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7/25/2019 Bài giảng Toán cao cấp II - Đặng Văn Vinh (Biên soạn), Đại Học Bách Khoa Tp.HCM, 2008 http://slidepdf.com/reader/full/bai-giang-toan-cao-cap-ii-dang-van-vinh-bien-soan-dai 1/249 Trườ ng Đại hc Bách khoa tp. H Chí Minh B môn Toán Ứ ng dng ------------------------------------------------------------------------------------- Đại s tuyến tính 1  Gi ng viên Ts. Đặ  ng V ă  n Vinh (9/2008) www.tanbachkhoa.edu.vn Môn hc cung cp các kiến thc cơ  bn ca đại s tuyến tính. Sinh viên sau khi k ết thúc môn hc nm vng các kiến thc nn tng và biết gii các bài toán cơ  bn: tính định thc, làm vic vớ i ma trn, bài toán gi i Mc tiêu ca môn hc Toán 2 2 h phươ ng trình tuy n tính, không gian véct ơ , ánh x tuy n tính, tìm tr riêng véc t ơ  riêng, đư a d ng toàn ph ươ ng v  chính t c.

Bài giảng Toán cao cấp II - Đặng Văn Vinh (Biên soạn), Đại Học Bách Khoa Tp.HCM, 2008

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    Trng i hc Bch khoa tp. HCh MinhBmn Ton ng dng

    -------------------------------------------------------------------------------------

    i stuyn tnh

    1

    Ging vin Ts.ng Vn Vinh (9/2008)www.tanbachkhoa.edu.vn

    Mn hc cung cp cc kin thc cbn ca i stuyn tnh. Sinh vinsau khi kt thc mn hc nm vng cc kin thc nn tng v bit giicc bi ton cbn: tnh nh thc, lm vic vi ma trn, bi ton gii

    Mc tiu ca mn hc Ton 2

    2

    hphng trnh tuy n tnh, khng gian vct, nh xtuy n tnh, tm trring vc tring, a dng ton phng vchnh tc.

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    Sphc

    Ma trn

    nh thc

    Hphng trnh tuyn tnh

    3

    Khng gian vc t

    Php bin i tuyn tnh

    Trring, vctring

    Dng ton phng

    Khng gian Euclide

    Nhim vca sinh vin.

    i hc y .

    Lm tt ccc bi tp cho vnh.

    c bi mi trc khi n lp.

    4

    nh gi, ki m tra.

    Thi gia hc k: hnh thc trc nghim (20%)

    Thi cui k: tlun (80%)

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    Ti liu tham kho

    1.Cng Khanh, Ng Thu Lng, Nguyn Minh Hng.i stuyn tnh. NXBi hc quc gia

    2. Ng Thu Lng, Nguyn Minh Hng. Bi tp ton caocp 2.

    5

    11.www.tanbachkhoa.edu.vn

    3.Cng Khanh.i stuyn tnh. NXBH quc gia

    Ni dung--------------------------------------------------------------------------------------------------------------------

    0.1 Dngi sca sphc

    0.2 Dng lng gic ca sphc

    0.3 Dng mca sphc

    6

    0.4 Nng sphc ln ly tha

    0.5 Khai cn sphc

    0.6 nh l cbn cai s

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    0.1 Dng i sca sphc-------------------------------------------------------------------------------------------------------------------

    Khng tn ti mt s thc no m bnh phng ca n l mt sm. Hay, khng tn ti s thcx sao chox2 = -1.

    Bnh phng ca mt s o l mt s m. K tic chn k

    th k th 17, ngi tanh ngha mt s o.

    7

    nh ngha si

    Si,c gi ln v o, l mt s sao cho

    i2

    = -1

    hiu mt s m bnh phng ca n b ng1.

    0.1 Dng i sca sphc-----------------------------------------------------------------

    nh ngha s phcChoa vb l hai s thc vi ln v o, khi z = a + bi

    c gi l s phc. S thca c gi l phn thc v sthcbc gi l phno ca s phcz.

    8

    Tp s thc l tp hp con ca tp s phc, bi v nu cho b = 0,tha + bi = a + 0i = a l mt s phc.

    Phn thc ca s phcz = a + bic k hiu lRe(z).Phno ca s phcz = a + bic k hiu lIm(z).

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    0.1 Dng i sca sphc-----------------------------------------------------------------

    Tt c cc s c dng 0 + bi, vi b l mt s thc khc khng

    c gi l s thuno. V d:i, -2i, 3i l nhng s thuno.

    9

    S phc ghidng z = a + bic gi l dngi s ca sphcz.

    0.1 Dng i sca sphc

    -----------------------------------------------------------------

    nh ngha php nhn hai s phc.Cho z1= a + bi vz2= c + di l hai s phc, khi

    z1.z2= (a + bi) (c + di) = (ac bd) + ( ad + bc)i

    V dTm d n i s ca s hc

    10

    z = (2 + 5i).(3+ 2i)

    Gii

    z = (2 + 5i)(3 + 2i)

    = 6 + 4i + 15i + 10 i2

    Vy dngi s ca s phc l: z = -4 + 19i.

    = 2.3 + 2.2i + 3.5i + 5i.2i

    = 6 + 19i + 10(-1) = -4 + 19i

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    0.1 Dng i sca sphc-----------------------------------------------------------------

    Hai sphc c gi l bng nhau nu chng c phn thc v phno tng ng bng nhau.

    Ni cch khc, hai sphcz1= a1+ ib1 vz2= a2+ib2 bng nhaukhi v chkhi a1= a2 v b1= b2.

    nh ngha sbng nhau

    11

    Choz1= 2 + 3i; z2= m + 3i.

    Tm tt c cc s thcmz1= z2.

    Gii

    1 2 2 3 3z z i m i= + = +2

    23 3

    mm

    = =

    =

    0.1 Dng i sca sphc

    -----------------------------------------------------------------nh ngha php cng v php tr ca hai s phc.Choa + bi vc + di l hai s phc, khi

    Php cng: (a + bi) + (c + di) = (a + c) + (b + d) i

    Php tr: (a + bi) - (c + di) = (a - c) + (b - d) i

    V d

    12

    Tm ph n thc v ph no ca s phcz = (3 + 5i) + (2 - 3i).

    Giiz = (3 + 5i) + (2 - 3i)

    Re( ) 5; Im( ) 2.z z = =

    = (3+2) + (5i 3i) = 5 + 2i.

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    0.1 Dng i sca sphc-----------------------------------------------------------------

    Choz vw l hai s phc; v l hai s phc lin hptngng. Khi:

    z w

    1. l mt s thc.z z+

    2. l mt s thc.z z

    Tnh cht ca s phc lin hp

    13

    3. khi v ch khiz l mt s thc.z z=

    4. z w z w+ = +

    5. z w z w =

    6. z z=

    7. vi mi s t nhinn( )n nz z=

    0.1 Dng i sca sphc

    -----------------------------------------------------------------

    Cng, tr, nhn hai sphc:Khi cng (tr) hai sphc, ta cng (tr) phn thc v

    phn o tng ng.

    14

    N n a s p c, ta t c n g ng n n n a uthc i svi ch i2 = 1.

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    0.1 Dng i sca sphc-----------------------------------------------------------------

    V d.

    nh ngha s phc lin hp

    S phc c gi l s phc lin hp ca sphcz =a + bi.

    z a bi=

    15

    Tm s phc lin hp ca s phcz = (2 + 3i) (4 - 2i).

    Gii.

    Vy s phc lin hp l 14 8 .= z i

    z = (2 + 3i) (4 - 2i) =2.4 2.2i + 3i.4 3i.2i

    =8 4i + 12i 6i2

    = 8 4i + 12i 6(-1) = 14 + 8i.

    Lu : So snh vi sphc.Trong trng sphc khng c khi nim so snh. Ni mt cchkhc, khng thso snh hai sphcz1 = a1 + ib1 v z2 = a2 + ib2

    0.1 Dng i s ca s phc------------------------------------------------------------------

    16

    . 1 2 2 1

    ngha trong trng sphc C ngoi trchng ta nh ngha khinim so snh mt cch khc.

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    0.1 Dng i sca sphc-----------------------------------------------------------------

    Php chia hai s phc.

    1 1 12 2 2

    z a ib

    z a ib

    +

    =+

    17

    2 2 2 2 2( )( )z a ib a ib=

    +

    1 1 2 1 2 1 2 2 12 2 2 2

    2 2 2 2 2

    z a a b b b a a bi

    z a b a b

    + = +

    + +

    Mun chia sphc z1 cho z2, ta nhn tv mu cho sphc lin hpca mu. (Gis )2 0z

    0.1 Dng i s ca s phc-----------------------------------------------------------------

    V d.

    Thc hin php toni

    i

    +

    5

    23

    Gii.Nhn t v mu cho s phc

    18

    )5)(5(5 iii +=

    125

    210315 2

    +

    +++=

    iii

    ii

    2

    1

    2

    1

    26

    1313+=

    +=

    n p c a m u + .

    Vitdngi s

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    0.2 Dng lng gic ca sphc---------------------------------------------------------------------------------------------------------------------------

    ( , ) = +M a b z a bib

    y

    trco

    19

    r

    ao x

    2 2 mod( )= + =r a b z

    cos

    :sin

    =

    =

    a

    r

    b

    r

    trc thc

    0.2 Dng lng gic ca sphc---------------------------------------------------------------------------------------------------------------------------

    2 2mod( ) | |= = +z z a b

    nh ngha Mdun ca s phcMdun ca s phcz = a + bi l mt s thc dngcnh nghanh sau:

    V d

    20

    Tm mun ca s phcz = 3 - 4i.

    Gii

    Vy mod(z) =|z| = 2 2 2 23 ( 4) 5.+ = + =a ba = 3; b = -4.

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    0.2 Dng lng gic ca sphc---------------------------------------------------------------------------------------------------------------------------

    Ch :

    Nu coi s phcz = a + bi l mtim c ta (a, b), th2 2 2 2| | ( 0) ( 0)= + = + z a b a b

    l khong cch t im (a, b)n gc ta.

    21

    Choz = a + bi v w = c + di.

    l khong cch gia haiim (a, b) v (c,d).

    2 2| | ( ) ( )z w a c b d = +

    0.2 Dng lng gic ca sphc---------------------------------------------------------------------------------------------------------------------------

    V d

    Tm tt c cc s phc z t ha

    | 2 3 | 5 + =z i

    22

    | 2 3 | 5z i + =

    | (2 3 ) | 5z i =

    ng trn tm (2,-3) bn knh bng 5.

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    0.2 Dng lng gic ca sphc---------------------------------------------------------------------------------------------------------------------------

    V d

    Tm tt c cc s phc z t ha

    | | | | 4z i z i + + =

    23

    | | | | 4 + + =z i z i

    Tp hp tt ccc im trong mt phng sao cho tngkhong cch t n hai im cho trc (0,1) v (0,-1)khng thay i bng 4 chnh l ellipse.

    0.2 Dng lng gic ca sphc

    ----------------------------------------------------------------------------nh ngha argument ca s phc

    Gc c gi l argument ca s phcz vc k hiu larg( ) .=z

    Gc c gii hn trong khongLu .

    24

    0 2 < hoc <

    Cng thc tm argument ca s phc.

    2 2

    2 2

    cos

    sin

    = =

    + = = +

    a a

    r a b

    b b

    r a b

    hoc tg = b

    a

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    0.2 Dng lng gic ca sphc---------------------------------------------------------------------------------------------------------------------------

    GiiMun:

    V d

    Tm dng lng gic ca s phc 1 3.= +z i

    1; 3.= =a b 2 2| | 2.= = + =r z a b

    25

    1 1os =

    23 1

    = =

    +

    ac

    r

    3 3sin =

    23 1 = =

    +

    b

    r

    Suy ra2

    3

    =

    Dng lng gic:

    Argument:

    2 21 3 2(cos sin )3 3

    = + = +z i i

    0.2 Dng lng gic ca sphc---------------------------------------------------------------------------------------------------------------------------

    V d

    Tm tt c cc s phc z t ha

    | 2 | | 2 |z z = +

    26

    | 2 | | 2 | = +z z

    Tp hp tt c ccim trong mt phng sao cho khong

    cch t n haiim (2,0) v (-2,0) bng nhau.

    y chnh l trc tung.

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    0.2 Dng lng gic ca sphc---------------------------------------------------------------------------------------------------------------------------

    Gii

    V d

    Tm argument ca s phc 3 .= +z i

    3; 1= =a b . Ta tm gc tha:

    27

    3 3os =

    23 1 = =

    +

    ac

    r

    1 1sin =

    23 1 = =

    +

    b

    r

    Suy ra6

    =

    Vy arg(z) =6

    0.2 Dng lng gic ca sphc---------------------------------------------------------------------------------------------------------------------------

    2 2; 0= + + >z a bi a b

    2 2

    2 2 2 2( )= + +

    + +

    a bz a b i

    a b a b

    28

    cos s n = +z r

    Dng lng gic ca s phc(cos sin )z r i = +

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    0.2 Dng lng gic ca sphc-----------------------------------------------------------------------------------------------------

    1 1 1 1 2 2 2 2(cos sin ); (cos sin )z r i z r i = + = +

    S bng nhau gia hai s phcdng lng gic

    1 2

    1 2 1 2 2

    r rz z

    k

    ==

    = +

    29

    1 2 1 2 1 2 1 2(cos( ) sin( ))z z r r i = + + +

    Nhn hai s phc dng lng gic: mun nhn vi nhau v

    argument cng li.

    0.2 Dng lng gic ca sphc---------------------------------------------------------------------------------------------------------------------------

    Gii

    V d

    Tm dng lng gic, mun v argument ca s phc

    (1 )(1 3).= + z i i

    30

    =

    Dng lng gic:

    2( os in ) 2( os in )4 4 3 3

    = + +z c is c is

    2 2[ os( ) in( )]4 3 4 3

    = + + +z c is

    2 2( os in ).12 12

    = +z c is

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    0.2 Dng lng gic ca sphc-----------------------------------------------------------------------------------------------------

    Php chia hai s phcdng lng gic1 1 cos sin

    z ri= +

    1 1 1 1 2 2 2 2(cos sin ); (cos sin )z r i z r i = + = +

    2 20 0. >z r

    31

    2 2z r

    Chia hai s phc dng lng gic: mun chia cho nhau v

    argument tr ra.

    0.2 Dng lng gic ca sphc---------------------------------------------------------------------------------------------------------------------------

    Gii

    V d

    Tm dng lng gic, mun v argument ca s phc

    2 12.

    3

    =

    +

    iz

    i

    - -

    32

    2 2 33

    = +

    izi

    Dng lng gic:7 7

    2( os in ).6 6

    = +z c is

    cos s n

    3 35 5

    2(cos sin )6 6

    +

    =

    + i

    - 5 - 52[cos( - ) sin( - )]

    3 6 3 6

    = +z i

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    0.3 Dng mca sphc---------------------------------------------------------------------------------------------------------------------------

    cos sini

    e i

    = +

    nh l Euler (1707-1783)

    33

    z a bi= +

    (cos sin )z r i = +

    iz re

    =

    Dngi s ca s phc z

    Dng lng gic ca s phc z

    Dng m ca s phc z

    0.3 Dng mca sphc---------------------------------------------------------------------------------------------------------------------------

    V d

    Tm dng m ca s phc sau

    3= +z i

    34

    Dng lng gic: 2(cos sin )6 6z i= +

    Dng m:5

    62 i

    z e

    =

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    0.3 Nng sphc ln ly tha-----------------------------------------------------------------------------------------------------

    V d. C h o z = 2 + i . T n h z5.

    =+= 55 )2( iz

    35

    =++++++= 555555 22222 iCiCiCiCiCC

    =+++++= iii 1.2.5).(4.10)1.(8.10.16.532

    i4138 +=

    0.3 Dng mca sphc---------------------------------------------------------------------------------------------------------------------------

    V d

    Biu din cc s phc sau ln mt phng phc

    2 ;i

    z e R

    +

    =

    36

    Mun khng thayi, suy ra tp hp ccim lng trn.

    2(cos sin )z e i = +

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    0.3 Dng mca sphc---------------------------------------------------------------------------------------------------------------------------

    V d

    Biu din cc s phc sau ln mt phng phc

    3;a iz e a R

    +=

    37

    (cos3 sin3)az e i= +

    Argument khng thay i, suy ra tp hp cc im l nangthng nm trong gc phn t th 2.

    0.4 Nng sphc ln ly tha

    -----------------------------------------------------------------------------------------------------

    nh ngha php nng s phc ln ly tha bc n

    z a bi= +

    2 2 2( )( ) ( ) (2 )z z z a bi a bi a b ab i= = + + = +

    38

    3 3 3 2 2 3

    ( ) 3 3 ( ) ( ) ...= + = + + + =z a bi a a bi a bi bi0 1 1 2 2 2( ) ( ) ( ) ... ( )n n n n n n nn n n nz a bi C a C a bi C a bi C bi

    = + = + + + +

    nz A iB= +

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    0.3 Nng s phc ln ly tha--------------------------------------------------------------

    Ly tha bc n ca s phc i:

    ii =1

    12 =i

    23

    iiiii === 145

    1)1(1246

    === iii

    iiiii === 1347

    39

    ===

    1)1()1(224 === iii 111448

    === iii

    Ly tha bc n ca i

    Gi sn l s t nhin, khiin = ir, virl phn d ca n chiacho 4.

    0.3 Dng mca sphc---------------------------------------------------------------------------------------------------------------------------

    V d

    Tnh 1987

    =z i

    40

    1987 4 496 3= +

    1987z i=

    4 496 3 3i i i

    += = =

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    0.3 Nng sphc ln ly tha-----------------------------------------------------------------------------------------------------

    Cho z = 1 + i.

    a) Tm z3;

    b) Tm z100.

    V d

    41

    3 3) (1 )a z i= + 2 31 3 3i i i= + + +

    1 3 3i i= +

    2 2z i= +

    ) Tnh tng t rat phc tap. Ta s dung cach khacb

    0.3 Nng sphc ln ly tha-----------------------------------------------------------------------------------------------------

    z a bi= + (cos sin )r i = +

    2 2(cos2 sin2 )z z z r i = = +

    3 2 3(cos3 sin3 )z z z r i = = +

    1n n n= =

    42

    [ (cos sin )] (cos sin )n nr i r n i n + = +

    Cng thc De MoivreCho r > 0, cho n l s t nhin. Khi

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    0.3 Nng sphc ln ly tha-----------------------------------------------------------------------------------------------------

    V d. S dng cng thc de Moivres, tnh:

    a) (1 + i)25 200)31( i+b)

    20

    17

    )212()3(i

    i+c)

    43

    Gii. a) Bc 1. Vit 1 + idng lng gic

    )4

    sin4

    (cos21

    iiz +=+=

    Bc 2 . S dng cng thc de Moivres:

    )4

    25sin

    4

    25(cos)2()]

    4sin

    4(cos2[ 252525

    iiz +=+=

    Bc 3. n gin )4

    sin4

    (cos221225

    iz +=

    0.4 Khai cn sphc-----------------------------------------------------------------------------------------------------

    nh ngha cn bc n ca s phc

    Cn bc n ca sphc z l sphc w, sao cho wn =z, trong n l stnhin.

    = =

    44

    2 2(cos sin ) (cos sin )n nn kk kz r i z r i

    n n

    + += + = = +

    vi k = 0, 1, 2, , n 1.

    Cn bc n ca sphc z c ng n nghim phn bit.

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    0.4 Khai cn sphc-----------------------------------------------------------------------------------------------------

    V d. Tm cn bc n ca cc sphc sau. Biu din cc nghim lntrn mt phng phc.

    3 8a) 4 3 + ib) 8

    16

    1

    i

    i+c)

    6 1

    3

    i

    i

    +

    d) 5 12i+e) 1 2i+f)

    45

    Gii cu a)

    Vit sphc dng lng gic: 8 8(cos0 sin0)i= +

    Sdng cng thc:

    3 0 2 0 2

    8(cos0 sin 0) 2(cos sin )3 3k

    k ki z i

    + ++ = = +

    0,1,2.k =

    0.5 nh l cbn ca i s---------------------------------------------------------------------------------------------------------------------------

    nh l c bn ca i s cho bit c s nghim ca phngtrnh m khng ch cch tm cc nghim nh th no.

    Nua thc vi h s thc, chng ta c mt h qu rt quan trng

    46

    H qu

    Nua + bi l mt nghim phc caa thcP(z) vi h s thc, tha bi cng l mt nghim phc.

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    0.4 Khai cn sphc-----------------------------------------------------------------------------------------------------

    Gii cu b)

    Vit sphc dng lng gic:

    Sdng cng thc:

    442 26 62(cos sin ) 2(cos sin )

    6 6 4 4

    + ++ = = +k

    k ki z i

    3 2(cos sin )6 6

    + = +i i

    47

    , , , .=

    0z

    1z

    2z

    3z

    0.5 nh l cbn ca i s---------------------------------------------------------------------------------------------------------------------------

    Nh bc hc ngi c Carl Friedrich Gauss (1777-1855) chngminh rng mia thc c t nht mt nghim.

    48

    nh l cbn cai sa thcP(z) bc n cng n nghim k c nghim bi.

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    0.5 nh l cbn ca i s---------------------------------------------------------------------------------------------------------------------------

    (s dng h qu canh l cbn)

    1) Tm a thc bc 3 vi hsthc nhnz1= 3i vz2= 2+i

    V d

    lm nghim.2) Tma thc bc 4 vi h s thc nhnz1= 3i vz2= 2+i

    49

    .

    1) Khng tn ti a thc tha yu cu bi ton.

    2)a thc cn tm l:

    1 1 2 2( ) ( )( )( )( )P z z z z z z z z z=

    ( ) ( 3 )( 3 )( (2 ))( (2 ))P z z i z i z i z i= + + 2 2( ) ( 9)( 4 5)P z z z z= + +

    0.5 nh l cbn ca i s---------------------------------------------------------------------------------------------------------------------------

    V d. Gii cc phng trnh sau trong C.

    015 =+ iza)

    0122 =++ izzd)

    0224 =++ zzc)

    012 =++ zzb)

    50

    Gii. Gii phng trnh 02 =++ cbzaz

    acb 42 =Bc 1. Tnh

    Bc 2. Tm2,1

    2 4 == acb

    Bc 3. 1 21 2;2 2

    b bz z

    a a

    + += =

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    0.5 nh l cbn ca i s---------------------------------------------------------------------------------------------------------------------------

    Gii. Bi va thc vi h s thc v2 + i l mt nghim, theo hqu ta c2 i cng l nghim.

    (s dng h qu canh l cbn)

    Tm tt ccc nghim cabit 2 + i l mt nghim.

    V d

    4536144)(234

    ++= zzzzzP

    51

    z c p n c n z + z - =

    = z2 4z + 5

    P(z) c th ghidng

    P(z) = (z2 4z + 5)(z2 + 9)

    z2 + 9 c hai nghim 3i v3i. Vy ta tmc c 4 nghimcaP(z) l2 + i, 2 i, 3i, -3i.

    0.5 nh l cbn ca i s---------------------------------------------------------------------------------------------------------------------------

    Gii phng trnh sau trong C.

    9 0z i+ =

    V d

    52

    9z i= 9z i = 9 cos sin

    2 2z i

    = +

    2 22 2cos sin

    9 9k

    k k

    z i

    + +

    = +

    0,1,...,8.k =

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    Kt lun---------------------------------------------------------------------------------------------------------------

    2. Dng Lng gic ca s phc)sin(cos irz +=

    1. Dngi s ca s phcbiaz +=

    53

    3. Nng ln ly tha)sin(cos)]sin(cos[ ninrirz nnn +=+=

    4. Cn bc n ca s phc

    )2sin2(cos)sin(cosn

    ki

    n

    krzirz nknn

    ++

    +==+=

    .1,...,3,2,1 = nk

    Thc hin php ton

    Bi tp 1

    )2(

    )32(5

    2

    ii

    iz

    +=

    54

    Vit s phc saudng lng gic.

    Bi tp 2

    )3)(1( iiz ++=

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    Vit s phc saudngi s

    Bi tp 3

    5)32( iz =

    Bi t 4

    55

    Tm tt c cc s phc z t ha

    1|21| + iz

    Cho |z| = 2. Chng t

    Bi tp 5

    6 8 13z i+ +

    Bi t 6

    56

    Cho |z| = 1. Chng t

    21 | 3 | 4z

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    Tm s phc z t ha

    Bi tp 7

    2z z i = +

    Bi t 8

    57

    Tm s phc z t ha2

    1 12 6z i z+ + =

    Cho z l mt s phc khc 0. Tm mun ca s phc sau

    Bi tp 9

    2008z i

    z

    58

    Xcnh tp hp ccim trong mt phng phc biu din cc sphc z t ha mn

    Bi tp 10

    zk

    z i=

    vi k l s thc dng cho trc.

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    Tm s phc z t ha mnng thi

    Bi tp 11

    11

    z

    z i

    =

    v

    31

    z i

    z i

    =

    +

    59

    Tm s phc z t ha mn

    Bi tp 12

    4

    1z i

    z i

    + =

    Tm phn thc v phno ca s phc

    Bi tp 13

    3 2

    1

    i iz

    i i

    +=

    +

    60

    Gii phng trnh .

    Bi tp 14

    2 | | 0z z+ =

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    Vit dng lng gic ca mi s phc

    Bi tp 15

    2) sin 2sin

    2

    a i

    + ) cos (1 sin )b i + +

    61

    Tm s cn bc hai ca s phcz = - 8 + 6i.

    Bi tp 16

    Vit s phc saudngi s

    Bi tp 17

    5)32( iz =

    62

    Tm tt c cc s phc z t ha

    1|21| + iz

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    Xcnh phn thc ca s phc

    Bi tp 19

    1

    1

    z

    z

    +

    bit rng |z| = 1 v 1.z

    63

    Chng minh rng nu l mt s o th |z| = 1.

    Bi tp 20

    1

    1

    z

    z

    +

    Bi tp 21

    Tnh v5cos 5sin

    Tnh vncos nsin

    64

    Bi tp 22

    Tnh , vi31

    1

    i

    iz

    +

    =6 z

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    Bi tp 23

    Tnh , vi 16=z4 z

    Bi tp 24

    65

    Tnh 3 22 i+

    Bi tp 25

    Tnh i41 +

    Bi t 26

    66

    Gii phng trnh 07

    =+iz

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    Bi tp 27

    Gii phng trnh 012 =++ izz

    Bi tp 28

    67

    Chng t rng s phc 2 + 3 i l mt nghim ca phng trnh

    05216174 234 =++ zzzz

    v tm tt c cc nghim cn li.

    Bi tp 29

    Gii phng trnh

    02)22()2( 23 =+++ iziziz

    bit rng phng trnh c mt nghim thuno.

    68

    Bi tp 30

    Phn tch x3 + 27 ra tha s bc nht v bc hai.

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    Bi tp 31

    Tnh 0 2 4 2006 2008

    2008 2008 2008 2008 2008A C C C C C = + +

    Bi tp 32Tnh

    nA cos3cos2coscos ++++=

    69

    Bi tp 33

    Tnh

    cos cos( ) cos( ) cos( )A b b b b n = + + + + + + + +2

    Vit dng lng gic ca mi s phc

    Bi tp 34

    2) sin 2sin2

    a i

    + ) cos (1 sin )b i + +

    70

    Tm s phc z sao cho|z| = |z 2| v mt argument ca z 2 bng

    Bi tp 35

    mt argument ca z + 2 cng vi .2

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    Chng minh rng nu b a s phc tha mn

    Bi tp 36

    th mt trong ba s phi bng 1.

    1 2 3, ,z z z

    1 2 3

    1 2 3

    | | | | | |

    1

    z z z

    z z z

    = =

    + + =

    71

    Xcnh tp hp ccim trong mt phng phc biu din cc s

    Bi tp 37

    2

    2

    z

    z

    +phc z sao cho c mt argument bng .3

    72

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    Trng i hc Bch khoa tp HCh MinhBmn Ton ng dng

    -------------------------------------------------------------------------------------

    Chng 1: Ma tran

    Giang vien: Ts. ang Van Vinh (9/2008)

    www.tanbachkhoa.edu.vn

    NOI DUNG---------------------------------------------------------------------------------------------------------------------------

    I. nh ngha ma tran va v du

    II. Cac phep bien oi s cap

    III. Cac phep toan oi vi ma tran

    IV. Hang cua ma tran

    V. Ma tran nghch ao

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    I. Cc khi nim cbn v v d.---------------------------------------------------------------------------------------------------------------

    nh ngham at rn

    Ma trn c mxn l bng s (thc hoc phc) hnh ch nht c mhng v n ct .

    Ma trn A cmxnCt j

    =

    mnmjm

    iniji

    nj

    aaa

    aaa

    aaa

    A

    ......

    ......

    ......

    1

    1

    1111

    Hng i

    I. Cc khi nim cbn v v d.---------------------------------------------------------------------------------------------------------------

    V d1.

    32502

    143

    =A

    y l ma trn thc c2x3.

    Ma trn A c 2 hng v 3 ct.5;0;2;1;4;3 232221131211 ====== aaaaaaPhn tca A:

    V d2

    223

    21

    +=

    ii

    iA

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    Tp hp tt ccc ma trn cmxn trn trng K c k hiul Mmxn[K]

    Ma trn A c m hng v n ct thng c k hiu bi

    nmijaA

    =

    I. Cc khi nim cbn v v d.---------------------------------------------------------

    Ma trn c tt c cc phn t l khngc gi l ma trn khng,k hiu 0, (aij = 0 vi mi i v j ) .

    nh ngha ma trn khng

    =

    000

    000A

    I. Cac khai niem c ban va v du---------------------------------------------------------------------------------------------------------------------------

    Phn t khc khng u tin ca mt hng k t bn tric gi l phn t csca hng.

    nh ngha ma trn dng bc thang1. Hng khng c phn tcs(nu tn ti) th nm di cng

    2. Phn tcsca hng di nm bn phi (khng cngct) so vi phn tcsca hng trn.

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    I. Cc khi nim cbn v v d.---------------------------------------------------------------------------------------------------------------------------

    V d

    52140

    62700

    23012

    =A Khng l ma trnbc thang

    =

    5000

    3000

    2112

    B Khng l ma trnbc thang

    54

    I. Cc khi nim v v dcbn.---------------------------------------------------------------------------------------------------------------------------

    V d

    52000

    41700

    22031

    =A

    L ma trn dng bcthang

    L ma trn dngbc thang

    54

    =

    7000

    3100

    2021

    B

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    I. Cc khi nim cbn v v d----------------------------------------------------------

    Chuyn vca l ma trn cnXmthu c tA bng cch chuyn hng thnh ct.

    ( )mnij

    TaA

    =

    nh ngha ma trn chuyn v

    nmijaA

    =

    2393

    01

    42

    =T

    A32

    904

    312

    =A

    I. Cc khi nim cbn v v d.----------------------------------------------------------

    Nu s hng v ct ca ma trn A bng nhau v bng n, th Ac gi l m a t rn vung cp n.

    nh ngha ma trn vung

    23

    12

    =A

    Tp hp cc ma trn vung cp n trn trng s Kc k hiubi [K]nM

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    I. Cc khi nim cbn v v d.----------------------------------------------------------

    Cc phn t a11, a22,,ann to nnng cho chnh ca ma trnvung A.

    2 3 1 1

    3 4 0 5

    2 1 3 7

    2 1 6 8

    I. Cc khi nim cbn v v d.----------------------------------------------------------

    Ma trn vung c gi l m a t rn tam gic trn nunh ngha ma trn tam gic trn

    =

    200

    630

    312

    A

    ( )ij n nA a =ij 0,a i j= >

    Ma trn vung c gi l ma trn tam gic dinu

    nh ngha ma trn tam gic di

    2 0 0

    4 1 0

    5 7 2

    A

    =

    ( )ij n nA a =ij 0,=

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    I. Cc khi nim cbn v v d.---------------------------------------------------------------

    Ma trn vung Ac gi l ma trn cho nu cc phn t nmngoing chou bng khng, c ngha l ( aij = 0 , ij).

    nh ngha ma trn cho

    = 030

    002

    D

    200

    Ma trn cho vi cc phn t ng chou bng 1c gi lma trnn v, tc l ( aij = 0 , ij; v aii = 1 vi mi i).

    nh ngha ma trn n v

    =

    100

    010

    001

    I

    I. Cc khi nim cbn v v d.---------------------------------------------------------------

    Ma trn ba ng cho l ma trn cc phn t nm ngoi bang cho (ng cho chnh, trn n mtng, di n mtng)u bng khng.

    nh ngha ma trn ba ng cho.

    =

    9500

    1840

    0713A

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    I. Cc khi nim cbn v v d.---------------------------------------------------------------

    Ma trn vung thc A t ha aij = aji vi mi i = 1,.n v j =1,,nc gi l m a t rni xng (tc l , nu A = AT)

    nh ngha ma trn i xng thc

    = 741

    312

    A

    Ma trn vung A tha aij = - aji vi mi i v j (tc l A = -AT)c gi l m a t rn phni xng.

    nh ngha ma trn phn i xng

    1 3

    1 7

    3 7

    0

    0

    0

    A

    =

    II. Cc php bin i scp.---------------------------------------------------------------------------------------------------------------------------

    Cc php bini scpi vi hng

    ; 0 i ih h1. Nhn mt hng ty vi mt skhc khng

    ; + i i jh h h

    2. Cng vo mt hng mt hng khcc nhn vi mt sty

    i jh h3.i ch hai hng ty

    Tng t c ba php bini scpi vi ct.

    Ch : cc php bin i s cp l cc php bin i c bn,thng dng nht!!!

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    II. Cc php bin i scp.---------------------------------------------------------------------------------------------------------------------------

    Mi ma trn u c tha vma trn dng bc thang bng ccphp bin i scp i vi hng.

    nh l 1

    Khi dng cc php bin i scp i vi hng ta thu cnhiu ma trn bc thang khc nhau

    Ch

    II. Cc php bin i scp.---------------------------------------------------------------------------------------------------------------------------

    Dng cc php bin i s cp i vi hng a ma trn sauy v ma trn dng bc thang.

    1 1 1 2 1

    2 3 1 4 5

    3 2 3 7 4

    1 1 2 3 1

    V d

    Bc 1. Bt u tct khc khng u tin tbn tri. Chnphn tkhc khng ty lm phn tcs.

    1 1 1 2 1

    2 3 1 4 5

    3 2 3 7 4

    1 1 2 3 1

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    1 1 1 2 1

    0 1 1 0 3

    0 1 0 1 1

    0 2 1 1 2

    4 4 1 +

    h h h

    2 2 12 h h h

    3 3 13 h h h

    Bc 2. Dng bsc i vi hng, khtt ccc phn tcn li cact.

    II. Cc php bin i scp.---------------------------------------------------------------------------------------------------------------------------

    1 1 1 2 1

    2 3 1 4 5

    3 2 3 7 4

    1 1 2 3 1

    A

    =

    4 4 3

    1 1 1 2 1

    0 1 1 0 3

    0 0 1 1 4

    0 0 0 0 0

    +

    h h h

    Bc . Che tt ccc hng thng cha phn tcsv nhnghng trn n. p dng bc 1 v 2 cho ma trn cn li

    3 3 2

    4 4 22

    1 1 1 2 1

    0 1 1 0 30 0 1 1 4

    0 0 1 1 4

    +

    h h hh h h

    II. Cc php bin i scp.---------------------------------------------------------------------------------------------------------------------------

    Nu dng cc bin i scp a A vma trn bc thangU, th U c gi l dng bc thang ca A.

    nh ngha

    Ct ca ma trn bc thang A c gi l ct csnu ct

    nh ngha

    c a p n cs

    1 2 0 2

    0 0 1 3

    0 0 0 7

    A

    =

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    III. Cc php ton i vi ma trn---------------------------------------------------------------------------------------------------------------------------

    Hai ma trn bng nhau nu: 1) cng c; 2) cc phn t nhngv tr tngng bng nhau (aij = bij vi mi i v j ) .

    Sbng nhau ca hai ma trn

    Php cng hai ma trn

    Tng A + B :Cng c

    Cc phn t tngng cng li

    =

    =

    741

    623;

    503

    421BA

    =+

    1244

    1002BA

    V d

    III. Cc php ton i vi ma trn---------------------------------------------------------------------------------------------------------------------------

    Php nhn ma trn vi mt s.Nhn ma trn vi mt s, t a ly s nhn vi tt c cc phn

    t cam at rn.

    =503

    421A

    =842

    2 A

    V d

    Tnh cht:a) A + B = B + A; b) (A + B) + C = A + ( B + C);

    c) A + 0 = A; d) k(A + B) = kA + kB;

    e) k (mA) = (km) A; f) (k + m)A = kA + mA;

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    III. Cc php ton i vi ma trn---------------------------------------------------------------------------------------------------------------------------

    Php nhn hai ma trn vi nhau

    ( ) ; ( )pij m i pj nA a B b = =

    nmijcCAB == )( vi pjipjijiij bababac +++= ...2211

    1*

    * *

    jb

    b

    tm phn tc2,3ma trn tch: ly hng 2 ca A nhn vi ct 3ca B (coi nhnhn tch v hng hai vctvi nhau)

    1 2 ... ... ...

    *

    i i ip

    pj

    ijAB a a a

    b

    c= =

    11 12 13

    1 2 22 1 4 c c c

    III. Cc php ton i vi ma trn---------------------------------------------------

    =

    =

    342

    103

    221

    ;014

    412BA

    V d

    Tnh AB

    12 137 c c

    21 22 234 1 0 2 4 3 c c c

    11c = ( )2 1 4

    1

    3

    2

    2 1 ( 1) 3 4 2 7= + + =

    21 22 23c c c

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    III. Cc php ton i vi ma trn---------------------------------------------------------------------------------------------------------------

    2 1 1;

    4 1 3

    = =

    A B

    V d

    Tm ma trnX

    , thaAX = B

    .

    a

    .

    AX=B

    b

    2 1 1

    4 1 3

    a

    b

    =

    2 1

    4 3

    a b

    a b

    =

    +

    2 1

    4 3

    a b

    a b

    =

    + =

    2 1,

    3 3a b = =

    2 / 3Vay

    1/ 3X

    =

    III. Cc php ton i vi ma trn---------------------------------------------------

    a. A(BC) = (AB)C; b. A(B + C) = AB + AC;

    e. k (AB) = (kA)B = A(kB).

    d. ImA = A = A Im

    Tnh cht ca php nhn hai ma trn

    c. (B + C)A = BA + CA;

    Ch :1. Ni chung BAAB

    2. ACAB = CB =

    0=AB 00 == BA3.

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    III. Cc php ton i vi ma trn---------------------------------------------------------------------------------------------------------------------------

    Nng ma trn ln ly tha.

    nA A A A A=

    0Qui c: A I= 2A A A=

    3A A A A=

    nnijn

    nn

    n aAaxaxaxaxf

    =++++= )(;...)( 011

    1

    n

    11 1 0( ) ... .

    n nn nA a A a A a IA a

    = + + + +

    III. Cc php ton i vi ma trn---------------------------------------------------------------------------------------------------------------

    22 1 ; ( ) 2 4 33 4

    A f x x x

    = = +

    V d

    Tnh f(A).

    2( ) 2 4 3A A IA= +

    2 1 2 1 2 1 1 0( ) 2 4 33 4 3 4 3 4 0 1

    A

    = +

    1 6 8 4 3 0( ) 2

    18 13 12 16 0 3A

    = +

    3 8( )

    24 13

    A

    =

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    III. Cc php ton i vi ma trn---------------------------------------------------------------------------------------------------------------

    1 3.

    0 1A

    =

    V d

    Tnh A2; A3, t suyraA200

    2 1 3 1 3

    1 6

    0 1 0 1 10

    3 2 1 6 1 3

    0 1 0 1A A A

    = =

    1

    1

    9

    0

    =

    200 1

    0 1

    200 3A

    =

    III. Cc php ton i vi ma trn---------------------------------------------------------------------------------------------------------------

    2 3.

    0 2

    =

    A

    V d

    Tnh A200

    2 3 1 3/ 2

    1 a

    0 2 0 1 0 1

    1 1Ta co:

    0 1 0 1

    na na

    =

    200200200

    200

    3002 2

    0 2A

    =

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    III. Cc php ton i vi ma trn---------------------------------------------------------------------------------------------------------------

    1 1.

    1 1

    =

    A

    V d

    Tnh A200

    2 1 1 1 1 2 2 1 1

    1 1 1 1 2 2 1 1

    1Suy ra: A 2n n A=

    199 199200

    199 1992 2

    2 2A =

    3 2.

    2 3

    =

    AV d

    Tnh A200

    1 1 1 02 2

    1 1 0 1A B I

    = + = +

    V B v I giao hon nhau nn ta dng nhthc Newton

    12n nB B=

    ( ) ( ) ( )200 200 1990 1 200 200200 200 2002 2 2 ...B I C B C B C I+ = + + +

    0 200 200 1 1 199 199 1 200 200200 200 2002 .2 2 .2 ...C B C B C I

    = + + +

    ( )0 200 1 199 199 200200 200 200 2004 4 ... .42

    BC C C C I = + + + +

    ( )( )200

    4 1 1 . 2

    B

    I= + +

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    IV. Hng ca ma trn-----------------------------------------------------------------------------------------------------

    nh ngha hng ca ma trn

    Gi s Amxn tngng hng (ct) vi ma trn bc thangE. Khi ta gi hng ca ma trn A l s cc hng khc

    r(A) = s hng khc khng ca m a t rn bc thang E

    IV. Hng ca ma trn-----------------------------------------------------------------------------------------------------

    V dTm hng ca m a t rn sau

    1 2 1 1

    2 4 2 2

    3 6 3 4

    A

    =

    Gii. 1 2 1 1

    2 4 2 2

    3 6 3 4

    =

    A

    1 2 1 1

    0 0 0 0

    0 0 0 1

    2 3

    1 2 1 1

    0 0 0 1

    0 0 0 0

    h h( ) 2r A =

    2 2 1

    3 3 1

    2

    3

    h h h

    h h h

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    IV. Hng ca ma trn-----------------------------------------------------------------------------------------------------

    V dS dng bini scp, tm hng cam at rn sau

    1 2 3 3

    2 4 6 9

    2 6 7 6

    A

    =

    V dTm hng ca m a t rn sau

    2 3 1 4

    3 4 2 9

    2 0 1 3

    A

    =

    IV. Hng ca ma trn-----------------------------------------------------------------------------------------------------

    V dTm tt c cc gi tr thc m sao cho r(A) =3

    1 1 1 2

    2 3 4 1

    3 2 1

    =

    +

    A

    m m

    1 1 1 2 1 1 1 2

    2 3 4 1 0 1 2 3

    3 2 1 0 1 3 5

    =

    +

    A

    m m m m

    1 1 1 2

    0 1 2 3

    0 0 1 8

    m m

    r(A) = 3 vi mi gi trm.

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    IV. Hng ca ma trn-----------------------------------------------------------------------------------------------------

    V dTm tt c cc gi tr thc m sao cho r(A) =2

    1

    11

    m m

    A m mm m

    =

    V dTm tt c cc gi tr thc ca m cho r(A) = 3.

    1 1 1 1

    2 3 1 4

    3 3 1

    A

    m m

    =

    +

    IV. Hng ca ma trn-----------------------------------------------------------------------------------------------------

    Tnh cht ca hng ma trn

    1. r (A) = 0 A = 0

    2. A = (aij)mxn r(A) min{m, n}

    BSC3. Nu A B, th r (B) = r (A)

    2 2 2

    2 2 2

    2 2 2

    A

    =

    2 2 2

    0 0 0

    0 0 0

    ( ) 1.r A =

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    V. Ma trn nghch o-----------------------------------------------------------------------------------------------------

    nh ngha m a t rn nghcho

    Ma trn vungAc gi l ma trn kh nghch nu tn tima trnIsao cho AB = I =BA. KhiBc gi l nghcho caA v k hiu lA-1.

    2 1 3 1 2 2

    5 3

    =

    2 25 2

    2 1 3 1 1 0

    5 3 5 2 0 1AB I

    = = =

    3 1 2 1 1 0

    5 2 5 3 0 1BA I

    = = =

    1 3 1

    5 2A B

    = =

    V. Ma trn nghch o-----------------------------------------------------------------------------------------------------

    Khng phi bt kma trn vungA no cng khnghch. Crt nhiu ma trn vung khng khnghch.

    Ch

    Ma trn khnghch c gi l ma trn khng suy bin

    n ng a

    Ma trn khng khnghch c gi l ma trn suy bin

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    V. Ma trn nghch o-----------------------------------------------------------------------------------------------------

    Cho ma trn vung A, cc mnh sau y tng ng

    Stn ti ca ma trn khnghch.

    1. Tn tiA-1 (A khng suy bin)

    .

    3. AX = 0 suy raX = 0.

    4. A ITng ng hng

    V. Ma trn nghch o-----------------------------------------------------------------------------------------------------

    Ma trn thu c tI bng ng 1 php bin i scp cgi l ma trn scp.

    nh ngha ma trn scp

    V d1 0 0 1 0 0

    2 2 12

    2

    1 0 0 1 0 0

    0 1 0 2 1 0

    0 0 1 0 0 1

    h h hI E

    +

    = =

    3 33

    10 1 0 0 1 00 0 1 0 0 3

    h h

    I E

    = =

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    V. Ma trn nghch o-----------------------------------------------------------------------------------------------------

    Mt php bin i scp i vi hng ca ma trn A ng

    3 1

    3

    1 0 0 0 0 1

    0 1 0 0 1 0

    0 0 1 1 0 0

    h hI E

    = =

    ngha vi nhn bn tri A vi ma trn sc p tng ng.

    Mt php bin i scp i vi ct ca ma trn A ng

    ngha vi nhn bn phi A vi ma trn scp tng ng.

    V. Ma trn nghch o-----------------------------------------------------------------------------------------------------

    3 1

    2 1 1 3 2 1

    1 1 0 1 1 0

    3 2 1 2 1 1

    h hA B

    = =

    3 2 1 0 0 1 2 1 1

    1 1 0 0 1 0 1 1 0

    2 1 1 1 0 0 3 2 1

    =

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    V. Ma trn nghch o-----------------------------------------------------------------------------------------------------

    1 1bsc hang

    ...n nA I I E E E A =

    11 1...n nA E E E I

    =

    1 trenbsc hangI A

    V. Ma trn nghch o-----------------------------------------------------------------------------------------------------

    Cch tm A-1

    [ A|I ] [ I|A-1 ]Bsc i vi hng

    V dTm nghcho (nu c ) cam at rn

    =

    111

    321

    =

    101

    011

    001

    210

    110

    111

    100

    010

    001

    321

    221

    111

    ]|[ IA

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    V. Ma trn nghch o-----------------------------------------------------------------------------------------------------

    110

    121

    111

    100

    010

    011

    110

    011

    001

    100

    110

    111

    ]|[

    110

    121

    100

    010 1=

    AI

    =

    110

    121

    0121

    A

    V. Ma trn nghch o-----------------------------------------------------------------------------------------------------

    Tnh bng cc php scp i vi hng ca ma trn[ A|I ] ta cn sdng

    phc tp ca thut ton tm A-1

    n3 php nhn hoc chia

    n3 2n2 + n php cng hoc tr

    1

    nnA

    i vi hai ma trn khnghch A v B, cc khng nh sau yng.

    Tnh ch t ca ma trn nghch o

    (A-1)-1 = A

    Tch AB l hai ma trn khnghch.

    (AB)-1 = B-1A-1

    (AT)-1 = (A-1)T

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    IV. Ma trn nghch o-----------------------------------------------------------------------------------------------------

    V dTm tt c cc gi tr thc m ma trn sau kh nghch

    1 1 2

    2 13 2 1

    =

    A m

    V dTm tt c cc gi tr thc ca m choAkh nghch.

    1 1 1 1

    2 3 1 4

    3 3 1

    A

    m m

    =

    +

    VI. Kt lun------------------------------------------------

    Ma trn l g? Ma trn vung ? Ma trn bc thangMa trn khng? Ma trn cho? Ma trn chuyn v?

    Ma trn n v? Ma trn i xng?

    Cc php ton i vi ma trn: Sbng nhau Php cng

    Nhn ma trn vi mt s Nhn hai ma trn vi nhau

    Hng ca ma trn l g?

    Lm thno tm hng ca mt ma trn cho trc?

    Ma trn khnghch l g?

    Lm thno tm nghch o ca mt ma trn cho trc?

    Nghch o ca ma trn A l g?

    Nng ln ly tha

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    Thc hin php tonBi tp 1

    2 3 21 2 1

    1 2 33 0 4

    Tm f(A), bit

    Bi tp 2.

    23 4 5x x x= + v 2 3

    A

    =

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    Bi tp 3.

    Tm ma trnX , saochoA X =B,vi2 1

    =2

    =3 1 3

    Cho

    Bi tp 4

    2 12 3 4

    ; 1 31 2 7

    3 2

    A B

    = =

    3 2 T

    A B+

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    a m a t rn s a u v dng bc thang bng bini scp

    Bi tp 5.

    1 1 2 1 1

    2 1 3 4 2

    3 4 7 3 1

    1 3 4 7 3

    Bi tp 6Tm ma trn nghcho, nu c 1 1 1

    2 3 1

    3 4 1

    A

    =

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    Bi tp 7a v ma trn bc thang, tm hng cam at rn

    1 1 1 0

    2 0 1 3

    Bi tp 8

    Tm ma trn A , nu

    1 0 5 25 3A A

    =

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    Bi tp 9

    Tm cc gi tr ca s v t, sao cho ma trn sau li xng

    2s s st

    2t s s

    Bi tp 10

    Cho , cho A l ma trn c3xn , B cnx3.1 0 0

    0 0 1P

    =

    a )Mt PA theo ngha bini scpi vi hngb)Mt BP theo ngha bini scpi vi ct

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    Bi tp 11

    Cho A, B, C l cc ma trn,n gin biu thc sau

    Bi tp 12

    Tm ma trn nghcho ca A

    2 7 1

    1 4 1A

    =

    1 3 0

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    Bi tp 13

    Tnh A43, bit

    2 1 =

    3 2

    Cho l ma trn vung.

    Bi tp 14

    cos sin

    sin cosA

    =

    a) Tnh A2.

    b) Tnh An.

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    Bi tp 15

    Cho hai ma trn A v B

    1 1 1

    3 2

    =

    0 0 2

    = 0 1

    Tm tt c ma trn X, sao cho AX = B.

    Bi tp 16

    Tm tt c cc gi tr msaocho(A)=2

    1 1

    1 1

    m

    A m

    =

    1 1 m

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    Bi tp 17

    Bin lun t h e o m hng ca m a t rn A

    1 1 2

    2 1 5

    m

    A m

    =

    1 10 6 m

    Bi tp 18

    Tm tt c s thc m, sao cho ma trn A k h nghch

    1 1 1

    2 3 1A

    =

    3 5 m

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    Bi tp 19

    Tm tt c cc s thc m, sao cho ma trn A k h nghch

    1 1 1 2

    2 3 1 4A

    =

    3 2 1m m +

    Bi tp 20

    Gi s A l ma trn kh nghch cp 5. Tm r(A) v r (A-1)

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    Trng i hc Bch khoa tp. HCh MinhBmn Ton ng dng

    ---------------------------------------------------------------

    i stuyn tnh

    Chng 2: nh thc

    Ging vin Ts. ng Vn Vinh (9/2008)

    www.tanbachkhoa.edu.vn

    NI DUNG---------------------------------------------------------------------------------------------------------------------------

    I nh nghanh thc v v d.

    II Tnh cht canh thc

    III Khai trin Laplace

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    I. nh ngha v v d---------------------------------------------------------------------

    Cho l ma trn vung cp n.

    nh thc ca A l mt sk hiu bi detnnij

    aA

    =

    AaAnnij

    ==

    )(

    K hiu lnh thc thuc tA b n cch b i hnMthi v ct th j ca ma trn A;

    ij( 1)i j

    ijA M+

    = Bi sca phn taij li lng

    nh ngha bi sca phn taij

    I. nh ngha v v d---------------------------------------------------------------------------------------------------------------------------

    b) k =2: 11 12

    11 22 12 21 11 11 12 1221 22

    a aA A a a a a a A a A

    a a

    = = = +

    a) k =1: [ ] 1111 aAaA ==

    nh nghanh thc bng qui np

    c) k =3:

    11 12 13

    21 22 23 11 11 12 12 13 13

    31 32 33

    a a a

    A a a a A a A a A a A

    a a a

    = = + +

    d) k =n: 11 12 1

    11 11 12 12 1 1*

    nn n

    a a aA A a A a A a A

    = = + + +

    ...............

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    I. nh ngha v v d---------------------------------------------------------------------------------------------------------------------------

    1 2 3A A A A= + +

    Tnh det (A), vi

    =

    423

    032

    321

    A

    V d

    Gii

    23

    32)1()3(

    43

    02)1(2

    42

    03)1(1

    312111 +++++=A

    11151612 =+=A

    1 1 1 111

    1 2 33 0

    2 3 0 ( 1) 122 4

    3

    ( )

    2 4

    1A + +

    == =

    II. Tnh cht ca nh thc---------------------------------------------------------------------

    1. C th tnh nh thc bng cch khai trin theo bt khng hoc ct ty no

    *

    a a a a A a A a A= = + + +

    1

    2

    1 1 2 2

    * *

    j

    j

    j j j j nj nj

    nj

    a

    aa A a A a A

    a

    = = + + +

    *

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    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    Tnhnh thc det (A), vi

    =

    004

    225

    313

    A

    V d

    Khai trin theo hng th3

    322231)1(4

    004

    225

    313

    )1(4

    004

    225

    3131313

    ==

    =

    = ++

    A

    Gii.

    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    2 3 3 2

    3 0 1 4

    V d

    n n c e , v

    2 0 3 2

    4 0 1 5

    =

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    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    Khai trin theo ct thhai

    12 22 32 42 12

    2 3 3 23 0 1 4

    ( 3) 0 0 0 3A A A A A A

    = = + + + =

    Gii

    4 0 1 5

    3 1 4

    3 2 3 2 87

    4 1 5

    A = = =

    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    nh thc ca ma trn tam gic bng tch cc phn t nmtrnng cho.

    V d

    120145)3(2

    10000

    94000

    82500

    1763040312

    ==

    =A

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    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    Sdng bini scpi vi hngtnhnh thc

    1.Nu thi ih hA B

    | | | |B A=

    i i jh h h + | | | |B A=

    3. Nu thi jh h

    A B

    | | | |B A=

    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    V d

    Sdng cc php bini scp, tnhnh thc

    0532

    1211

    =

    13122623

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    17301010

    2110

    1211

    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    Gii

    1312

    2623

    0532

    1211

    ||

    =A

    2 2 12 h h h

    3 3 13 h h h

    4 4 12 +h h h

    1504101

    211

    ||

    =A

    Khai trin theo ctu tin||A

    173

    101

    211

    )1(1 11

    +

    19154

    11

    )1(1 21

    =

    =

    +

    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    Bc 1. Chn 1 hng (hoc mt ct) ty ;

    Bc 2. Chn mt phn tkhc khng ty ca hng (hay ct)bc 1. Dng bini scp, khtt ccc phn tkhc.

    Nguyn tc tnhnh thc sdng bini scp

    Bc 3. Khai trin theo hng (hay ct) chn.

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    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    V d

    Sdng bini scp, tnhnh thc

    =02321123

    A

    1314

    2413

    0411

    0253

    0232

    1123

    1314

    2413

    0232

    1123

    ||

    =A

    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    Gii

    3 3 12 +h h h

    4 4 1 h h h

    2 3 2

    | | 5 8 0

    5 5 0

    A

    =

    411

    253232

    )1(1 41

    +

    1 35 8( 2) ( 1) 30

    5 5

    += =

    Khai trin theo ct s4||A

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    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    det (AT) = det (A)

    det(AB) = det(A) det(B)

    Ma trn c mt hng (ct) b ng khng, th det (A) = 0

    Ma trn c hai hng (ct) tlnhau, th det (A) = 0

    Ch :det(A+B) det(A) + det(B).

    II. Tnh ch t ca nh thc---------------------------------------------------------------------------------------------------------------------------

    GisAl ma trn khnghch nxn. Khi tn ti matrn khnghch A-1, sao cho AA-1 = I. Suy ra

    Chng minh

    Ma trn vungA khnghchkhi v chkhi det(A) 0.

    nh l

    det(AA-1) = det (I) det(A).det(A-1) = 1 det(A) 0

    1 1AA P

    A

    = , vi

    11 12 1

    21 22 2

    1 2

    Tn

    nA

    n n nn

    A A A

    A A AP

    A A A

    =

    Gisdet(A) 0. Khi

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    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    | |A i =

    =

    *

    *

    *

    111

    111

    iii

    jjj

    aaa

    aaa

    A

    =

    *

    *

    *

    111

    111

    jjj

    jjj

    aaa

    aaa

    B

    1 1 2 20,

    i j i j in jna A a A a Ai j

    + + + =

    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    Tnh cht ca ma trn nghcho

    1. 1 1det( )

    det( )A

    A

    =

    1n=

    Chng minh.

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    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    ChoA l ma trn khnghch. Khi

    11 12 1T

    nA A A

    Cng thc tnh ma trn nghcho A-1

    1 1AA P

    A

    = , vi 21 22 2

    1 2

    nA

    n n nn

    A A AP

    A A A

    =

    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    V d. Tm ma trn nghcho ca

    =

    043

    132

    111

    A

    Gii. 02)det( =A Akhnghch

    Tnh 9 bi sca cc phn t

    1 111 ( 1) 4;4 0A +

    = = 1 212 ( 1) 3;3 0A +

    = = 1 313 ( 1) 13 4A +

    = =

    21 22 23 31 32 334; 3; 1; 2; 1; 1A A A A A A= = = = = =

    1

    4 4 21

    3 3 12

    1 1 1

    A

    =

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    Tnh det(A), nu

    V d1

    2 1 1 3

    3 2 1 2

    4 1 0 1

    3 3 2 2

    =

    Tnh det(A), vi

    V d2

    4 1 1 0

    3 2 4 1

    2 1 3 1

    5 1 2 3

    =

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    Khngnh no sauyng?

    V d3

    2

    3

    2 1 3

    3 2 3 1( )

    3 5 2 1

    6 3 2 1 9

    x

    xf x

    x x

    x

    +=

    +

    +

    a) Bc ca f(x)l 5.

    b) Bc ca f(x)l 4.

    c) Bc caf(x)l 3.

    d) Cc cu khcu sai.

    Tnhnh thc ca ma trn sau

    V d4

    1 0 1 i+

    0 11 1

    A ii i

    =

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    Tnhnh thc

    V d5

    1 2 2 2 2

    2 1 2 2 2

    2 2 1 2 2

    2 2 2 1 2

    2 2 2 2 1

    I =

    Gii phng trnh, vi a, b, c l cc sthc.

    V d6

    2 3

    2 3

    1

    1

    x x x

    a a a

    2 3

    2 3

    1

    1

    b b b

    c c c

    =

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    Gii phng trnh

    V d7

    1 1 1 11 1 1 1x

    1 1 1

    x

    n x

    =

    Tnhnh thc

    V d8

    1 1 1 1

    1 0 1 1

    1 1 0 1

    1 1 1 0

    I =

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    Tnhnh thc

    V d9

    1 2 3 n

    1 2 0 0

    1 2 3 0

    nD

    =

    Tnhnh thc

    V d10

    3 2 2 2

    2 2 3 2

    2 2 2 3

    nD =

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    Gii phng trnh trong C

    V d11

    2 2 3x

    0

    0 0 7 6

    0 0 5 3

    =

    Tnhnh thc

    V d12

    7 5 0 0

    0 2 7 0

    0 0 0 7

    nD =

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    Khai trin theo hng 1, ta cGii v d9 11 127 5nD A A= +

    1 1 1 1

    7 5 0 0 2 5 0 0

    2 7 5 0 0 7 5 0

    7( 1) 5( 1)0 2 7 0 0 2 7 0

    0 0 0 7 0 0 0 7

    nD + +

    = +

    1 11

    7 5 0 0

    2 7 5 0

    7 5.2( 1) 0 2 7 0

    0 0 0 7

    n nD D +

    =

    1 2

    7 10n n n

    D D D

    =

    1 1 25 2( 5 )n n n nD D D D =

    1 2 2 35 2( 5 )n n n nD D D D =

    21 2 35 2 ( 5 )n n n nD D D D =

    21 2 15 2 ( 5 ) *( )

    nn nD D D D

    =

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    1 27 10n n nD D D =

    1 1 22 5( 2 )n n n nD D D D =

    1 2 2 32 5( 2 )n n n nD D D D =

    1 2 12 5 ( 2 ) * )*(n

    n nD D D D

    =

    21 2 15 2 ( 5 ) *( )

    nn nD D D D

    =

    1 2* **( ) & ( ) theo vanD D D

    Tnhnh thc

    V d13

    5 3 0 0

    0 2 5 0

    0 0 0 5

    nD =

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    Tnhnh thc

    V d14

    9 5 0 0

    0 4 9 0

    0 0 0 9

    nD =

    Tm ma trn nghcho bng cch tnhnh thc

    V d15

    1 2 1

    2 3 1A

    =

    3 5 2

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    Tm ma trn nghcho ca ma trn sau

    V d16

    1 0 0 0

    2 1 0 0

    5 4 1 0

    1 2 3 2

    =

    Tm tt ccc gi trca mma trn sau khnghch

    V d17

    1 1 2 1

    2 1 5 3A

    =

    5 0 7

    1 2 3 3

    m

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    Tm tt ccc gi trthc ca mma trn sau khnghch.

    V d18

    1 2 1 1 1 12 3 2 3 2A m

    =

    Cho . 1) Tnh det (A-1).

    V d19

    1 1 1

    2 3 1

    3 3 5

    A

    = 2) Tnh det (5A)-1.

    3) Tnh det (PA).

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    Cho

    V d20

    3 3[ ]; [ ];det( ) 2;det( ) 3.A M R B M R A B = =

    1) Tnh det (4AB)-1.

    2) Tnh det (PAB).

    Cho k l s t nhin nh hn hoc bng n; i1, i2, , ikv j1, j2,, jkl nhng stnhin tha

    1 ... ;1 ...i i i n j j j n < < < < < <

    III. Khai trin Laplace

    -----------------------------------------------------------------------------

    nh thc con cp k, k hiu bi , lnh thc thuc tA bi nhng phn tgiao ca k hng i1, i2, , ikv k ct j1, j2, , jk.

    1

    1

    ,...,

    ,...,k

    k

    i i

    j ja

    nh nghanh thc con cpk

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    III. Khai trin Laplace

    -----------------------------------------------------------------------------

    i lng 1 111

    1

    1

    . . . . . . ,, . . . ,

    ,

    . . . ,

    . . . , , . . . ,( 1 )n

    n

    n n n

    n

    i ii i

    j j

    j j i i

    j jMA + + + + +

    =

    c gi l bi scpkca 11

    ,...,

    ,...,k

    k

    i i

    j ja

    nh l (Khai tri n Laplace)nh thc ca ma trn vung A bngtng tt c cc tchcanhthc con cp krt ra t k hng (hoc k ct) no vi bi sca chng.

    III. Khai trin Laplace

    -----------------------------------------------------------------------------

    Tnhnh thc bng khai trin Laplace.

    bc 1. Chn k hng (hoc k ct) ty

    bc 2. Tnh tt c ccnh thc con cp k thuc t k hngchn. Tng cng c nh thc con cpk.k

    nC

    bc 3. Tm tt ccc bi scp k tngng ca ccnh thccon cpkbc 2.

    bc 4.nh thc ca ma trn A bngtng tt ccc tch canhthc con cp k vi bi sca chng.

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    III. Khai trin Laplace

    -----------------------------------------------------------------------------

    V d21

    Tnhnh thc ca A bng cch sdng khai trin Laplace.

    2 3 1 13 0 1 0

    A

    =

    1 0 2 0

    III. Khai trin Laplace

    -----------------------------------------------------------------------------

    Gii

    Chn k = 2, chn 2 hng: hng 2 v hng 4.

    3 0 1 0

    1 0 2 0

    T n t i nh thc con c 2 nhn chc 1 khc khn .2 =

    2,41,3

    3 15

    1 2a

    = =

    2,4 2 4 1 31,3

    3 1( 1) 1

    2 1A

    + + += =

    2,4 2,41,3 1,3det( ) . 5.1 5A a A= = =

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    Tnh det(A) sdng khai trin Laplace

    V d22

    2 1 2 3 5

    1 0 3 0 2

    3 4 2 5 1

    2 0 1 0 4

    3 2 5 2 1

    A =

    III. Khai trin Laplace

    ----------------------------------------------------------------------------

    Chn k = 2, chn 2 hng: hng 2 v hng 4.1 0 3 0 2

    2 0 1 0 4

    Tn ti nh thc con cp 2 nhng chc 2 khc khng.25 10C =

    2,41 3

    1 35a = = 2,4 1 3 2 4

    1 3 5+ + +

    1,3

    2 2 1

    2,4 2,4 2,4 2,4 2,4 2,41,3 1,3 1,5 1,5 3,5 3,5det( ) . . .A a A a A a A= + +

    2,41,5

    1 20

    2 4a = =

    2,43,5

    3 210

    1 4a = =

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    II. Tnh cht ca nh thc---------------------------------------------------------------------------------------------------------------------------

    Tnhnh thc bng bi scn n! php ton.

    Nu mt my tnh siu tc c th tnh t t php tontrong mt giy thtnh mtnh thc cp 25 cn 500.000nm (cn 25! , khong 1.5x1025 php ton).

    Phn ln cc my tnh s dng bin i s cp tnhdet (A).

    Cc php bin i s cp cn (n3+2n-3)/3 php nhn v

    chia. Bt kmy tnh no cng c thtnhnh thc cp 25trong vng phn ca 1 giy, chcn khong 5300 php ton.

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    Trng i hc Bch khoa tp. HCh MinhBmn Ton ng dng

    ---------------------------------------------------------------

    i stuyn tnh

    Chng 3: Hphng trnh tuyn tnh

    Ging vin Ts. ng Vn Vinh (9/2007)

    www.tanbachkhoa.edu.vn

    Ni dung---------------------------------------------------------------------------------------------------------------------------

    I H phng trnh tuyn tnh tng qut

    II H hn trnh tu n tnh thu n nh t

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    I. Hphng trnh tuyn tnh tng qut---------------------------------------------------------------------------------------------------------------------------

    11 1 12 2 1 1

    21 1 22 2 2 2

    n n

    n n

    a x a x a x ba x a x a x b

    + + + =

    + + + =

    H phng trnh tuyn tnh gm m phng trnh, n n cdng:

    nh ngha h phng trnh tuyn tnh.

    a11, a12, , amnc gi l h s ca h phng trnh.

    1 1 2 2m m mn m ma x a x a x b

    + + + =

    b1, b2, , bmc gi l h s t do ca h phng trnh.

    I. Hphng trnh tuyn tnh tng qut---------------------------------------------------------------------------------------------------------------------------

    H phng trnh tuyn tnh c gi l thun nht nu tt ccc h s t do b1, b2, , bmu bng 0.

    nh ngha h thun nht.

    nh ngha h khng thun nht.

    Nghim ca h l mt b n s c1, c2, , cm sao cho khi thayvo tng phng trnh ca h tac nhngng thcng.

    p ng r n uy n n c g ng u n n n u

    nht mt trong cc h s t do b1, b2, , bm khc 0.

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    I. Hphng trnh tuyn tnh tng qut---------------------------------------------------------------------------------------------------------------------------

    H tng thch

    H khng tng thch

    Mt h phng trnh tuyn tnh c th:

    1. v nghim,

    2. c duy nht mt nghim3. C v s nghim

    Hai h phng trnhc gi l tngng nu chng cngchung mt tp nghim.

    gii h phng trnh ta dng cc php bini h vh tngng, m h ny giin gin hn.

    I. Hphng trnh tuyn tnh tng qut---------------------------------------------------------------------------------------------------------------------------

    C 3 php bini tngngi vi h phng trnh .

    Mt php binic gi l tngng nu bin mt hphng trnh v mt h tngng.

    nh ngha php bini tngng

    1. Nhn hai v ca phng trnh vi mt s khc khng.

    3.i ch hai phng trnh.

    2. Cng vo mt phng trnh mt phng trnh khc c nhn vi mt s ty .

    Ch : Chng ta c th kim tra d dng rng cc php bini trn l cc php bini tngng.

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    I. Hphng trnh tuyn tnh tng qut---------------------------------------------------------------------------------------------------------------------------

    0y+ =

    Gii h phng trnh:

    0

    2 3 3

    2 3

    x y

    x y z

    x y z

    + =

    + =

    =

    V d

    1 2

    1 3h h

    + 3 3 3

    3 3

    y z

    y z

    + =

    =

    2 3h h +

    0

    3 3 3

    4 0

    x y

    y z

    z

    + =

    + = =

    Phng trnh c nghim duy nht : x = 1 ; y = - 1 ; z = 0

    I. Hphng trnh tuyn tnh tng qut---------------------------------------------------------------------------------------------------------------------------

    1 1 0

    2 1 3

    1 2 1

    Ma trn h s:

    Ma trn mrng:1 1 0 02 1 3 3

    1 2 1 3

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    I. Hphng trnh tuyn tnh tng qut---------------------------------------------------------------------------------------------------------------------------

    1 1 0 0

    2 1 3 3

    1 2 1 3

    1 1 0 0 1 2

    1 3

    2h h

    h h

    +

    +

    2 3h h +

    0 3 3 3

    0 3 1 3

    1 1 0 0

    0 3 3 3

    0 0 4 0

    I. Hphng trnh tuyn tnh tng qut

    n csln tngng vi ct cha phn t cs.

    n t d o l tngng vi ct khng c phn t cs.

    nh nghan csvn t do.

    1 1 1 2 12 2 3 5 6

    3 3 4 1 1

    BSC HNG1 1 1 2 10 0 1 1 4

    0 0 0 6 8

    x1, x3, x4: l cc n cs

    x2: n tdo

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    N u , th h AX = b v n him.( | ) ( )r A b r A

    I. Hphng trnh tuyn tnh tng qut

    nh l Kronecker Capelli

    Nu hai ma trn m rng ca hai h phng trnh tuyn tnhtngng hng vi nhau th hai h tngng.

    Nu , th h AX = b c nghim.( | ) ( )r A b r A=

    Nu = s n, th h AX = b c nghim duynht.

    ( | ) ( )r A b r A=

    Nu < s, th h A X = b c v s nghim.( | ) ( )r A b r A=

    I. Hphng trnh tuyn tnh tng qut--------------------------------------------------------------------------------------------------

    2. Dng bin i scp i vi hng a ma trn mrngvma trn dng bc thang. Kim tra hc nghim haykhng

    Sdng bin i scp i vi hng gii h

    1. Lp ra ma trn mrng ( | )A A b=

    3. Vit hphng trnh tng ng vi ma trn bc thang

    4. Gii hphng trnh ngc tdi ln, tm n xn, sau xn-1, ., x1.

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    I. Hphng trnh tuyn tnh tng qut----------------------------------------------------------------------------------------------------------------------

    Gii cc h phng trnh sau y vi cc ma trn m rng chotrc.

    1 5 2 6 1 1 1 3

    V d

    . ,

    0 0 5 0

    a

    . 0 1 2 4 ,

    0 0 0 5

    b

    1 1 1 0

    . 0 1 2 5 ,0 0 0 0

    c

    1 1 1 0

    . 0 3 1 0 .

    0 0 0 0

    c

    I. Hphng trnh tuyn tnh tng qut--------------------------------------------------------------------------------------------------------------------

    V d

    5 2 1

    4 6

    3 3 9

    x y z

    x y z

    x

    + + =

    + =

    + =

    Gii h phng trnh:

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    I. Hphng trnh tuyn tnh tng qut---------------------------------------------------------------------------------------------------------------------------

    3

    3 5 9 2

    2 3 3

    y z

    x y z

    x

    + =

    + + =

    + + =

    V d

    Gii h phng trnh

    I. Hphng trnh tuyn tnh tng qut---------------------------------------------------------------------------------------------------------------------------

    Tm nghim tng qut ca hphng trnhV d

    2 3 4 5

    1 2 3 4 5

    1 2 3 4 5

    3 6 6 4 5

    3 7 8 5 8 9

    3 9 12 9 6 15

    x x x x

    x x x x x

    x x x x x

    + + =

    + + =

    + + =

    n cs: 521 ,, xxx n tdo: 43 ,x

    Nghim tng qut:

    1

    2

    3

    4

    5

    24 2 3

    7 2 2

    4

    x

    x

    x

    x

    x

    = +

    = +

    =

    =

    =

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    I. Hphng trnh tuyn tnh tng qut---------------------------------------------------------------------------------------------------------------------------

    Tm nghim tng qut ca hphng trnh bit ma trn mrng

    V d

    1 1 1 1

    2 3 4 1

    3 4 2 1

    I. Hphng trnh tuyn tnh tng qut-----------------------------------------------------------------------------------------------------------

    ---

    Tm nghim tng qut ca hphng trnh bit ma trn mrng

    V d

    1 1 2 0

    2 1 5 0

    3 4 5 0

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    I. Hphng trnh tuyn tnh tng qut-----------------------------------------------------------------------------------------------------------

    --

    Tm nghim tng qut ca hphng trnh bit ma trn mrng

    V d

    1 1 1 1 2

    2 1 3 0 1

    3 4 2 2 5

    2 3 1 1 3

    I. Hphng trnh tuyn tnh tng qut---------------------------------------------------------------------------------------------------------------------------

    Tm nghim tng qut ca hphng trnh bit ma trn mrng

    1 1 2 0 1

    V d

    2 3 1 2 43 4 5 1 3

    1 2 3 1 0

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    I. Hphng trnh tuyn tnh tng qut------------------------------------------------------------------------------------------------------------

    -

    Tm tt ccc gi trca tham sm phng trnh sau c nghim

    V d

    1 1 1m

    2

    1 1 ,

    1 1

    m m

    m m

    I. Hphng trnh tuyn tnh tng qut---------------------------------------------------------------------------------------------------------------------------

    1 1 1 1

    Tm tt ccc gi trca tham sm phng trnh sau c nghimExample

    2 3 1 4

    3 4 1m m

    +

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    I. I. Hphng trnh tuyn tnh tng qut

    V dTm tt ccc gi trca tham sm phng trnh sau c nghim

    duy nht1 1 1 1 1

    2 1 3 1 2,

    3 4 2 0 6

    2 1 0 1m m

    I. Hphng trnh tuyn tnh tng qut---------------------------------------------------------------------------------------------------------------------------

    V dTm tt ccc gi trca tham sm phng trnh sau c nghim

    duy nht

    2 3 1 4 0

    23 2 1 5 71 1 1m m

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    II. Hthun nht.---------------------------------------------------------------------------------------------------------------------------

    H phng trnh tuyn tnh c gi l thun nht nu tt ccc h s t do b1, b2, , bmu bng 0.

    nh ngha h thun nht.

    x1 = x2 = = xn = 0.

    Nghim nyc gi l nghim tm thng.

    H thun nht ch c nghim duy nht bng khng khi v chkhi r (A) = n = s n .

    II. Hthun nht.---------------------------------------------------------------------------------------------------------------------------

    H thun nht AX = 0 c nghim khng tm thng khi v chkhi r(A) < n.

    H thun nht A X = 0 , vi A l m a t rn vung c nghim khngtm thng (nghim khc 0) khi v ch khi det(A) = 0.

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    II. Hthun nht.---------------------------------------------------------------------------------------------------------------------------

    Tm nghim tng qut ca h phng trnh.

    V d

    1 2 3 42 2 0x x x+ + + =

    1 2 3 4

    1 2 3 43 6 4 0

    x x x x

    x x x x

    + + + =

    + + + =

    II. Hthun nht.---------------------------------------------------------------------------------------------------------------------------

    Gia nhng nghim ca h

    V d

    2 0

    2 4 0

    y z

    x y z

    + + =

    + + =

    2 0x y z+ =

    tm nghim tha biu thc y x y = 2 z

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    II. Hthun nht.---------------------------------------------------------------------------------------------------------------------------

    Gi s Almat rn ca h thun nht c 4 p hng trnh v 8n,

    gi s c 5n t do. Tm r(A)?

    V d

    Gii thch v sao h phng trnh thun nht c m p hng trnh,nn vim

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    Trng H Bch khoa tp HCh Minh

    Khoa Khoa hc ng dng - Bmn Ton ng dng------------------------------------------------------

    ai so tuyen tnh

    Chng 4: KHONG GIAN VECT

    Giang vien TS. ang Van Vinh

    www.tanbachkhoa.edu.vn

    Ni dung---------------------------------------------------------------------------------------------------------------------------

    I nh ngha va V du

    II oc lap tuyen tnh, phu thuoc tuyen tnh

    III Han cua ho vect

    V Khong gian con.

    IV C s va so chieu

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    I. nh ngha va cac v du---------------------------------------------------------------------------------------------------------------------------

    ==

    Tp khc rngV Hai php ton

    Nhn vctvi 1 sCng

    8 tin

    KHNG GIAN VCTV3. Tn ti vc tkhng, k hiu 0 sao chox + 0 = x

    4. Mixthuc V, tn ti vect, k hiu xsao cho x + (-x) = 0

    8.1x = x

    5. Vi mi s v mi vectorx:, K ( ) x x x + = +

    6. Vi mi s , vi mi :K x , y V ( x y ) x y + = +

    7.( ) x ( x ) =

    I. nh ngha v cc v d---------------------------------------------------------------------------------------------------------------------------

    Tnh cht ca khng gian vct

    1) Vctkhng l duy nht.

    2) Phn t i xng ca vctxl duy nht.

    3) 0x = 0

    5) -x = (-1)x

    Vi mi vectxthucVv mi s :K

    4) 0 0 =

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    I. nh ngha v cc v d----------------------------------------------------------------------------------------------------------

    --

    }{ RxxxxV i = ),,( 3211

    ),,(),,(),,( 332211321321 yxyxyxyyyxxxyx +++=+=+

    V d1

    nh ngha php cng hai vctnhsau:

    nh ngha php nhn vctvi mt sthc nhsau:

    ),,(),,( 321321 xxxxxx ==

    =

    =

    =

    =

    33

    22

    11

    yx

    yx

    yx

    yx

    V1 - Khng gian vct trn trngsthc3R

    nh ngha sbng nhau:

    I. nh ngha v cc v d---------------------------------------------------------------------------------------------------------------------------

    RcbacbxaxV ++= ,,2

    2

    V d2

    nh ngha php cng hai vct: l php cng hai a thcthng thng, bitphthng.

    nh ngha php nhn vctvi mt s: l php nhna thc

    V2 - Khng gian vct ][2 xP

    vi mts thcthng thng, bi tph thng.

    nh ngha sbng nhau: hai vc tbng nhau nu hai athc bng nhau, tc l cc hstngng bng nhau).

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    I. nh ngha v cc v d---------------------------------------------------------------------------------------------------------------------------

    = Rdcba

    dc

    baV ,,,3

    V d3

    nh ngha php cng hai vct: l php cng hai ma trnbit trong chng ma trn.

    V3 - Khng gian vct ][2 RM

    vi mt s bit.

    nh ngha sbng nhau ca hai vct: hai vc tbng nhauhai ma trn bng nhau.

    I. nh ngha v cc v d----------------------------------------------------------------------------------------------------------

    --

    }{4 1 2 3 1 2 32 3 0ix x x x R x x x= + + =( , , )

    Php cng hai vctv nhn vctvi mt sging nhtrong v d1.

    V d4

    V4 - l KGVT

    CH : C nhiu cch khc nhau nh ngha hai phpton trn V1, ( hocV2, hocV3) sao cho V1 ( hocV2, hocV3) l khng gian vct.

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    I. nh ngha v cc v d----------------------------------------------------------------------------------------------------------

    --

    }{5 1 2 3 1 2 32 1i( x ,x ,x ) x R x x x= + =

    Php cng hai vctv nhn vctvi mt sging nh

    V d5

    .

    V4 -KHNGl KGVT

    4 4(1,2,1) , (2,3,2)= = x V y V

    4)3,5,3( Vyx =+

    II. c lp tuyn tnh---------------------------------------------------------------------------------------------------------------------------

    V- KGVT trn K

    1 2{ , ,..., }mM x x x=

    Tp con

    MPTTT1 2, , , m K khngng thi bng 0

    1 1 2 2 0m mx x + + + =

    M c lp tuyn tnh1 1 2 2 0m mx x + + + =

    1 2 0m = = =

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    II. c lp tuyn tnh---------------------------------------------------------------------------------------------------------------------------

    V- KGVT trn K

    1 2{ , ,..., }mM x x x=

    Tp con

    1 2, , , m K

    1 1 2 2 m mx x x x = + + +

    Vector x thuc Vc gi lThp tuyn tnhcaM,nu

    II. c lp tuyn tnh---------------------------------------------------------------------------------------------------------------------------

    { (1,1,1) ; ( 2 ,1, 3 ) , (1, 2 , 0 ) }M =

    Trong khng gian R3cho hvc tV d5

    1. Hi Mc lp tuyn tnh hay phthuc tuyn tnh?

    2. Vctx = (2,-1,3) c l thp tuyn tnh ca hM?

    c u . s , , , , , , =

    2 2 3 0 0 0( , , ) ( , , ) + + + + + =

    2 0

    2 0

    3 0

    + + =

    + + = + =

    1 2 1

    1 1 2

    1 3 0

    A

    =

    2r( A ) =

    Hc v snghim, suy ra M phthuc tuyn tnh

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