Ch 0.I.ii Tap Hop Anh Xa
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Transcript of Ch 0.I.ii Tap Hop Anh Xa
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CHNG 0. TP HP - NH X
A. TP HP
1. Tp hp Phn t- Tp rng.1.1.
Tp hp, phn tl nhng t ban u khng nh ngha c. Tp hp cnc gi tt l tp.
Chng hn ta ni: Tp hp 3 mu tu ca bng ch ci la tinh. Khi ta ni a l phn
t ca tp hp ny; b, c cng vy. Nhng d khng phi phn t catp ny.
Tp hp cc s nguyn. Khi 11, 2 l trong s rt nhiu phn t
ca tp ny. Nhng 32
, 5 khng l phn t ca tp ny.
Cc tp s bit , , ,
Ta ghi tp hp bng cc ch k t in , , , ,...B X Y v ghi phn t ca tpbng cc k t thng , , , ,...a b x y
Nu a l phn t ca tpXth ta ghi a X (c l a thuc X).Nu a khng l phn t ca tpXth ta ghi a X (c l a khng thuc X).
Trong th d trn, ta ghi 2 , 5 .
1.2.
Cch ghi tp hp.
Lit k: K ra tt c phn t ca tp gia hai du { , }.Th d:
{ , , }, {0,1,2,3,4,....}A a b c Dng biu :
Th d:
Nu c trng:Tp hp c vit di dng{ / ( )}x P x (c l tp nhng phn tx sao cho x c inh cht P)hay
Aa
b
c X
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{ / ( )}x A P x (c l tp nhng phn tx thuc A sao cho x c tnh cht P)
Th d: 2{ / 3 2 0}M x x v x x
Trong cch ghi
{ / ( ) }phan tu dactrung
A x P x , Pgi l c trng ca tpA.
1.3.Tp hp rng.Tp hp khng c phn t gi l tp rng. K hiu: , { } .Th d:
Tp 2{ / 1 0}x x l tp rng.
2. Tp hp con Tp hp bng nhau.2.1.Tp hp con:
o Cho 2 tpA vB. TpA gi l tp hp con caB nu mi phn t caAul phn t caB.K hiu B (c lA nm trong B)NuA khng phi l tp con caB th ta ghi B (c lA khng nmtrong B)
o Qui c: Tp rng l tp con ca bt k tp noo Tnh cht:
A B v B C A C .
2.2.Tp hp bng nhau.Hai tpA vB gi l bng nhau nuA l tp con caB vB l tp con caA.K hiu B .Nu tpA khng bng tpB th ta ghi B .Tm li:
B A B B A Th d:
2{ / 2 3 1 0}A x x x , 12
{1}, { , 1}B C .
A
B
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Ta c: ,B A C .
2.3.Tp cc tp con.Cho tpA. Tp hp gm tt c cc tp con caAc gi l tp cc tp con caA (hay tp cc phn ca A). K hiu ( )P .
Vy: ( ) { / }X X A P .Th d:Cho {1, }a th ( ) ,{1}, { }, {1, }a a P .
3. Php ton gia cc tp hp.3.1.Php giao:
Giao ca 2 tpA vB l 1 tp hp, k hiu B (c lA giao B), gm vch gm tt c nhng phn t va thucA va thucB.
{ / }B x x A x B
Tnh cht:
( ) ( ) :
A A
A B B AA
A B B B A
B C A B C A B C
3.2.Php hp: Hp ca 2 tpA vB l 1 tp hp, k hiu B (c lA hp B), gm v
ch gm tt c nhng phn t thuc v t nht 1 trong 2 tp A hayB.
{ / }B x x A x B
A
B
B
A B
B
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Tnh cht:A A
B B A
A
( ) ( ) :B B A BB C A B C A B C
Ngoi ra, gia php giao v php hp c tnh phn phi:
( ) ( ) ( )
( ) ( ) ( )
B C A B A C
B C A B A C
3.3.Php hiu: Hiu ca 2 tpA vB l 1 tp hp, k hiu \B (c lA trB), gm v ch
gm tt c nhng phn t thuc v tp A m khng thuc tpB.
\ { / }B x x A x B
Tnh cht:\
\\
\ \ ( )
A A
A AB A B
B A A B
Phn b.o Cho tpE v ( )EP .
Khi , hiu \E A c gi lphn b caA trongE, k hiu EC A .
AA
\B
E
A
EC A
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o Tnh cht:, ( )
( )E E
E
E
A B E
C C A A
C A A E
C A E A
P
( )
( )E E E
E E E
C A B C A C B
C A B C A C B
4. Tch Descartes cc tp.4.1.Tch 2 tp.
Tp { ,{ , }}a a b gi l cp th t ca 2 phn ta v b.K hiu ( , )a b .
Hin nhin tnh ngha ta thy:( , ) ( , )
a ca b c d
b d
.
Ta gi tch Descartes ca 2 tpA vB, k hiu B , l tp tt c cc cpth t ( , )a b vi a A v b B .Vy {( , ) / }B a b a A b B .Th d:
Vi {1,2}, { ,2, }B a c , th{(1, ),(1,2),(1, ),(2, ),(2,2),(2, )}B a c a c .
4.2.
Tch n tp ( 2n
) Bn th t:
B 3 th t ca 3 phn t , ,a b c l cp (( , ), )a b c , k hiu ( , , )a b c Tng qut : Bn th t ca n phn t 1,..., na a l cp
1 1(( ,..., ), )n na a a , trong 1 1( ,..., )na a l b 1n th t ca 1n phn t 1 1,..., na a . K hiu 1( ,..., )na a .Lu :
1 1 1 1( ,..., ) ( ,..., ) ,...,n n n na a b b a b a b . TchDescartes ca n tp 1 2, ,..., nA A , k hiu 1 2 nA A , l tp tt
c cc bn tht 1( ,..., )na a , trong i ia A vi mi 1,...,i n .Vy:
1 2 1( ,..., ) / , 1,n n i iA A a a a A i n
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Qui c:1
2
n
n
A A
A A A
A A
BI TPTP HP
1. Tm cc s thcx, y 2 tpA,B bng nhau:a. { 2, { 2}}, { ,{2 }}x y B y x
b. 2 2 3 2{ , {2 }}, {3 4,{ +3}}A x x B x x x 2. Cho tp {1,2,3,4,5,6,7}E v cc tp
{1, 2,3, 4,5}, {1,2,3}, {2, 4, 6}A B C .a. Tnh \ , , ( \ ) ,EB A C B A C C A C .b. Vit ( \ ( )), ( \ ), ( )B C C B C .c. Gii trong ( )E phng trnh X B .d. Gii trong ( )E phng trnh B X A .
3. ChoA, B, Cl cc tp con ca mt tpE. Chng minha. B A B A A B B b. \
E
B A C B c. ( \ )B B A B d. ( \ ) \ \ ( )B C A B C e. \ ( \ ) ( \ ) ( )B C A B A C f. ( \ ) ( )B A B A g. \A B A A B
4. Cho tpA c m phn t, tpB c n phn t. Bit tp B cp phn t.Hi s phn t ca tp \ ,B A B .
5. So snh cc tp:
a. v ( ) ( )C B D b. ( ) ( )B C D v ( ) ( )C B D
Chng minhc. B A B
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B. NH X
1. nh x.1.1.nh ngha 1. Cho 2 tp hpA vB khc rng.
Mt nh xftAnB l mt b 3 th t ( , , )f A B F , vi F A B , tha:i) ( , )x A y B x y F ii) ( , ), ( , )x y x y F x x y y
Nu ( , )x y F th ta niy l nh cax qua nh xf,x l to nh cay quanh xfv ta ghi ( )y f x .Tc l ( , ) ( )x y F y f x .
TpFc gi l thca nh xf. TpA gi l tp i,B gi l tp n ca nh xf.
1.2.Th d.a) Cho { , , }, {1,2,3,4}A a b c B v
{( ,1), ( ,2), ( ,5)}F a b c {(1, ), (2, ), (4, )}G b c a {( ,1), ( ,2), ( ,2), ( ,1)}H a b c a {( ,4), ( ,3), ( ,4), ( ,1)}K a c b c
Ta thy: ( , , )f A B F khng phi nh x v F A B ( , , )g B A G khng phi nh x v 3 khng c nh. ( , , )h A B H l nh x tA dnB.
( , , )k A B K khng l nh x v c c 2 nh !
b) Cho ( , , ) , vi 2{( , ) / 1}x y y x . l nh x t n .
c) Cho ( , , ) , vi 2{( , ) / 1}x y x y . khng l nh x t n .
1.3.Nhn xt. Ta thng hay cho mt nh xkhng phi bng th ca n (nh trong
cc v d trn) m bngc trng ca th ca n. c trng thngl qui tc tm nh y ca phn tx thuc tp i. Chng hn trong th d b) trn y, nh x c thc cho nh sau:
:
1y x
Do , ta c thnh ngha nh x mt cch khc nh sau.
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1.4.nh ngha 2:
Mt nh xftAnB l mt qui tc t tng ng mi phn tx ca tp iA vi mt phn t duy nhty thuc tp nB m ta gi l nh cax quaf.
K hiu
:( )
f A Bx f x
hay f B .
Th d: y l nh xf:
nh xf cn c thc m t nh sau:
x a b cy 3 1 1
1.5.nh xng nht.nh x
:( )
f A Aa f a a
c gi l nh xng nht trnA.K hiu , 1A AId .
1.6.nh x bng nhau.Cho 2 nh x ( , , )f A B F v ( , , )g C D G .Tnh ngha nh x ta thy:
C
f g B D
F G
Ni cch khc, cho :f A B v :g C D l 2 nh x, ta thy:
( ) ( )
A C
f g B D
A f x g x
A B
a
b
c
1
2
3
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2. Tch cc nh x2.1.nh ngha.
Cho 2 nh xf gB C .
Khi , qui tc
:[ ( )]
h A Ca g f a
l 1 nh x tAn Cv c gi l nh x tch ( hay nh x hp) ca 2 nhxfvg.K hiu g f ( c l g trn f )Tm li:
g f C [ ]( ) [ ( )]a A g f a g f a
Tnh cht:
Vi nh x :f A B ta c:i) Af Id f ii) BId f f
Qui c.
Cho nh x :f A A , ta k hiu:2f f f , 3f f f f ,..., n
n
f f f
2.2.Mnh .Tch cc nh x c tnh kt hp (nu chng tn ti), tc l
( ) ( )h g f h g f .Chng minh:
f g hB C D [( ) ]( ) ( )[ ( )] ( [ ( )]) ([ ]( ))x A h g f x h g f x h g f x h g f x
[ ( )]( )h g f x Do ( ) ( )h g f h g f .
3. nh v nh ngc ca cc tp hp3.1.nh ca tp.
Cho nh x :f A B Xt X A , tp hp
( ) { ( ) / }f X f x x X B gi l nh ca tp Xqua nh xf.
Nhn xt:( ) { / ( )}f X y B x A y f x
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3.2.nh ngc ca tp.
Cho nh x :f A B Xt Y B , tp hp
1( ) { / ( ) }f Y x A f x Y gi l nh ngc ca tp conYqua nh xf.
Nhn xt:1( ) { / ( )}f Y x A y Y y f x
Th d:
Cho nh x :f A B v X A .
Chng minh 1( ( ( ))) ( )f f f X f X .Gii:
1 1( ( ( ))) ( ( )) ( )y f f f X x f f X y f x .
V1
( ( ))f f X
nn ( ) ( )f x f X . Suy ra ( )y f X . ( ) ( )y f X x X f x y
V ( ) ( )f x F X nn 1( ( ))x f f X v do 1( ) ( ( ( )))y f x f f f X .
4. n nh Ton nh Song nh4.1.n nh
Cho nh x :f A B .
nh xfc gi l n nh
,
( ) ( )
x x A
x x f x f x
,
( ) ( )
x x A
f x f x x x
Ni cch khc,nh xfl n nh Mi phn t ca tp n c nhiu nht1 to nh qua f.Th d:
nh x :f , ( ) 2f n n l mt n nh v, ( ) ( ) 2 2n n f n f n n n n n .
nh x
AB
a
b
c
d
1
2
3
4
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khng phi l n nh v phn t 1 ca tp n c n 2 to nh!
Xt 2 nh x :[0;1] [0;1]f v :[0;1] [0;1]g xc nh bi thsau:
nh x no l n nh ? (Tr li:f).
4.2.Ton nhCho nh x :f A B .
nh xfc gi l ton nh ( )y B x A y f x Ni cch khc,
nh xfl ton nh Mi phn t ca tp n c t nht1 to nh qua f.Th d:
nh x : {1,2,3,4} { , , }f a b c x 1 2 3 4y a c a b
l mt ton nh.
4.3.Song nhCho nh x :f A B .
nh xfc gi l song nh fva l n nh, va l ton nh.Ni cch khc,nh xfl song nh Mi phn t ca tp n c 1 v ch1 to nh qua f.
Th d:
y l mt song nh tAnB.
0 1
1f
0 1
1
g
A
B
ab
c
d
12
3
4
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nh xng nht AId lun l song nh.
4.4.Tnh cht Tch 2 n nh (nu c) l mt n nh Tch 2 ton nh (nu c) l mt ton nh
Tch 2 song nh (nu c) l mt song nh.Chng minh tnh cht u:
Cho 2 n nhf gB C .
Khi
,
[ ]( ) [ ]( ) [ ( )] [ ( )] ( ) ( )
x x A
g f x g f x g f x g f x f x f x x x
Vy g f n nh.
4.5.nh x ngc ca mt song nh.
Cho song nh :f A B .Khi , qui tc
:g B A y x , sao cho ( )y f x ,
l mt nh x tBnA, c gi l nh x ngc ca song nhf.
K hiu: 1f Tm li:
1fB A
1( ) ( )y B f y x f x y
Hin nhin nh x ngc lun l mt song nh. Th d:
1
A AId Id
Song nhf
c nh x ngc l 1f
AB
a
b
c
1
2
3
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Song nh
:
2
f
x
c nh x ngc l
1
2
:x
f
x
.
Tnh cht:
Cho song nh :f A B .Ta c:
i) 1 Bf f Id
ii) 1 Af f Id
iii) 1 1( )f f
Bi tp NH X
6. Vit tt c nh x tAnB:
a. {1,2}, { , }B a b b. {1,2,3}, { , }B a b c. {1,2}, { , , }B a b c
7. Xt tnh n nh, ton nh, song nh ca cc nh x sau, tm nh x ngcnu c:
a. 1 :f
3x
b. 2 :f 1n n
c. 3 : \{2}f
2 1xx
x
d. 4 : \ {1} \{1}f
31
xx
x
e. 5 :f
xx e
BA
1
2
3
a
b
c
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f. 2 3:f
( , ) ( , , )x y x x x y
g.3 3:
( , , ) ( , , )
g
x y z x y z x y z
8. Tnh 1 3 5 1 1 5, , ( 1), ( 1)n nf f f f f n f n (xem cu 7).
9. Tnh g f v f g vi
khi
2 khi
:
2( )
2 2
x
f
e xf x
x x x
v
2
:
ln 0( )
0
g
x xg x
x x
khi
khi
10.Cho cc nh x : , :f A B g B C . Chng minha. Nu g f n nh th f n nh.
b. Nu g f ton nh th gton nh.
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I STUYN TNH
15
Chng I. MA TRN NH THC
I. MA TRN1. Cc nh ngha:
ChoKl trng s ( , hay ) Cho {1,2,..., }I m v {1,2,..., }n , vi ,m n l 2 s nguyn dng.Mt nh x tI J n K gi l mt ma trn cp m n vi phn tthuc K.
Tp cc ma trn cp m n vi phn t thuc Kc k hiu l( , )Kat m n hay gn hn: ( , )at m n .
Cho ( , )Mat m n , ta thng ghi ma trnA di dng 1 bng c mdngv n ct:
11 12 1
21 22 2
1 2
n
n
m m mn
a a a
a a aA
a a a
vi qui c ija l nh ca cp ( , )i j qua nh xA.
Th d:
Ma trn : {1,2} {1,2,3}A
(1,1) 2
(1,2) 0
(1,3) 2(2,1) 1
(2,2) 1
(2,3)
c ghi l:
2 0 2
1 1A
Vy ta c th ghi ma trn di dng tng qut nh sau:
1,1,i mij j n
A a
hoc ija , nu khng snhm ln v cp caA.
Cho ma trn 1,1,
i mijj n
A a
.
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I STUYN TNH
16
Phn t ija gi l phn tdng i, ct j ca ma trn
A. 1 2{ , ,..., }i i ina a a gi l dng i ca ma trnA. 1 2{ , ,..., }j j mja a a gi l ct j ca ma trnA.
Cho 2 ma trn ( , )ija Mat m n v
( , )klB b Mat p q . Tnh ngha nh x bngnhau, ta thy ngay:
, ,ij ij
m p
A B n q
a b i j
Ma trn dng-Ma trn ct:Ma trn ch c 1 dng gi l ma trn dng.
Ma trn ch c 1 ct gi l ma trn ct. Ma trn khng:
Ma trn ( , )ija Mat m n vi 0 , ,ija i j , c gi l
ma trn khng, k hiu 0mn hay vn tt l 0 ( nu khng snhm ln v cp).
Ma trn chuyn v.
Cho ma trn 1,1,
( , )i mijj n
a Mat m n
.
Khi ma trn 1,1,
( , )i nijj m
B b Mat n m
xc nh bi
, 1,..., , 1,...,ij jib a i n j m
c gi l ma trn chuyn vca ma trnA, v k hiu l t.
Th d:
1 2
0 1
1 0 2
x
A a
v
1 1
0 0
2 1 2
ta
A x
.
Ma trn vung. Nu ma trn c s dng m bng vi s ct n th ta
gi l ma trn vung cp n. Khi ta ghi( , ) ( )at n n Mat n .
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I STUYN TNH
17
Cho ma trn vung ( )ija Mat n .
11 22{ , ,..., }nna a a gi l ng cho chnh ca matrn.
Ma trn vung cp nc bit n ijI , trong
1
0ijkhi i j
khi i j
( k hiu Kronecker), gi l
ma trn n vcp n.Th d:
2 3 4
1 0 0 01 0 0
1 0 0 1 0 0, 0 1 0 ,
0 1 0 0 1 00 0 1
0 0 0 1
I I I
,
2. Php ton gia cc ma trn.
2.1. Php cng 2 ma trn cng cp:
Cho 2ma trn , ( , )ij ija B b Mat m n .
Khi , ma trn ( , )ijC c Mat m n xc nh bi
, ,ij ij ijc a b i j
c gi l tng ca 2 ma trnA vB, k hiu B .
Th d:1 2 0 1 1 3
2 1 1 1 3 2
2 0 2 1 0 1
Tnh cht:
, ( , )B Mat m n A B B A
, , ( , ) ( ) ( )B C Mat m n A B C A B C ( , ) 0Mat m n A A
( , ) ( , )ij ija Mat m n A a Mat m n
0A A A A
, ( , ) ( )t t tB Mat m n A B A B
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I STUYN TNH
18
2.2.Php nhn mt svi mt ma trn.
Cho s K v ma trn ( , )ij Ka Mat m n .
Ma trn ( , )ij Ka Mat m n c gi l tch ca vi ma trn v
c k hiu l .
Th d:
Cho
1 3 0
1 2
0 1
x
b
. Khi
2 6 0
2 2 4 2
0 2 2
x
b
Tnh cht: , ( , ) ( )KB Mat m n K A B A B , ( , ) ( ) ( )KK A Mat m n A A , ( , ) ( )KK A Mat m n A A A
( , ) ( )t t
KK A Mat m n A A
2.3.Php nhn 2 ma trn thch hp.Cho 2 ma trn:
( , )ij KA a Mat m p ,
( , )ij KB b Mat n p .
Ta nh ngha mt ma trn ( , )ij KC c Mat m n nh sau:
1 1 2 21
p
ij ik kj i j i j ip pjkc a b a b a b a b
.Ma trn C ny c gi l tch ca ma trnA v ma trnB.K hiu B .
1
21 2 1 1 2 2
j
ji i ip ij i j i j ip pj
pj
b
ba a a c a b a b a b
b
Th d:
0 1 11 1 1 2 0 2
1 1 12 0 2 2 6 2
1 2 0
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I STUYN TNH
19
1 0 0 0 0 0
02 0 1 3 0 0
Tnh cht:
( , ) ( , ) ( , )
( ) ( )
Mat m p B Mat p q C Mat q n
AB C A BC
( , ) , ( , ) ( )Mat m p B C Mat p n A B C AB AC , ( , ) ( , ) ( )B C Mat m p A Mat p n B C A BA CA ( , ) ( , ) ( ) ( ) ( )K A Mat m p B Mat p n A B AB A B
( , ) ( , ) ( )t t tMat m p B Mat p n AB B A ( , ) n mMat m n AI A I A
II. NH THC1. nh ngha:
Cho ma trn vung ( )ija Mat n
1.1. S ( )D A nh ngha theo qui np nh sau c gi l nh thcca ma trnA:
Nu 1n th nh thc 11( )D A a . Gi s ta nh ngha c nh thc ca ma trn cp 1n .
Ta nh ngha nh thc ca ma trn cp n ( )ija Mat n
l:
11 11 21 211
1 1( 1) )( n nnD A a D a D a D
trong ijD l nh thc ca ma trn cp 1n nhn c t ma trnA bng cch b
dng i v ctj caA.
Ta cn ghi ( )D A l det( ) hay | | .
Nu ( )ija Mat n
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I STUYN TNH
20
th ta ghi
11 12 1
21 22 2
1 2
( )
n
n
n n nn
a a a
a a aD A
a a a
nh thc c n dng (tt nhin cng c n ct) gi l nh thc cp n.
1.2. Th d:
1n :
11| |a 11a 2n :
11 1221 22
a aa a 11 11 21 21 11 22 21 12a D a D a a a a
11 22 21 12a a a a . 3n :
11 12 13
21 22 23 11 11 21 21 31 31
31 32 33
a a a
a a a a D a D a D
a a a
22 23 12 13 12 1311 21 3132 33 32 33 22 23
a a a a a aa a aa a a a a a
11 22 33 23 32 21 12 33 13 32 31 12 23 13 22( ) ( ) ( )a a a a a a a a a a a a a a a
11 22 33 12 23 31 13 32 21 11 23 32 13 22 31 12 21 33a a a a a a a a a a a a a a a a a a
Qui tc Sarrus tnh nh thc cp 3:
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I STUYN TNH
21
nh thc11 12 1
22 20
0 0
n
n
nn
a a a
a aD
a
(nh thc tam gic)
22 2
11 11 11
0
n
nn
a a
a D a
a
11 22 11 11 22 nna a D a a a .
1.3.nh thc chuyn v
o nh thc ( )tD A gi l nh thc chuyn v ca nh thc ( )D A .o Th dnh thc
1 1 0
0 1 1
2 2 3
c chuyn v l nh thc
1 0 2
1 1 2
0 1 3
.
o
o Mnh : nh thc v chuyn v ca n l bng nhau.
Nhn xt: Do kt qu ny, nhng tnh cht no sau y ng cho dng ca nh thccng sng cho ct ca nh thc.
2. Tnh cht2.1 Tnh thay phin:i ch 2 dng th nh thc i du.
Th d:1 2 3 2 1 3
1 2 3
a b c
x y z x y z y x z
a b c b a c
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I STUYN TNH
22
H qu: nh thc c 2 dng ging nhau th bng 0.
0
a b
D x y z x y z
a b c
a b c
D
c
a b c
D .
2.2. Tnh tuyn tnh:
Nhn 1 s vi 1 nh thc l nhn s vi 1 dng no ca nh thc.
12
2 2 2 2
2 2 2 2 4 2 2
2
a x m a x m a x m a x m
D b y n D b y n b y n b y n
c z p c z p c z p c z p
H qu:
nh thc c dng bng 0 th bng 0. nh thc c 2 dng t l th bng 0.
0 0 0 0
a b c
m n p
,
1 1 2
2 2 4 0
3 1 3
,
1 1 1
0 2 2 0
3 1 1
2.3. Tnh cng tnh:
Ta c th cng 2 nh thc cp n thnh mt nh thc cp n (m khng cntnh gi trnh thc) nu 2 nh thc c 1n dng, theo cng th t dng, i mtging nhau.
Khi , ta cng cc phn t tng ng trn dng cn li v gi nguyn 1n dng kia c nh thc tng.
1 2 3 1 2 3
a b c a b c a b c
x y z x y z
m n p m n p m n p
H qu:Nu ta ly kt qu nhn 1 dng ca nh thc vi 1 s cng vo 1 dng khc (cngcc phn t cng th t vi nhau) th nh thc khng thay i ga tr.
( hay ni cch khc: Nu cng vo 1 dng ca nh thc mt dng khc c nhn vi 1s th nh thc khng thay i gi tr.)
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I STUYN TNH
23
1 2 3 1 2 3 1 2 3
2 4 6
1 2 3
a b c a b c a b c
x y z x y z x a y b z c
ng dng tam gic ha mtnh thc:
2 1 3 1 1 2 1 2
1 2 1 2 2 1 3 1
1 0 0 1 1 0 0 1
0 1 1 2 0 1 1 2
1 2 1 2
0 5 5 5
1 0 0 1
0 1 1 2
1 2 1 2
0 5 5 5
0 2 1 3
0 1 1 2
1 2 1 2
0 1 1 1
5 0 2 1 3
0 1 1 2
1 2 1 2
0 1 1 1
5 0 0 1 1
0 0 0 3
15
3. Khai trin nh thc3.1.nh thc con Phn b i s:
Cho nh thc, 1,ij i j n
D a
.
nh thc cp knhn c tD bng cch bi n k dng v n k ct no caDc gi l nh thc con caD.Ta cng gi nh thc l nh thc con ca ma trn ija .
Th d:
2
1 1 3
3 32 2
1
3 4
45 2 53
1 6
4
2
6
3
15 2
6 1
a x
b y
c z
c 1 trong cc nh thc cp 3 l
a x
b y
c z
.
nh thc con cp 1n nhn c bng cch b dng thi v ct thkcgi l nh thc con ghp theo phn tdng i ct j. K hiu ijD .
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I STUYN TNH
24
Th d:Cho nh thc
2 1 0
0 2 3
1 1 2
D
.
nh thc con caD ghp theo phn t dng 2 ct 3 l 232 1
1 1D
.
S ( 1)i jij ijD c gi l phn b i s ng vi phn t dng i ct j.
3.2.nh l:
Cho ijD a v {1,2,..., }i n . Khi :
1 1
( 1)n n
i k
ik ik ik ik k k
D a A a D
.
Ta gi y l cng thc khai trin nh thcD theo dng thi.Tng t, ta c cng thc khai trin nh thcD theo ct thj:
1 1
{1,2,..., }
( 1)n n
k jkj kj kj kj
k k
j n
D a A a D
GHI CH: nh ngha nh thc nu chnh l khai trin nh thc theo ct 1.
Th d:
1 1 1 2 1 31 0 2 0 2 12 1 0 ( 1) 3 ( 1) ( 2) ( 1
3 2 1
) ( 1)1 2 0 2 0 1
0 1 2
6 8 2 0
3.1.nh l:( ) ( ) ( )D AB D A D B
III. HNG CA MA TRN1.nh thc con ca ma trn
Cho ma trn ( , )ija Mat m n .
nh thc cp rnhn c tA bng cch bi m r dng v n r ct no
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I STUYN TNH
25
caAc gi l nh thc con caA.Th d:
Cho ma trn
2 4 0 2
1 3 2 5
0 1 1 0
A
.
nh thc4 2
1 0D
l 1 nh thc con cp 2 caA.
nh thc
2 0 2
1 2 5
0 1 0
E
l 1 nh thc con cp 3 caA.
nh thc chnh:Ta gi nh thc chnh ca ma trnA l mt nh thc conD cp rno caAm:
o 0D o Mi nh thc con caA c cp ln hn ru bng 0.
Th d:
Ma trn0 0 0
0 1 0
c 1 nh thc chnh duy nht l 1 .
Ma trn
1 0 2 3
1 1 2 3
1 1 2 3
1 2 2 3
c 1 trong cc nh thc chnh l
1 0
1 1.
2.Hng ca ma trnnh ngha:Hng ca mt ma trnA l cp ca nh thc chnh ca ma trn .K hiu: ( ), ( )rg A r A Th d:
o (0) 0rg
o
1 0 2 31 1 2 3
21 1 2 3
1 2 2 3
rg
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I STUYN TNH
26
o Nhn xt: Nu ( )ija Mat n th :
( ) ( ) 0rg A n D A
3.Php bin i scp dng3.1.Cc php bin i sau y p dng cho mt ma trn (c mt ma trn khc)gi l php bin i scp dng:
Loi 1: i ch 2 dng ca ma trn cho nhau Loi 2: Nhn 1 dng ca ma trn vi 1 s khc 0. Loi 3: Ly kt qu nhn 1 dng ca ma trn vi 1 s cng vo 1 dng
khc (cng cc phn t cng th t vi nhau).
Th d:
2 0 0 1 1 2 2 2 1 2 2 2
0 1 1 1 0 1 1 1 0 1 1 11 2 2 2 2 0 0 1 0 4 4 5
1 2 2 2
0 1 1 1
0 0 0 9
3.2.nh l:Php bin i scp dng khng lm thay i hng ca ma trn.
Th d:
2 0 0 1 1 2 2 2 1 2 2 2
0 1 1 1 0 1 1 1 0 1 1 1
1 2 2 2 2 0 0 1 0 4 4 5
1 2 2 2
0 1 1 1 3
0 0 0 9
rg rg rg
rg
3.3.Ma trn rt gn theo dng tng bc:
Mt ma trn ( , )ija Mat m n c gi l c dng rt gn theo
dng tng bc nu tha 2 iu kin sau:o Tt c cc dng bng 0 (nu c) u nm di nhng dng khc 0
(nu c).o Nu rdng u c cha phn t khc 0 (dng 1r n dng m
u bng 0) v nu phn t khc 0 u tin (t tri sang) ca dng
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I STUYN TNH
27
thi nm ct ik ( 1,...,i r ) th:
1 2 rk k k .Th d:
0 2 0 1 1
1 0 0 0 20 0 2 1 3
A
0 3 2 1 1 0
0 0 1 1 3 4
0 0 0 0 2 1
0 0 0 0 0 0
0 0 0 0 0 0
B
0 3 2 1 1 0
1 0 1 1 3 4
0 0 0 2 2 1
0 0 0 0 0 4
0 0 0 0 0 0
C
nh l:Nu ma trn A c dng rt gn theo dng tng bc vi s dng khc 0 lrth ( )rg A r .
nh l:Lun c th bin i mt ma trn v dng rt gn theo dng tng bcnhmt s hu hn php bin i scp dng.
Th d:
Tm hng ca ma trn:
1 1 2 3 1
3 1 2 2 4
2 3 1 1 6
1 2 3 1 4
A
.
Ta c
1 1 2 3 1 1 1 2 3 13 1 2 2 4 0 4 8 11 7
2 3 1 1 6 0 1 5 7 8
1 2 3 1 4 0 1 1 4 5
rgA rg rg
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I STUYN TNH
28
1 1 2 3 1 1 1 2 3 1
0 1 1 4 5 0 1 1 4 54
0 0 6 3 3 0 0 2 1 1
0 0 4 27 27 0 0 0 25 25
rg rg
IV. MA TRN KHO MA TRN O1. nh ngha:
Mt ma trn vung cp n ( )ija Mat n c gi l kho
(hay kh nghch) nu ta tm c mt ma trn vung cng cp( )B Mat n sao cho:
nB BA I ( nI l ma trn n v cp n).
Ma trnB ni trn c gi l ma trn o caA v k hiu l 1 . Ma trn kho cn gi l ma trn khng suy bin. Th d:
o Cho2 1
3 2A
v
2 1
3 2B
.
V 2B BA I nnA kh nghch v1 2 1
3 2A B
.
o Tm , , ,y z t tha
2 1 1 1
3 2 0 2
x z
y t
2 1 2 1 2 1 1 1
3 2 3 2 3 2 0 2
2
2 0 3
3 1 0
1
x z
y t
x
x z y
y t z
t
2. nh l:Cho ( )ija Mat n .
A kho ( ) 0D A
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I STUYN TNH
29
Lu :
Vi ( )ija Mat n :
A kho A khng suy bin ( ) 0D A ( )rg A n
3. Cch tm ma trn o.3.1. Dng phn b i s:
Cho ( )ija Mat n .
Khi :
11 21 1
12 22 21 1( )
1 2
n
nD A
n n nn
A A A
A A AA
A A A
3.2. Dng php bin i scp dng:
Cho ( )ija Mat n .
Ta vit mt ma trn nB A I cp 2n n (na bn tri l matrnA, na bn phi l ma trn n v) v p dng php bin i scp dng trn matrnB sao choA trthnh nI . Khi ma trn thu c na bn phi l ma trn o caA.
Th d:
Chng minh ma trn
0 1 1
1 0 1
1 1 0
C
kho v tm ma trn o ca n.
Gii:
Ta c
0 1 1 0 1 11 1
( ) 1 0 1 0 1 1 2 01 1
1 1 0 1 1 0
D C
nn Ckh nghch.
v Cch 1:
11 21 311 1 1
12 22 32( ) 2
13 23 33
1 1 1
1 1 1
1 1 1D C
C C C
C C C C
C C C
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I STUYN TNH
30
Cch 2:
0 1 1 1 0 0 1 0 1 0 1 0
1 0 1 0 1 0 0 1 1 1 0 0
1 1 0 0 0 1 1 1 0 0 0 1
A
1 0 1 0 1 00 1 1 1 0 0
0 1 1 0 1 1
1 0 1 0 1 00 1 1 1 0 0
0 0 2 1 1 1
1 1 12 2 2
1 0 1 0 1 0
0 1 1 1 0 0
0 0 1
1 1 12 2 21 1 12 2 21 1 12 2 2
1 0 0
0 1 0
0 0 1
Vy
1 1 1
2 2 21 1 1 1
2 2 21 1 12 2 2
C
= 12
1 1 11 1 1
1 1 1
BI TP CHNG I.TNH TON TRN MA TRN
1. Cho cc ma trn2 0 10 1 2
1 1 0
A
v1 0 1 11 1 2 0
1 1 3 1
B
.
Tnh cc ma trn , , ( )t t tB B A B A A .
2. Chocos sin
sin cosA
.
a. Chng minh , ( )n
nA A A A
.
b. Tnh
20053 1 1 1
,1 11 3
n
3. Tm ma trn 2( )X M sao cho X XA vi:
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I STUYN TNH
31
a.1 1
0 1A
b.
1 2
1 1A
NH THC
4. Tnh cc nh thc:
2 1
3 1,
13547 13647
28423 28523,
0 1 2
2 2 0
1 0 3
,
246 427 327
1014 543 443
342 721 621
,
3 0 1 1
1 3 0 1
1 1 3 0
0 1 1 3
,
1 2 3 4
2 1 4 3
3 4 1 2
4 3 2 1
5. Chng minh:
a.
1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 2
3 3 3 3 3 3 3 3 3
2
a b b c c a a b c
a b b c c a a b c
a b b c c a a b c
b.
2 2 2 2
2 2 2 2
2 2 2 2
2 2 2 2
( 1) ( 2) ( 3)
( 1) ( 2) ( 3)0
( 1) ( 2) ( 3)
( 1) ( 2) ( 3)
a a a a
b b b b
c c c c
d d d d
6. Tnh nh thc cp n :
1 2 2 2 22 2 2 2 2
2 2 3 2 2
2 2 2 1 2
2 2 2 2
n
n
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I STUYN TNH
32
7. Gii phng trnh:
2 31
1 1 1 10
1 2 4 8
1 3 9 27
x x x
HNG MA TRN MA TRN KH NGHCH
8. Tm hng ca ma trn:
1 3 4 5 2
0 1 3 4 6
3 5 2 3 4
2 3 5 6 4
,
3 1 3 2 5
5 3 2 3 4
1 3 5 0 7
7 5 1 4 1
,
2 0 3 1
1 2 2 3
3 2 5 4
5 2 8 5
,
1 1 2
2 1 5
1 10 6 1
a
a
,
3 1 1 4
4 10 1
1 7 17 3
2 2 4 3
a
,
1 1 1
1 1 1
1 1 1
1 1 1
m
m
m
m
9. Cho ma trn
2
3 2 3
1 0 2 3
2 1 3
5 3 3 7
m
mA
m m
m m
.
Tm m sao cho
a.
0 4 10 1
4 8 18 7
10 18 40 17
1 7 17 3
rankA rank
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I STUYN TNH
33
b.
2 1 1 1
1 3 1 1
1 1 4 1
1 1 1 5
1 2 3 4
1 1 1 1
rankA rank
c.
1 0 0 1 4
0 1 0 2 5
0 0 1 3 6
1 2 3 14 32
4 5 6 32 77
rankA rank
10.Tm ma trn o ca:
a.1 2
2 2
b.
2 0 0
1 1 0
2 1 1
c.
1 0 3
2 1 1
3 2 2
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I STUYN TNH
34
Chng II.H PHNG TRNH TUYN TNH
I.nh ngha1. H phng trnh11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
( )
n n
n n
m m mn n m
a x a x a x b
a x a x a x bL
a x a x a x b
,
vi , , ( 1, , 1, )j i ijx b a i m j n v jx l n s, c gi l hphng trnh tuyn tnh.
Ma trn
11 12 1
21 22 2
1 2
n
n
m m mn
a a a
a a aA
a a a
gi l ma trn h sca h phng
trnh tuyn tnh (L).
Ma trn
11 12 1 1
21 22 2 2
1 2
n
n
m m mn m
a a a b
a a a bA
a a a b
gi l ma trn bsungca h
phng trnh tuyn tnh (L). Nu ta chn 1 nh thc chnh ca ma trn h sA th cc n ca h phng
trnh tuyn tnh (L) ng vi cc ct ca nh thc chnh ny sc gi l ccn chnh, cc n cn li l n khng chnh.
2. Dng ma trn ca h phng trnh tuyn tnh
Nu t
1
2
n
x
x
Xx
v
1
2
m
b
b
Bb
th ta thy h (L) c vit thnh
( )L AX B Ta gi y l dng ma trn ca h phng trnh tuyn tnh.
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I STUYN TNH
35
Nghim ca h phng trnh tuyn tnh (L) l nhng bn th t
1( ,..., )n
nc c sao cho tt cng thc trong (L) l ng khi thay
j jx c .
Gii h phng trnh tuyn tnh l tm tt c nghim ca n. Hai h phng trnh tuyn tnh gi l tng ng nu chng c cng
tp nghim.NHN XT:
p dng php bin i scp dng ln cc dng ca 1 h phng trnh tuyntnh ta c 1 h phng trnh tuyn tnh khc tng ng.
II. H phng trnh Cramer.1. H phng trnh tuyn tnh (L) c gi l h Cramer nu n tha 2 iu kin
sau: m n ( ) 0D A
2. nh l (qui tc Cramer):H phng trnh Cramer lun c nghim v c duy nht 1 nghim xc nh bi:
jDj D
x , trong :
( )D D A jD l nh thc nhn c bng cch thay ct thj caD bi ct t do
1,...,
nb b .
3. Th d:Gii h phng trnh tuyn tnh:
1
1
2
x z
y z
x y
Gii:H cho c dng ma trn:
11 0 1 1 1 0 1 1
0 1 1 1 0 1 1 1
1 1 0 2 1 1 0 2
x x
y y
z z
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I STUYN TNH
36
12
1 1 1 1 2
1 1 1 1 0
1 1 1 2 1
x
y
z
III.iu kin c nghim ca h phng trnh tuyn tnh.1. nh l (Kronecker-Capelly):
Cho h phng trnh tuyn tnh (L).
H (L) c nghim rgA rgA .
2. Gii h phng trnh tuyn tnhChng minh trn cho ta mt phng php gii h phng trnh tuyn tnh (L), gil phng php Gauss.
Vit ma trn v a n v dng rt gn theo dng tng bc.
Nu rgA rgA th h v nghim.
Nu rgA rgA r , th ta xc nh 1 nh thc chnh caA. Sau chuyn cc n khng chnh sang v phi lm tham s ly gi tr ty . Ta tnhc cc n chnh nhqui tc Cramer. Lu rng trong trng hp ny h sc v s nghim ph thuc vo n r tham s ly gi tr ty trong . Vnu 1 1,..., { ,..., }n r ny x x cc n khng chnh, th mt n chnh kx no s
c dng tng qut: 1
n r
k k kj jjx y
. Do , h phng trnh tuyn tnh (L) c nghim duy nht khi v ch khi
rgA rgA r v khng c n chnh, tc l khi v ch khi rgA n .
IV.H phng trnh tuyn tnh thun nht1. nh ngha:
H phng trnh tuyn tnh
11 1 12 2 121 1 22 2 2
1 1 2 2
00( )
0
n nn n
m m mn n
a x a x a xa x a x a xH
a x a x a x
c gi l h phng trnh tuyn tnh thun nht.
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I STUYN TNH
37
H thun nht lun lun c nghim, chng hn (0,0,...,0) n l 1 nghimca h. Ta gi nghim (0,0,...,0) ny l nghim tm thng.
Theo kt qu trn, h (H) c nghim duy nht khi v ch khi rgA n .
2. H nghim cbn ca h thun nht:Xt h phng trnh tuyn tnh thun nht (H) vi rgA r n .Gi s 1 2, ,...,r r nx x x l cc n khng chnh ca h.Suy ra cc nghim ca h (H) c dng tng qut l:
1 2 11 1 1
, ,..., , ,...,n n n
j j j j rj j r nj r j r j r
u c x c x c x x x
.
Khi , n r nghim c th:
1 1 1 2 1 1( , ,..., ,1,0,...,0)r r rr u c c c ,
2 1 2 2 2 2( , ,..., ,0,1,...,0)
r r rr u c c c
,
1 2( , ,..., ,0,0,...,1)n r n n rnu c c c (nhn c bng cch gn cho cc tham s 1( ,..., )r nx cc gi tr thuc tp
{(1,0,..,0),(0,1,0,..,0),...,(0,0,..,0,1)} n r )c gi l 1 h nghim cbn ca h (H).
Th d:Tm mt h nghim cbn ca h
1 2 3 4
1 2 3 4
1 2 3 4
2 5 7 0( ) 4 2 7 5 0
2 5 0
x x x xH x x x x
x x x x
Gii:
2 1 5 7 2 1 5 7 2 1 5 7
4 2 7 5 0 0 3 9 0 0 1 3
2 1 1 5 0 0 4 12 0 0 1 3
2 1 5 7
0 0 1 3
0 0 0 0
Vy 1 2 3 4
3 4
2 5 7 0( )
3 0
x x x xH
x x
2 3 1 4
3 4
5 2 7
3
x x x x
x x
1 1
2 1 2
3 2
4 2
2 8
3
x t
t t
x t
x t
Nghim tng qut ca (H) l:
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I STUYN TNH
38
1 1 2 2 2 1 2( , 2 8 , 3 , ) , ,u t t t t t t t
Cho 1 2( , ) {(1,0), (0,1)}t t , ta c mt h nghim cbn l:
1
2
(1,2,0,0)
(0, 8, 3,1)
u
u
BI TP CHNG II.
H PHNG TRNH TUYN TNH
11.Gii cc h phng trnh sau:
a.1 2 3
1 2 3
1 2 3
2 4
3 4 2 11
3 2 4 11
x x x
x x x
x x x
b.
1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
2 3 1
3 2 4
2 3 6
2 3 4
x x x x
x x x x
x x x x
x x x x
c.
1 2 3 4
2 3 4 5
1 2 32 3 4
3 4 5
0
0
2 3 22 3 2
2 3 2
x x x x
x x x x
x x xx x x
x x x
12.Gii cc h sau:
a.
1 2 3
1 2 31 2 3
1 2 3
3 1
2 2 1
3
2 3 1
x x x
x x x
x x x
x x x
b.
1 2 3
1 2 3
1 2 3
1 2 3
2 2
3 5
5 7
2 3 3 14
x x x
x x x
x x x
x x x
c.
1 2 3 4
1 2 3 41 2 3 4
1 2 3 4
2 1
3 2 2 3 2
5 2 1
2 3 4
x x x x
x x x x
x x x x
x x x x
d.
1 2 3 4
2 3 4
1 2 4
2 3 4
2 3 4 4
3
3 3 1
7 3 3
x x x x
x x x
x x x
x x x
e.
1 2 4 5
1 2 3 4
1 2 3 4 5
1 2 3 4 5
2 3 0
2 04 2 6 3 4 0
2 4 2 4 7 0
x x x x
x x x xx x x x x
x x x x x
f.
1 2 3 4 5
1 2 3 4 5
2 3 4 5
1 2 3 4 5
7
3 2 3 22 2 6 23
5 4 3 3 12
x x x x x
x x x x xx x x x
x x x x x
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8/8/2019 Ch 0.I.ii Tap Hop Anh Xa
39/39
I STUYN TNH
13.Tm nghim tng qut v h nghim cbn (nu c) ca cc h thun nht sau:
a.1 2 3 4
1 2 3 4
1 2 3 4
2 4 5 3 0
3 6 4 2 0
4 8 17 11 0
x x x x
x x x x
x x x x
b.
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
6 2 2 5 7 09 3 4 8 9 0
6 2 6 7 0
3 4 4 0
x x x x xx x x x x
x x x x x
x x x x x
c.
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
2 0
2 2 3 0
3 2 2 0
2 5 2 2 0
x x x x x
x x x x x
x x x x x
x x x x x
d.
1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
0
2 3 4 0
3 6 10 0
4 10 20 0
x x x x
x x x x
x x x x
x x x x