
=+ + + 5 : 6 5 7 05 : 6 5 7 05 : 6 5 7 02 : 1 1 3 13 : 5 4 4 11
: 3 2 2 31 : 4 3 1 22 : 1 1 3 13 : 5 4 4 11 : 3 2 2 32 : 1 1 3 11 :
4 3 1 2A~2 14 13 12 1d d 2d dd d 3d d (((((
(((((
+ + + + + 0 : 0 0 0 00 : 0 0 0 05 : 6 5 7 02 : 1 1 3 15 : 6 5 7
05 : 6 5 7 05 : 6 5 7 02 : 1 1 3 13 24 22 14 13 1d dd dd d 2d dd d
3 Khi r(A) = r( A~) = 2 < n = 4. Do h c v s nghim ThS Phng Duy
Quang Trng Khoa C bn Trng i hc Ngoi Thng 42 Khi h cho = + = + 5 x 6
x 5 x 72 x x x 3 x3 3 24 3 2 1 Chn 4 2 = 2 n t do l x3, x4 v x1, x2
khi l n c bn.H tr thnh R x , x ;7x 6 x 5 5x7x 11 x 8 1x5 x 6 x 5 x
72 x x x 3 x4 34 324 313 3 24 3 2 1 += += = + = Vy h c v s nghim) R
; ( ; ;76 5 5;711 8 1X ||
\| + +=V d 6. Gii v bin lun h phng trnh sau = + += + += + +1 x x
x1 mx mx x1 mx x mx3 2 13 2 13 2 1 Gii Nhn xt: y l h c 3 phng trnh,
3 n s c ma trn b sung l((((
=1 : 1 1 11 : m m 11 : 1 1 mA~ Ta c 2) 1 m (1 1 1m m 1m 1 mA =
=* Nu1 m th A 0 nn h l h Cramer nn c nghim duy nht. Mt khc ta li
c232221) 1 m (1 1 11 m 11 1 m; ) 1 m (1 1 1m 1 1m 1 m; ) 1 m (1 1
1m m 1m 1 1 = = = = = = Do , nghim ca h l |||
\| == == == 1Ax ; 1Ax ; 1Ax312211 * Nu m = 1 khi ma trn b sung c
dngThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 43
((((
((((
=0 : 0 0 00 : 0 0 01 : 1 1 11 : 1 1 11 : 1 1 11 : 1 1 1A~ Suy ra
r(A) =3 n 1 ) A~( r = < =nn h c v s nghim Hx1 + x2 + x3 = 1R x ,
x ; x x 1 x2 1 2 1 3 = Vy h c nghim X =( ) R , ; ; ; 1 4. H phng
trnh tuyn tnh thun nht a) Khi nim nh ngha 6. H phng trnh tuyn tnh
thun nht l h phng trnh tuyn tnh c cc h s t do bng khng, tc l h c
dng 11 1 12 2 1n n21 1 22 2 2n nm1 1 m2 2 mn na x+ a x+ ... + a x=
0a x+ a x+ ... + a x=
0..................................................a x+ a x+ ... +
a x= 0 (IV)
Nhn xt.+ H (IV) lun c nghim l x1 = x2 = = xn = 0. Nghim ny c gi
l nghim tm thng ca h (IV).+ H (IV) c nghim duy nht ( l nghim tm
thng) r(A) = n. + H (IV) c nghim khng tm thng r(A) < n. + H thun
nht vung (h c s phng trnh bng s n) c nghim khng tm thng khi v ch
khi det(A) = 0. b)Cu to nghim ca h phng trnh tuyn tnh thun nht
nhl4.Tpttcccnghimcahthunnhtnn(IV)lmtkhnggian vect con ca khng gian
n .Hn na, nu h thun nht c ma trn h s l A th s chiu ca khng gian
vect con ny ln r, vi r l hng ca ma trn A.Khng gian vect ny c gi l
khng gian nghim ca h (IV).
nhngha7.Micscakhnggiannghimcahphngtrnhtuyntnh thun nht c gi l mt h
nghim c bn ca n. ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi
Thng 44 Nhn xt.Mt h nghim ca h thun nht (IV) l mt h nghim c bn ) +
H c ng n - r(A) nghim+) H c lp tuyn tnh
Gis{x(1)=(x(1)1,x(1)2,,x(1)n),x(2)=(x(2)1,x(2)2,,x(2)n),,x(n-r)=(x(n
r )1, x(n r )2, , x(n r )n)} l mt h nghim c bn ca h (IV).Khi mi
nghim x = (x1, x2, , xn) ca h (IV) u c dng l t hp tuyn tnh ca x(1),
x(2), , x(n-r):x = n r(k)kk 1x=(*) Cng thc (*) vi 1, 2, , n-r tu c
gi l cng thc nghim tng qut ca h thun nht (IV).V d 1. Gii v bin lun
h phng trnh sau 3x + y + 10z = 02x + ay + 5z = 0x + 4y + 7z = 0 Gii
y l h thun nht vung. Ta cdet(A) = 3 1 102 a51 47 = 11(a + 1)+ Vi a
-1, det(A) 0 nn h cho ch c nghim tm thung x = y = z = 0. + Vi a=
-1, h cho tr thnh 3x + y + 10z = 02x -y + 5z = 0x + 4y + 7z = 0 V 3
12-1 = - 5 0 nn h cho tng ng vi h3x + y + 10z = 02x -y + 5z = 03x +
y = -10z 2x -y = -5z 5x = -15z x = -3z y = 2x + 5z = - z Vy, vi a =
-1, h cho c v s nghim l ThS Phng Duy Quang Trng Khoa C bn Trng i hc
Ngoi Thng 45 x = -3ty = -tz = t; t V d 2. Tm m h sau c nghim khng
tm thng v tm cc nghim = + += + = + 0 x ) m 1 ( x 3 x0 x mx x0 x x )
m 2 (3 2 13 2 12 1 Gii Cch 1. Dng phng php Gauss Ta c, ma trn h s
l((((
+ + ((((
((((
=2 m 3 m 5 m 3 0m 2 3 m 0m 1 3 10 1 m 21 m 1m 1 3 1m 1 3 11 m 10
1 m 2A2 * Nu m + 3 = 03 m = , ta c 3 ) A ( r1 0 02 4 0m 1 3 12 4 01
0 0m 1 3 1A = ((((
((((
nn h ch c nghim tm thng * Nu3 m ta c (((((((
((((
+ m 34 m 3 m0 0m 3m 21 0m 1 3 12 m 3 m 5 m 3 0m 3m 21 0m 1 3 1A2
3 2 h c nghim tm thng th m3 3m2 4 = 02 ) A ( r1 m2 m=
== .Khi , h c nghim tm thng +) Nu m = 2 th ((((
0 0 00 1 01 3 1AKhi h 11 2 3 *223x x 3x x 0x 0; x 0x = + = =
==
ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 46 +) Nu m
= - 1 th ((((
0 0 0431 01 3 1AKhi h *11 2 322 33x4x 3x x 03x ; 34 x x 04x =+ =
= + = =
Cch 2.V h c s phng trnh bng s n nn ta tnh nh thc ma trn h s A. )
1 m ( ) 2 m (m 1 3 11 m 10 1 m 2A2+ = =* Nu 1 m2 m0 Ath h c nghim
tm thng duy nht *Nu
== =1 m2 m0 A .Khihcnghimkhngtmthngvsdng phng php Gauss ta tm c
nghim ca h . V d 3. Tm h nghim c bn ca h sau = = + = + += + + + 0 x
x 3 x 2 x0 x x 4 x x 30 x 2 x x 3 x 20 x x 3 x 2 x4 3 2 14 3 2 14 3
2 14 3 2 1 Gii Ta c ma trn h s cc n (((((
(((((
(((((
=+ +++ 0 0 0 00 0 0 04 5 7 01 3 2 10 0 0 04 5 7 04 5 7 01 3 2 11
3 2 11 4 1 32 1 3 21 3 2 1A2 1 2 14 13 1d d d d 2d dd d 3 Suy ra
r(A) = 2 < n = 4 nn h c v s nghim. Chn x3, x4 lm n t do v x1, x2
l n c bn. Khi , h tr thnh ThS Phng Duy Quang Trng Khoa C bn Trng i
hc Ngoi Thng 47 =+ = + + == + += + + + 4 3 24 3 4 3 2 14 3 24 3 2
1x74x75xx73x711x x 3 x 2 x0 x 4 x 5 x 70 x x 3 x 2 x (x3, x4 tu )
VkhnggiannghimScahcdimS=2nnhnghimcbnc2vct nghim Chn x3 = 7, x4 = 0
ta c 1 vc t nghim : X1 = (11 ; - 5 ; 7 ; 0) Chn x3 = 0, x4 = 7 ta c
1 vc t nghim : X2 = (3 ; - 4 ; 0 ; 7) Vy mt h nghim c bn ca h l
{X1, X2} ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 48
TON CAO CP 2 Chuyn 3. Gii hn, lin tc, vi tch phn hm mt bin s 1. Gii
hn ca dy s I. Cc khi nim c bn v hm mt bin s 1) nh ngha:nh x f : X R
xa y=f x ( )vi, X R Xc gi l hm s. * Ta gi X l tp xc nh ca hmf x ( )
, k hiu D(f); f(X) l tp gi tr ca hm f, k hiu R(f); xD(f) gi l bin c
lp. V d 1: Hmy = 24 x c= =f fD [- , ], R [ , ] 2 2 0 4 . * Hm chn :
Gi s X R , X nhn gc O lm tm i xng. Hm sf x ( )c gi l chn nu = f x f
x ( ) ( ) x X , l hm l nu = f x f x ( ) ( ) x X .
*Ch:thhmschnnhntrctunglmtrcixng.thhmsl nhn gc ta O lm tm i xng. *
Hm l :Hm s f x ( )c gi l hm s tun hon nu T > 0 sao cho ff( x T )
f ( x ), x D + = S T nh nht sao cho c ng thc trn c gi l chu k ca hm
sf x ( ) . V d 2: Cc hm s y = sinx, y = cosx l tun hon vi chu k 2 ;
cc hm s y = tgx, y = cotgx l tun hon vi chu k . Hm n iu:Hmy f x = (
)c gi l tng (tng ngt) trn khong I Df nu x1, x2 I, x1 < x2 th
f(x1) f(x2) (f(x1) < f(x2)); gim (gim ngt) trn I nux1, x2 I th
(x1) f(x2) (f(x1) > f(x2)). Hm s tng hoc gim trn I c gi l hm n
iu trn I. Hm b chn:Cho hm s f(x) xc nh trn X. Hm s f(x) c gi l b
chn trn nuM : f( x ) M x X ; b chn di nuM : f( x ) M x X ; b chn nu
M : f( x ) M , x X . 2) Hm s hp: nh ngha.Cho X, Y, Z R, cho hm s f
: X Y, g: Y Z. Khi , hm s h: X Z c nh ngha bi h(x):= g(f(x)), x X c
gi l hm s hp ca hm s f v g. K hiu l: hay o g[ f ( x )] ( g f )( x
), x X .ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 49 V
d 3. Xt cc hm s = + = + f ( x ) x , g( x ) x .22 1 4 Khi : = + = +
+= + = + +g[ f ( x )] f ( x ) ( x )f [g( x )] g( x ) ( x )2 224 2 1
42 1 2 4 1 3) Hm s ngc: * nh ngha. Cho hai tp X, Y R;cho hm sf: X Y
xay = f(x) Nu tn ti hm s g: Y X tho mn: + (go f)(x) = X1+ (f o
g)(y) = Y1th g(x) c gi l hm s ngc ca hm sf(x). K hiu: g = f-1
* Ch : 1.f: X Y l song nh g = f-1: Y X. Tc f c hm s ngc khi v ch
khi f l song nh. 2.Nu hm s y = f(x) n iu nghim ngt th n c hm ngc 3.
D1f = Rf , R1f = Df . 4.thcahmngcy=f-1(x)ixngvithhmsy=f(x)qua ng
phn gic ca gc th nht. V d 4. Tm hm ngc ca hm y = 4xGii. Ta c Df =
[0, +), Rf = [0, +). Hm y = 4xl hm tng nghim ngt trn Df nn n c hm
ngc. Rt x theo y, ta c: x y , y = 40 , i vai tr ca vx yta c: hm y
4x =c hm ngc l y = x4, x 0. 4) Cc hm s thng gp: . Cc hm s s cp c bn
* Hm s lu thay = x, l mt s thc cho trc
f fD R; R R = =* Hm s m: y = ax (a>0, a 1)
*f fD R; R R+= =* Hm s logarit: y = logax ( a > 0 v a1 )
*f fD R ; R R+= =* Cc hm s lng gic: +) Hm f(x) = sinx ThS Phng
Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 50 [ ]f fD R; R 1,1 =
= +)Hm y = cosx [ ]f fD R; R 1,1 = = +) Hm y = tgx
f fD x R/ x k , k Z , R R2 = + = ` ) +) Hm y = cotgx { }f fD x R
/ x k , k Z , R R = =* Cc hm s lng gic ngc: +)Hm y = arcsinx l hm
ngc ca hm s y = sinx trn,2 2 ( ( c - Min xc nh Dy = [-1, 1] - Min
gi tr Ry =,2 2 ( ( . +) Hm y = arccosx l hm ngc ca hm s y = cosx
trn [ ]0,c - Min xc nh Dy = [-1, 1] - Min gi tr Ry = [ ]0, . +) Hm
y = arctgx l hm ngc ca hm s y = tgx trong,2 2 | | |\ c - Min xc nh
Dy =R- Min gi tr Ry = ,2 2 | | |\ .+) Hm y = arccotgx l hm ngc ca
hm s y = cotgx trong( ) 0,c - Min xc nh Dy =R - Min gi trRy =( ) 0,
. * Hm s s cp: Ta gi cchm ss cp l nhng hm s c to thnh bimt s hu hn
cc php ton s hc v php ton hp trn cc hms s cp c bn, v cc hng s. 5)
Mt s hm s kinh t thng gp trong kinh t * Hm cung v hm cu: Cc nh kinh
t s dng khi nim hm cung v hm cu biu din s ph thuc ca lng cung v lng
cu ca mt loi hng ha vo gi ca hng ha . Hm cung v hm cu c dng: Hm
cung: QS = S(p) ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi
Thng 51 Hm cu: QD = D(p) Trong p l gi hng ha; QS l lng cung: tc l
lng hng ha ngi bn bnglngbnmimcgi;QDllngcu:tcllnghnghangimua bng lng
mua mi mc gi. Khi xem xtm hnhhm cung, hm cu ni trn ta gi thit rngcc
yu t khc khng i. Hm sn xut ngn hn Cc nh kinh t hc s dng khi nim hm
sn xut m t s ph thuc ca sn lng hng ha (tng s lng sn phm hin vt) ca
mt nh sn xut vo cc yu t u vo, gi l yu t sn xut.
Khiphntchsnxut,tathngquantmnhaiyutsnxutquan trng l vn v lao ng c k
hiu tng ng l K v L. Vd:HmsnxutdngCobbDouglasvihaiyutvn(K)vlao ng
(L): = L aK QHm doanh thu, hm chi ph v hm li nhun
Hmdoanhthulhmsbiudinsphthuccatngdoanhthu(TR) vo sn lng (Q): TR =
TR(Q). Hm chi ph l hm s biu din s ph thuc ca tng chi ph sn xut (TC)
vo sn lng (Q): TC = TC(Q). Hm li nhun l hm s biu din s ph thuc ca
tng li nhun ( ) vo sn lng (Q):) Q ( = . Hm li nhun c th xc nh bi) Q
( TC ) Q ( TR = . II. Gii hn ca dy s 1.nh ngha. Ta ni rng dy s {xn}
c gii hn l a (hu hn ) nu vi mi s > 0 nh tu , tn ti mt s t nhinn0
sao chonx a = = < 0 0 Vy theo nh ngha nlim c c+= . V d:( )knlim
kn+= >10 0v khi nnlim q q+= < 0 1 ThS Phng Duy Quang Trng
Khoa C bn Trng i hc Ngoi Thng 52 Ch : +) sao chon nnx A n N x A n
n=+ > > 0 00 lim ,+)sao chon nnx A n N x A n n=+ > k 0ta c
khi khi knAAnA+ >= 0, n0N*sao cho n n0, k N*thn k nx x .+ <
ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 53 2. Gii hn
ca hm s 1. nh ngha: nh ngha 1.Cho hm s f(x) xc nh trong khong (a,
b). Ta ni rng f(x) c gi hn l A (hu hn)khixx0,nu
vimidy{xn}trong(a,b)\{x0}mn 0x x n khi th dy gi tr tng ng {f(xn)}
hi t n A. K hiu:0xlimxf(x) = A hay f(x) A khi x x0. nh ngha 2. Cho
hm s f(x) xc nh trong khong (a, b). Ta ni rng f(x) c gi hn l A (hu
hn) khi x x0, nu vi bt k > 0, tn ti s > 0 sao cho vi mi x (a,
b) tho mn 0 < |x x0| < th |f(x) A| < .. Ch : nh ngha 1 tng
ng vi nh ngha 22. Tnh chtCho hm s f(x) xc nh trn tp D Tnh cht 1.
Gii hn ca hm s f(x) khi x x0 nu c l duy nht.Tnh cht 2. Gi s tn ti
0xlimxf(x) = A, 0xlimxg(x) = B. Khi i) x af x g x A B = lim[ ( ) (
)]ii) x af x g x A B= lim[ ( ). ( )] .iii) x af x Ag x B =( )lim(
), (B 0). Tnh cht 3 Nu hm s s cp f(x) xc nh ti 0xth 00x xf x f x=
lim ( ) ( ) . Tnh cht 4 (Nguyn l kp):Nu g(x) f(x) h(x), ( )0 0x x x
+ ;vi> 0 v0x xlim g(x)=0xlimxh(x) = Ath 0x xlimf(x) = A.Tnh cht
5.Nu( )0 0; f(x) g(x),x x x , v i > 0 no + v 0 0x x x xlim f (x)
A, lim g(x) B = =thA B Tnh cht 6. Nu 0 0x x x xlim f (x) A 0 lim
g(x) B v = > = th[ ]0g( x)Bx xlim f (x) A= . 3. Gii hn mt pha
Khi nim x x0 cn c xt trong hai trng hp: Th1. x x0, x > x0: tc l
x dn n x0 t bn phi ( x x0+). ThS Phng Duy Quang Trng Khoa C bn Trng
i hc Ngoi Thng 54 Th2. x x0, x < x0: tc l x dn n x0 t bn tri (x
x0- ).Gii hn ca hm s f(x) khi x x0+ v khix x0- c gi tng ng l gii hn
bn phi v gii hn bn tri ca hm s ti im x0: Gii hn bn phi: 0x
xlim+f(x) = 00x xx xlim>f(x); Gii hn bn tri: 0x xlimf(x) = 00x
xx xlim 0), sinx, tgx l cc VCB khi x 0Hm s tgx l VCL khix2b) So snh
cc VCB Cho f(x), g(x) l cc VCB khi x a. Gi s x af (x)lim kg(x)= i)
Nu k = 0 th f(x) c gi l VCB bc cao hn so vi g(x) khi x a. K hiu
f(x) = 0(g(x)),ii) Nu k 0 v hu hn th f(x) v g(x) c gi l nhng VCB
cng bc khi x a. K hiu f = 0*(g) khi x a. c bit, khi k = 1, th f(x)
v g(x) c gi l nhng VCB tng ng khi x a v vit: f(x) g(x), x a. Nhn
xt:Nup q > > 0 th p qx x = 0( ) . Nu[ ] f x h x = 0 ( ) ( )
,[ ] g x h x = 0 ( ) ( )v a l hng s th i)[ ] a f x h x = 0 . ( ) (
)ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 55 ii) [ ] f
x g x h x = 0 ( ) ( ) ( ) iii)[ ] f x g x h x = 0 ( ). ( ) ( )V d 3
: Vi k>0, khix 0 ta c: k k k k p kpa x a x a x a x x+ + + ++ + +
+ =1 2 31 2 30 ... ( )V d 4: Cc hm sinx v x l hai VCB tng ng khi x
0 v 0lim xsinxx = 1. V d 5:Cc hm tg2x v sinx l hai VCB cng bc khi x
0, v 0lim x 02lim . .2 22 stg2xsinx inx = =xtg x xx c) nh l: nh l
1. Hm s f(x) c gii hn l L khi x a khi v ch khif x L x = + ( ) ( ) ,
vi mix ( ) l hm v cng b khi x a. nh l 2. Gi skhi x x0, ta c cc cp
VCB tng ng: (x) 1(x), (x) 1(x) Khi , nu x x( x )lim( x ) 011 tn ti
(hu hn hoc v hn) thx x x x( x ) ( x )lim lim( x ) ( x ) = 0 011 .V
d 6. Tnh0lim x23x + 2xsinx + tg x Gii. V x + 2x2 x, x 0 v sin2x +
tg5 x sin2x, x 0. p dng nh l trn, ta c 0lim x25x + 2xsin2x + tg x =
0lim x2xsin2x = 2. ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi
Thng 56 3. Hm s lin tc 1. nh ngha nh ngha 1. Cho hm s f(x) xc nh
trn tp hp Df, Ta ni hm s f(x) -Lin tc ti im x0Dfnu: 0lim x xf(x) =
f(x0). - Lin tc phi ti x0 nuf(x0+)= 0lim+ x xf(x) = .f(x0)- Lin tc
tri ti x0 nuf(x0-) =0lim x xf(x) =f(x0) Hm s khng lin tc tix0 th ta
ni hm s gin on tix .0 VD 1: Xt tnh lin tc ca hm s x khi xf ( x )x a
khi x+ 0, S(M, r) D.Tp hp D c gi l m nu mi im ca n u l im trong. nh
ngha 2. Cho D Rn,nh x f : D R xc nh bi M = (x1, x2, ..., xn) Daf(M)
= f(x1, x2, ..., xn)c gi l hm s ca n bin s xc nh trn D. +) D c gi l
min xc nh ca hm s f +) x1, x2, ..., xn c gi l cc bin s c lp.+) { f
(D) , , ) : , , )1 2 n 1 2 nR|(xx..., xDf(xx..., x=} = cgilmin gi
tr ca f. ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 80 V
d 1. Hm s)2 2 2f(x, y, z) = 1 - (x+ y+ z c tp xc nh l: ( ) 3 2 22 3
2 2 2 D = { (x, y, z) R / 1 -x+ y + z 0}= { (x, y, z) R / x+ y+ z
1}V d 2. Hm s f(x, y) = xsin y,2 22 22 2 - ysinx nu x+ y0x+ y0, nu
x+ y0= c min xc nh l: D = R2. 2 Cc php ton *Cc php ton s hc Cho D1,
D2 Rn, f: D1 R, g: D2 R. Ta nh ngha cc hm mi nh sau:(f g)(M) = f(M)
g(M), c min xc nh l D = D1 D2. (f g)(M) = f(M).g(M), c min xc nh l
D = D1 D2. fg| | |\ (M) = f (M)g(M), c min xc nh l D = {M D1 D2|
g(M) 0} * Hm hp Chohmsf(u1,u2,...,um)lhmmbinviminxcnh lDRmvu1 =
u1(x1, x2, ..., xn), u2 = u2(x1, x2, ..., xn), ...,um = um(x1, x2,
..., xn) l cc hm n bin vi min xc nh l X Rn sao cho vi mi (x1, x2,
..., xn) X th (u1, u2, ..., um) D. Khi , ta c hm hp:F(x1, x2, ...,
xn) = f(u1(x1, x2, ..., xn), u2(x1, x2, ..., xn), ...,um(x1, x2,
..., xn)) Ch . Trong tnh ton, ngi ta khng phn bit f v F, tc l ta c
th vit f(x1, x2, ..., xn) = f(u1(x1, x2, ..., xn), u2(x1, x2, ...,
xn), ...,um(x1, x2, ..., xn)) V d 3. f(u, v)=e2u2 - 2v, u=cosx, v =
2x2 + y f(x, y) = ec x )2 2 2os- 2(x+ y II. Gii hn hm nhiu bin 1.
Gii hn kp Gi s hm s w = f(M) = f(x1, x2, , xn) xc nh trn tp D *nh
ngha . S A c gi l gii hn ca hm f(M) khi M dn n M0 nu > 0, > 0
sao cho vi mi M D (c th tr im M0) v 0 < d(M0, M) < th |f(M)
A| < K hiu: 10 1 022 0nn 01M M x xx x...x xlim f (x , lim f (M)2
n
x , ..., x ) = Ahay = A ThS Phng Duy Quang Trng Khoa C bn Trng i
hc Ngoi Thng 81 * Tnh cht Tnh cht 1. Gii hn ca hm nhiu bin nu c l
duy nht. Tnh cht 2. 0M Mlim f (M) = A dy {Mk}, Mk D\{M0}, Mk k M0
f(Mk) k A.Tnh cht 3. 0M Mlim f(M) = A < A < r>0,M ( ) {
}0\ M0S(M , r)D
th < f(M) < Tnh cht 4.( ) { }0M M0limr (M ,0f(M) = A >
0,MSr) D\M : a < f(M) < b
a < A < b Tnh cht 5.00M MM Mlim f (M)lim g(M) = A = B 000M
MM MM Mlim (f (M)limf (M)lim (g(M) Bg(M) g(M)) = A B[f(M).g(M)] =
A.BA = 0, 0)B Tnh cht 6. (Nguyn l kp)
{ }0 0M M M Mlim limr ,0 0 g(M) = h(M) = A > 0,M S(Mr) \M :
g(M) f(M) h(M) 0M Mlimf(M) =A. V d 1. Tnh gii hn sau3x 0y 0xlim32 2
+ yx+ y Gii.Min xc nh D =R2\{(0, 0)}.V0 3x32 2 + yx+ y 3x2 2 x+ y +
32 2yx+ y 3x2 x + 3y2 y |x| + |y| M (x,y) (0,lim (| x | 0) + |y|) =
0 Do , theo Nguyn l kp, ta c: 3(x,y) (0,xlim32 2 0) + yx+ y = 0. V
vy3( x,y) (0,xlim32 2 0) + yx+ y = 0. V d 2. Tnh 2x 0y 0xlim22 2 -
yx+ y Gii. Min xc nh D = R2\{(0, 0)}.Xt hai dy imk1M ,k2 k | | ` |\
) v k1M ,k1 k | | ` |\ ). Ta c ThS Phng Duy Quang Trng Khoa C bn
Trng i hc Ngoi Thng 82 Mk k (0, 0) vf(Mk) =35k35 Mk k (0, 0) vf(Mk)
=0 k 0Theo tnh cht 2, suy ra khng tn ti 2x 0y 0xlim22 2 - yx+ y. 2.
Gii hn lp n gin, ta ch xt khi nim ny cho hm hai bin f(x, y).Gi s hm
f(x, y) xc nh trong hnh ch nht D = {(x, y): 0 < |x x0| < d1,0
< |y y0| < d2}. C nh y bt k tho mn iu kin 0 < |y y0| <
d2 th hm f(x, y) tr thnh hm mt bin theo bin x, gi s tn ti gii hn 0x
xlimf(x, y) = (y) Tip theo, gi s 0y ylim(y) = b. Khi , ngi ta ni
rng tn ti gii hn lp ca hm f(x, y) ti im M0(x0, y0) v vit 0y ylim0x
xlimf(x, y) = b. Tng t, ta cng c th nh ngha gii hn lp 0x xlim0y
ylimf(x, y).Ch.Ccgiihnlp 0y ylim0x xlimf(x,y)v 0x xlim0y
ylimf(x,y)cthkhngbng nhau. Vd 3. Xt V d 2 trn. Ta chng minh c hm s
f(x, y) khng c gii hn kp khi x 0, y 0. Nhng f(x, y) li c gii hn lp
ti im (0, 0). Ta c(y) = x 0limf (x, y) =2x 0xlim22 2 - yx+ y = -1,
y 0 y 0lim x 0limf (x, y) = y 0lim(y) = -1. (x) = y 0limf (x, y) =
2y 0xlim22 2 - yx+ y= 1, x 0 x 0lim y 0limf (x, y) = x 0lim(y) =
1.Ch.Khinimgiihnvhncahmnhiubinscngcnhngha tng t nh i vi hm s mt bin
s. III. Tnh lin tc 1) nh ngha nh ngha 1. Cho hm s f(M) xc nh trong
min D, M0 D. Ta ni rng, hm s f(M) lin tc ti im M0 nu tn ti gii hn:
ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 83 0M
Mlimf(M) = f(M0). Khi , ta k hiu f C(M0). nh ngha 2. Hm s f(M) c gi
l lin tc trong min D nu n lin tc ti mi im thuc D. K hiu: f C(D). V
d 1. Xt s lin tc trn min xc nh ca hm s f(x, y) = 32x,x32 222 2 + y
nu x+ y0 y0, nu x+ y0+= Gii.+) Vi (x0, y0) (0, 0) bt k (x20 + y20
0), ta c 00x xy ylimf (x,y) = 3020xx3020 + y + y = f(x0, y0) hm s
cho lin tc ti (x0, y0) (0, 0) bt k. +) Vi (x0, y0) = (0, 0):V3x32 2
+ yx+ y 3x2 2 x+ y + 32 2yx+ y 3x2 x + 3y2 y |x| + |y| M (x,y)
(0,lim (| x | 0) + |y|) = 0 Do 3(x,y) (0,xlim32 2 0) + yx+ y = 0 =
f(0, 0) Suy ra hm s cho lin tc ti im (0, 0). Vy hm s cho lin tc trn
R2. V d 2. f(x, y) = 2s,x2 222 2in(xy) nu x+ y0 y0, nu x+ y0+= Gii.
+) Vi (x0, y0) (0, 0) bt k (x20 + y20 0), ta c 00x xy ylimf (x,y) =
0 020sin(x y )x20+ y = f(x0, y0) hm s cho lin tc ti (x0, y0) (0, 0)
bt k. +) Vi (x0, y0) = (0, 0): Ta xt hai dy im: {Mk(xk, yk)}, tho
mn xk = yk 0, khi k f(Mk) = 22s )2 )kkin(x(x kx 0 12 ThS Phng Duy
Quang Trng Khoa C bn Trng i hc Ngoi Thng 84 {Mk(xk, yk)}, tho mn xk
= 2yk 0, khi k f(Mk) = 22s ))kkin2(y5(y ky 0 25. Vy, hm s cho khng
c gii hn ti (0, 0), do n khng lin tc ti (0, 0). 2) Tnh cht
Hmsnhiubinslintccngcnhngtnhchtnhhmsmtbins lin tc. Chng hn:
Tnhcht1.Nucchmsf , glintcti 0M thcchms( )ff fg, )g0 g, g(M0 cng lin
tc ti 0MTnh cht 2.D ,fn ng, bchn (D bchn nur > 0, P R: D S(P,
r)) C(D) 1 1DMM , M ) f (M), ) f (M)+2 2D D, R : |f(M)| MD: f(=mi
nf(M=max
ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 85 2. o hm
ring v vi phn ca hm nhiu bin I. o hm ring v vi phn cp 1 1) nh
ngha.Cho hm s y = f(x1, x2, ..., xn)xc nh trong mt min D Rn M0 =
(x01,x02, ..., x0n) D. Cho x01 s gia x1, x02 s gia x2 , , ...,x0n s
gia xn sao cho im M(x01, x02, ..., x0n) D * S gia ton phn ca hm fti
im M0: f(M0) = f(M) f(M0) = f(x01 + x1, x02 + x2, ..., x0n + xn )
f(x01,x02, ..., x0n) * S gia ring ca f theo bin xi ( i = 1, n) ti
im M0: xif(M0) =f(x01, x02, ...,x0i 1 , x0i + xi , x0i 1 +, ...,
x0n) f(x01, x02, ...,x0i 1 , x0i, x0i 1 +, ..., x0n) * o hm ring
theo bin xi ca f ti im M: Ta ni hm s f c o hm ring theo bin xi ti
im M0 nuiix 0x 0if (M )limx (hu hn). K hiu: fi'x(M0)hay0if (M )x.
Nhnxt.ohmringcahmw=f(M)theobinxitiM0 chnhlohm ca hm mt bin xi khi
ta coi cc bin cn li l hng s. V d 1. Tm cc o hm ring ca hm s sau:
f(x, y) = yln(x2 y2) Gii. Ta c fx(x, y) = f (x,x y) = y22xx2 - y =
22xyx2 - y fy(x, y) = f (x,y y) = ln(x2 y2) + y22yx2 - y = ln(x2
y2)- 222yx2 - y * Vi phn ton phn ca hm s f ti M0: HmsfkhvitiM0
nusgiatonphncantiimM0cth biu din di dng f(M0) = A1x1 + A2x2 + ... +
Anxn +( )2 2 210 ( x ) ) )2 n+ ( x+ ... + ( x trong ( )i2 2 212 2 2
x 0, i 1,n10 ( x ) ) )lim 0( x ) ) )2 n2 n + ( x+ ... + ( x + ( x+
... + ( x = = = , ccAi (i = 1, n ) khng ph thuc vo cc x1, x2, , xn.
K hiu: f C1(M0).Khi , biu thc A1x1 + A2x2 + ... + Anxnc gi l vi phn
ton phn ca hm s fti M0. K hiu l df(M0):ThS Phng Duy Quang Trng Khoa
C bn Trng i hc Ngoi Thng 86 df(M0) = A1x1 + A2x2 + ... + Anxn. V d
2. Xt tnh kh vi ca hm s sau f(x, y) = x2 + y2 Gii. Ti im M0(x0, y0)
bt k, ta c f(x0, y0) = f(x0 + x, y0 + y) f(x0, y0) = (x0 + x)2 +
(y0 + y)2 x20 - y20 = 2x0x + 2y0y + x2 + y2
V 2 22 2 x 0y 0x ylimx y + + = 2 2x 0y 0lim x y + = 0 nnx2 + y2
= o(2 2x y + ), khi x 0, y 0. Vy,fkhvitiimM0vdf ( x , y ) x x y y =
+ 0 0 0 02 2 .DoM0limbtk nn f kh vi trn R2. 2) Tnh cht Tnh cht 1.
(iu kin cn hm s f(M) kh vi ti im M0) f C1(M0) f C(M0). H qu. f
C(M0) f C1(M0). Tnh cht 2. 00 if (M )df (M ) A x1nii= 1 C = fxi(M0)
= Ai(i = 1,n). H qu. Nu i0 {1, 2, , n} m00if (M )x th hm s f khng
kh vi ti M0. Ch . i vi hm s nhiu bin s f(M), s tn ti cc o hm ring
ti M0 cha suy ra hm s kh vi ti im . V d 3. Xt hm s f(x, y) = 2s,x2
222 2in(xy) nu x+ y0 y0, nu x+ y0+= Ta c f (0 +x, 0) - f(0, 0)x = f
( x, 0) - f(0, 0)x = 2sin( x.0)( x)x2 - 0 + 0 = 0 x 0 0 fx(0, 0) =
0 Tng t fy(0, 0) = 0 Nhng ta chng minh c hm s cho khng lin tc ti im
(0, 0) nn n khng kh vi ti im . ThS Phng Duy Quang Trng Khoa C bn
Trng i hc Ngoi Thng 87 Tnh cht 3. (iu kin hm s f(M) kh vi ti im M0)
Nu1 2 ni' ' 'x x x 0'xf f f (x , )f ) )0 00,, ..., trong mt ln cn
no ca im My C(M( i = 1, n th i00 x 0 if (M )df (M ) (M ). x1n'i= 1
C =f Ch . Cng ging nh i vi hm mt bin s, nu cc xi l cc bin s c lp th
dxi = xi, do i0 x 0 idf (M ) (M ).dxn'i = 1 =f II. o hm ring v vi
phn cp cao Cho hm sn bin s w = f(x1, x2, , xn) *o hm ring cp cao
nhngha:Ccohmfxi (i=1,n)cgilnhngohmgringcp mt. Cc o hm ring y li l
nhng hm ca n bin x1, x2, , xn, chng c th
cohmring.Ccohmringcaccohmringcpmtnutnti c gi l c gi l o hm ring cp
hai ca hm f(x1, x2, , xn) v c k hiu nh sau: 22ifx1 2 n(x , x , ...,
x ) = fx2i (x1, x2, , xn)=fi ix x(x1, x2, , xn) = 1i if (x , , )x
x2 nx..., x | | | \ (i = 1, n) 2i jfx x1 2 n(x , x , ..., x ) =fi
jx x(x1,x2,,xn)= 1j if (x , , )x x2 nx..., x | | | \ (i j) (gi l o
hm ring hn hp cp 2) Tng t nh vy, ta c th nh ngha cc o hm ring cp n
3. V d 1. Cho hm s f(x, y) = x2y3 + x4. Ta c: fx(x, y) = 2xy3 +
4x3, fy(x, y) = 3x2y2
f2x(x, y) = 2y3 + 12x2, fyx(x, y) = 6xy2 fxy(x, y) = 6xy2,
f2y(x, y) = 6x2y Trong v d trn, ta nhn thy rng fxy(x, y) = , fyx(x,
y). Liu iu c lun lun ng khng? Ta tha nhn nh l quan trng sau y: ThS
Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 88 nh l
(Schwarz). Nu trong mt ln cn no ca im M0(x0, y0), hm s f(x,
y)cccohmringfxy,fyxvnuccohmylintctiM0th fxy(M0) = fyx(M0). Ch . nh
l trn cng c m rngcho o hm ring cp cao hn v cho hm s n bin s vi n 3.
Tc l: i vi hm n bin,nu cc o hm ring hn hp cp k(k 2) ch khc nhau v
th t o hm v cng lin tc ti im no th ti im chng bng nhau.* Vi phn ton
phn cp cao nh ngha: Vi phn ton phn ca f(x1, x2, , xn)ix idf dxn'i =
1 =f (nu c), cng l mt hm s ca cc bin xi. Vi phn ton phn ca dfnu tn
ti, c gi l vi phn ton phn cp hai ca f v c k hiu l d2f. Vy d2f =
d(df) Tng qut, ta nh ngha c vi phn cp m (m 2) ca hm s f nh sau: dmf
= d(dm-1) By gi ta xt hm hai bin f(x, y). Gi s x, y l nhng bin s c
lp, khi y dx=x, dy = y, l nhng hng s khng ph thuc x, y. Gi s d2f tn
ti. Ta c d2f = d(df) = (fxdx + fydy)x .dx + (fxdx + fydy)y. dy =
f2xdx2 + (fyx + fxy)dxdy + f2ydy2 Nu fxy v fyx lin tc, khi chng bng
nhau, v vy d2f= f2xdx2 + 2 fxydxdy + f2ydy2 =22fxdx2 + 22fx y dxdy
+ 22fydy2Tip tc tnh ton nh vy, ta c kt qu sau: Nu cc o hm ring hn
hp n cp m ca hm f(x, y) l lin tc th ta c dmf = m mk m k km m k kk
0fC dx dyx y - - = V d 2. Tnh d3f nu f(x, y) = sinxcosy Gii. Ta c
fx = cosxcosy; fy = -sinxsiny; f2x = -sinxcosy; f2y = -sinxcosy;
fxy = -cosxsiny; f3(3)x = -cosxcosy; ThS Phng Duy Quang Trng Khoa C
bn Trng i hc Ngoi Thng 89 f3(3)y = sinxsiny; f2(3)x y = sinxsiny;
f2(3)xy = -cosxcosy; Vy, theo cng thc trn ta c d3f=-cosxcosydx3 +
3sinxsinydx2dy -3cosxcosydxdy2 + sinxsinydy3 3) o hm ca hm hp nh l.
Nu hm w = f(u1, u2, , um) c cc o hm ring jf)u (j = 1, mti im (u01,
u02, , u0m), cc hm u1 = u1(x1, x2, , xn), , um(x1, x2, , xn) c o hm
ring ti im. M0 = (x01,x02, ..., x0n) v u01 = u1(M0), u02 = u2(M0),
, u0m = um(M0), . Khi , hm hp(u1(x1,x2,...,xn), , um(x1, x2, , xn))
c cc o hm ring ifx
(i = 1,n) ti M0 v ta c: ifx(M0) = mjj 1 j iuf.u x= (M0) Cng thc
trn cn c th vit di dng ma trn 0 0 01 2 nf (M ) f (M ) f (M )x x x
... | | | \ =1 2 mf f fu u u ... | | | \ 1 11 nm m1 nu ux xu ux x |
| | | | | | | \ KM O ML(M0) Ma trn1 11 nm m1 nu ux xu ux x | | | |
| | | | \ KM O ML c gi l ma trn Jacobi ca u1, u2, , um i vi x1, x2,
, xn, cn nh thc ca ma trnycgilnhthcJacobicau1,u2,,umivix1,x2,,xn v
c k hiu l 11D(u , , )D(x , , )2 m2 n u..., u x..., x. V d 3. Tnh o
hm ca hm s hp sau f(u, v) = ln(u2 + v2), u = xy, v = xy (x 0, y 0)
Gii. Theo cng thc o hm ca hm s hp, ta c fx = f u.u x + f v.v x =
22uu2 + vy + 22vu2 + v.1y =ThS Phng Duy Quang Trng Khoa C bn Trng i
hc Ngoi Thng 90 = 222xyyxxy22y+ + 2 2x2yx y22x + y.1y = 2x Tng t,
ta c fy = f u.u y + f v.v y = 2 22 + vuux+ 2 22 + vvu.2xy- | | |\ =
442(yy(y - 1) + 1) ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi
Thng 91 4. Cc tr hm nhiu bin I. Cc tr khng c iu kin 1) nh ngha. Cho
hm s f(x1, x2, , xn) xc nh trong mt min D no , M0 = (x01,x02, ...,
x0n) l mt im trong ca D. Ta ni rng f(x1, x2, , xn) t cc tr ti M0 nu
vi mi im M trong mt ln cn no ca M0, hiu s f(M0) = f(M) f(M0) c du
khng i. Nu f(M0) 0vimiMkhcMothM0cgilimcctiuv f(M0) c gi l gi tr cc
tiu. 2) iu kin cn tn ti cc tr: nh l.iu kin cn hm s f(x1, x2, , xn)
t cc tr (cc i hoc cc tiu) ti im M0=(x01,x02,...,x0n) D l ti im tt c
cc o hm ring cp mt trit tiu: ( )( )i0 0 0 0x 1 2 3 nff ' x , x , x
,..., x 0x *i 1, n = ==
nh ngha. im M0 = (x01,x02, ..., x0n) tho mn iu kin (*) c gi l im
dng ca hm s f(x1,x2, , xn).im M0 = (x01,x02, ..., x0n) c gi l im
dng.ca hm s f(x1, x2, , xn), nu ti im cc o hm ring cp mt u trit
tiu: 01f (M )x = 02f (M )x = = 0nf (M )x = 0 3) Cc bc tm cc tr ca
hm nhiu bin w = f(x1, x2, , xn):Bc 1: Tm min xc nh ca hm s w =
f(M), M = (x1, x2, , xn), Bc 2: Tm cc im dng M0, Bc 3: Kho st xem
nhng im dng no l im cc tr. thc hin bc 3, ta c th s dng iu kin hm s
c cc tr sau: 4) iu kin c cc tr: a) Hm haibin: nhl.Gi s M0(x0,y0)lmt
imdngcahmf(x,y)vtrongmtlncn no ca im ny hm f(x, y) c cc o hm ring n
cp 2 lin tc v D = 1221aa1122a a trong ThS Phng Duy Quang Trng Khoa
C bn Trng i hc Ngoi Thng 92 a11 = fxx(M0); a12 = fxy(M0) = a21; a22
=fyy(M0);nh l: i) Nu D > 0 th im dng M0 l im cc tr ca hm s f(x,
y): +) M0 l im cc i nu a11 < 0. +) M0 l im cc tiu nu a11 > 0.
ii) Nu D < 0 th im M0 khng phi l im cc tr ca hm f(x, y). iii) Nu
D = 0 th M0 l im nghi vn, cn c nhng kho st b sung.V d 1. Tm cc tr
ca hm s f(x, y)= x2 + xy+ y2 -2x 3y Gii. Hm s cho xc nh khp ni trn
mt phng Oxy. Ta c fx = 2x + y -2; fy = x + 2y 3; fxx = 2; fxy = 1;
fyy = 2Gii h 2x + y -2 = 0x + 2y - 3 = 0 H c nghim duy nht M0 =
(1,34 3). l im dng v l im dng duy nht. V fx , fyxc nh khp ni.a11 =
2; a12 = 1; a22 = 2Do D = 3 > 0 v a11 >0 nn hm s cho c cc tiu
ti im M0(1,34 3) v fmin = f(1,34 3) = -73. V d 2. Tm cc tr ca hm s
f(x, y) = x4 + y4 Gii. Hm s xc nh vi mi (x, y) R2. Hm c im dng l
nghim ca h 334x 04y= 0 = M0 = (0, 0) D dng thy rng fxx(0, 0) =
fxy(0, 0) = fyy(0, 0) = 0 nn D = 0. iu ny c ngha l M0 = (0, 0) l im
nghi vn. Trong trng hp ny im M0 l im cc tiu ca hm s vf(0, 0) = f(x,
y) f(0, 0) = x4 + y4 > 0, (x, y) (0, 0). V d 3. Tm cc tr ca hm
f(x, y) = x3 + y3 Gii. Hm xc nh (x, y) R2. Hm c im dng l nghim ca h
23x2 = 03y= 0 M0 = (0, 0) Ti im M0, ta c f2x(M0) = fxy(M0) =
f2y(M0) = 0 Do D = 0, tc l ta gp trng hp nghi vn ThS Phng Duy Quang
Trng Khoa C bn Trng i hc Ngoi Thng 93 Xt f(0, 0) = f(x, y) f(0, 0)
= x3 + y3 Khi chuyn t im M0 n im M(x, 0) thuc ln cn b ca im M0 ta
thu c: f(0, 0) = x3 v f> 0nu x > 0, f < 0 nu x < 0
trong ln cn bt k ca im M0 s gia f khng bo ton du. Vy hm s cho khng
c cc tr. V d 4. Tm cc tr ca hm s2 2f(x, y) =1 x+ y b) Hm n bin (n
3) nh l. . Gi s M0 = (x01,x02, ..., x0n) l mt im dng ca hm w =
f(x1, x2, , xn) v trong mt ln cn no ca im ny hm f(x1, x2, , xn) c
cc o hm ring lin tc ti cp 2 v Hn = 1121aa12 1n22 2nn1 n2 nn aa
a...a
aa...a ( ( ( ( ( LM M O M(Ma trn Hess) trong aij = fi jx x(M0),
i, j = 1,n Hk = 1121k1aa.....................a12 1k22 2kk2 kk a...
a a... a a... a ( ( ( ( ( (ma trn to bi k dng u v k ct u ca ma trn
H). Khi i) Nu det(Hk) > 0, k = 1,n th M0 l im cc tiu ca hm s
f(x1, x2, , xn). ii) Nu (-1)kdet(Hk) > 0, k = 1,n th M0 l im cc
i ca hm s f(x1, x2, , xn). Ch . Vi gi thit v s tn ti cc o hm ring
lin tc ti cp 2, ta lun c aij = aji(i, j = 1,n, i j) V d. Tm cc tr
ca hm s sau f(x, y, z) = x+ 2y4x + 2zy +2z Gii. Trc ht ta tm cc im
dng ca f(x, y, z) 'x 2'yzf4xf222'2y = 1 - = 0y z = - = 02x y2z 2f=
- = 0y z 3y 2xy23 = = 2xzy = z xy1 = 2 =1z =1 ThS Phng Duy Quang
Trng Khoa C bn Trng i hc Ngoi Thng 94 M1 =1,2 1, 1| | |\ , M2 = 1,2
-1, -1| | |\ Ti im M11,2 1, 1| | |\
fxx = 23y2x a11 = 4; fxy = fyx = -2y2x a12 = a21 = -2; fxz = fzx
= 0 a13 = a31 = 0; fyy = 12x + 232zy a22 = 3; fyz =fzy =-22zy a23 =
a32 = -2 fzz = 2y34 + z a33 = 6 H3 3(M1) =4 -2 0-2 3 -20 -2 6 ( ( (
( det(H1 ) = |4| = 4 > 0 det(H2 ) = 4 -2-2 3 = 8 > 0 det(H3
)= 4 -2 0-2 3 -20 -2 6 = 32 > 0 *imM11,2 1, 1| | |\
limcctiucahmschovf(M1) = 4. * Ti im M21,2--1, -1| | |\ : ta lm tng
t nh trn. II. Cc tr c iu kin 1) nh ngha. Ngi ta gi cc tr ca hm sw =
f(M) = f(x1, x2, , xn),(1) trong cc bin s x1, x2, , xn b rng buc bi
h thc g(x1, x2, , xn) = b (2)l cc tr c iu kin. 2) Phng php nhn t
Lagrange tm cc tr c iu kin Bi ton: Tm cc tr ca hm s w = f(x1, x2, ,
xn) (1) vi iu kin ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi
Thng 95 g(x1, x2, , xn) = b (2) Bi ton ny c gii quyt theo phng php
Lagrange, gm cc bc sau: Bc 1: Lp hm s Lagrange L(x1, x2, , xn) =
f(x1, x2, , xn) + [b - g(x1, x2, , xn)](3)Bin ph c gi l nhn t
Lagrange. Bc 2: Tm cc im M0 = (x01, x02, ..., x0n) m ti hm s (1) c
th c cc tr vi iu kin rng buc (2) bng cch p dng iu kin cn sau nh l.
Gi s cc hm s f(x1, x2, , xn) v g(x1, x2, , xn) c cc o hm ring lin
tc trong mt ln cn ca im M0 = (x01, x02, ..., x0n) v ti im c t nht
mttrongccohmringcag(x1,x2,,xn)khckhng.Nuhms(1)vi iu kin (2), t cc
tr ti im M0 = (x01,x02, ..., x0n), th tn ti gi tr = 0 sao cho0 M
(0, x01,x02, ..., x0n) l nghim ca h phng trnh: i iL fx x1 2 nig(x ,
x , ..., x ) = bg = - = 0, i = 1, nx (4)
Bc3:KimtraxemimM0climcctrhaykhngbngcchdavo iu kin sau y: i) Nu
d2L(M0) = n0i, j 1 i jL(M )x x= dxidxj > 0, vi mi dx1, dx2, ,
dxn khng ng thi bng 0, th M0 l im cc tiu ca f(x1, x2, , xn).ii) Nu
d2L(M0) = n0i, j 1 i jL(M )x x= dxidxj < 0, vi mi dx1, dx2, ,
dxn khng ng thi bng 0, th M0 l im cc i ca f(x1, x2, , xn).iii) Nu
dx(1)1, dx(1)2, , dx(1)n khng ng thi bng khng sao cho d2L(M0)=n0i,
j 1 i jL(M )x x= dx(1)idx(1)j>0v dx(2)1, dx(2)2, , dx(2)n khng
ng thi bng khng sao cho d2L(M0) = n0i, j 1 i jL(M )x x=
dx(2)idx(2)j < 0 th im M0 khng l im cc tr ca ca f(x1, x2, , xn).
V d. Tm cc im cc tr ca hm s f(x, y) = 8x + 15y +30 vi iu kin 2x2 +
3y2 = 107. Gii. Hm s Lagrange l L = 8x + 15y +30 + (107 - 2x2 -
3y2) ThS Phng Duy Quang Trng Khoa C bn Trng i hc Ngoi Thng 96 H
phng trnh iu kin cn l h 2 2'x'y2x- 3y= 107L= 8 - 4 x = 0L= 15 - 6 y
= 0 22x 2 + 3y= 1072 5 = = x 2y Gii h ny, ta dc hai nghim x = 4y =
51 = 2vx = -4y = -51 = -2 Ta c M1 = (4, 5), ng vi 1 = 12 vM2 = (-4,
-5) ng vi 2 = -12. Lxx = -4, Lxy = 0, Lyy = -6. +) Ti im M1 = (4,
5), ng vi 1 = 12, ta c Lxx(M1) = -2, Lxy(M1) = 0, Lyy(M0) = -3
d2L(M1) = -2dx2 3dy2 < 0 im M1 = (4, 5) l im cc i ca hm s cho v
fmax = 135.+) Ti im M2 = (-4, -5), ng vi 2 = -12 ta c Lxx(M2) = 2,
Lxy(M1) = 0, Lyy(M0) = 3 d2L(M1) = 2dx2 + 3dy2 > 0 im M2 = (-4,
-5) l im cc tiu ca hm s cho v fmin = -77
